help-49

[pearl star]

0 idea on how to approach these type of questions

never done them before

[late rover]

why another guy also has name catloaf in his channel

[pearl star]

cuz hes trolling

[late rover]

Damn

[pearl star]

u hav any idea on how i can solve this

[late rover]

Pythagore

[pearl star]

true

but when do i use this

[late rover]

x(t)^2+y(t)^2=3.9^2

[pearl star]

which part would this help me in exactly

[late rover]

part a you have to find y(t) right?

and you already has x(t)

[pearl star]

dy/dt i think

[late rover]

yeah\

you find y and then dy/dt

[pearl star]

oh

3.9^2 = 1.5^2 + y^2 ?

3.9^2 - 1.5^2 = y^2

[late rover]

wait wait hang on a sec

[pearl star]

oke

[late rover]

SO

x(t) the function is unknown

We got this

[pearl star]

yus

[late rover]

what we know is

when x(t)=3.6

x'(t)=1.5

and we have to find y'(t) at that time

[pearl star]

oh i see

yes that is true

[late rover]

Do you know how to do implicit differentiation?

[pearl star]

yes

[late rover]

okay so just apply it here

What you get

[pearl star]

(2x) (dx/dt) + (2y) (dy/dt) = 0

[late rover]

great!

[pearl star]

I think we have dx/dt at this moment if am not mistaken

[late rover]

Now what is x at the time

[pearl star]

which is 1.5

[late rover]

yup

x is known

[pearl star]

and x is 3.6 ?

[late rover]

what about y

[late rover]

yeah

[pearl star]

y idk

do we get it from pythagorean?

no idea

[late rover]

YESSSS

[pearl star]

lets go

ok one second so lemme find it

3.9^2 - 1.5^2 = y^2

wait

is the x right

i think is supposed to be 3.6^2

3.9^2 - 3.6^2 = y^2

[late rover]

yeah

[pearl star]

y = 3/2

or 1.5

[late rover]

that's nice

[pearl star]

then solve for dy/dx ?

for a)

[late rover]

now plug everything back

Yeah

[pearl star]

(2(3.6))(1.5)+(2(1.5))(dy/dt))

wait

dy/dt = -3.6

and it makes sense cuz its goign down

b idk what to do

[late rover]

Area = y(t)*x(t)*1/2

Same idea

[pearl star]

not the derivative?

[late rover]

not sure what you meant

[pearl star]

like

2x dx/dt + 2y dy/dt

cuz its saying rate

but idk

[late rover]

Nah

If you want to find the rate of smth

You find derivative that

Area = y(t)*x(t)*1/2 so we're gonna find derivative of this

Same idea

[pearl star]

oh

wait so

y(t) and x(t) we dont have?

[late rover]

we do haha

[pearl star]

ah okay lol

so we just plug in thats it?

[late rover]

it's still

[late rover]

yeah basically

It's still asking about that specific time when x=3.6

[pearl star]

ohhh waitttttt

okok

so y(t)*x(t)

then derive

but this time since its not being added or whatever

[late rover]

yup

[pearl star]

we gotta do product rule

[late rover]

yes

[pearl star]

(1/2)(dx/dt) (y) + (1/2x)(dy/dt)

whoops

[late rover]

0?

[pearl star]

wdym?

[late rover]

the rate

what do you get

[pearl star]

oh one sec

x = 3.6
y = 1.5
dx/dt = 1.5
dy/dt = -3.6

so i think it is -5.4

[late rover]

bruh how

Oh yeah nvm

I'm blind

this's why calculator is so importance to me

[pearl star]

lol

i dint wanna spend the time typing it out here so yah

okay c

[late rover]

the angle same idea

[pearl star]

ill be honest i forgot how to find angles in right angle triangle

i remember sohcahtoa

[late rover]

sin(alpha)=y/3.9

[pearl star]

we gotta derive this?

pls help

[slate crater]

,tex .sohcahtoa

hm

[pearl star]

I think it didnt work

[grand pond]

Poly(^///^)

[pearl star]

I understood these

But I dont know where to move on from here

[pearl star]

This

[late rover]

just do implicit differentiation

and see what you get

[pearl star]

costheta dtheta/dt = 1/3.9(dy/dt)

im not too sure about this

whoops i meant alpha

[late rover]

It should be obvious

[pearl star]

sub in thats it?

[late rover]

cos(alpha) at that time is known

[late rover]

yeah basically

[pearl star]

so wait

cosalpha should be

3.6/3.9

[late rover]

yah

[pearl star]

and dy/dt = -3.6

[late rover]

yupppppp

[pearl star]

-1?

[late rover]

ehh, not sure I have done any calculation

[pearl star]

i think it makes sense

cuz the angle will decrease as the ladder falls down

is it ok if we can solve couple more that are similar if u have time

[late rover]

,ti

[grand pond]

The current time for fionna_fx is 12:39 AM (+08) on Mon, 27/10/2025.

[late rover]

[pearl star]

😳

thakn you for your time

[late rover]

np

[pearl star]

if anyone can help me with this 🙏

[astral canyon]

start by writing y as a function of theta

[midnight plank]

@pearl star Has your question been resolved?

[pearl star]

idk what to do

i did tantheta = y/150 and derived

[astral canyon]

that's the right track

what do you get after deriving?

[pearl star]

i think i got it but i got sec^2theta dtheta/dt = 1/150 dy/dt

then i realize it said at this moment the angle is pi/4 and rate of angle is 0.14

but now im stuck at sec^2(pi/4)(0.14) = 1/150 dy/dt

[astral canyon]

why are you stuck at this?

you wanna find dy/dt... you have an expression that you can solve for dy/dt...

[pearl star]

sec^2(pi/4)(0.14)/1/150 = dy/dt

[astral canyon]

,w 150*sec^2(pi/4)(0.14)

[grand pond]

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[pearl star]

Thanks

Yep think thats my answer

[astral canyon]

Looks realistic

[pearl star]

Oke I have a couple more is it ok if u can also help me with them sorry for taking ur time

[astral canyon]

Feel free to send it, no guarentee I can help, i have a paper I'm already a week late on, but I'm sure someone else would if I don't

[pearl star]

Dang

U are doing PHD?

[astral canyon]

nah, just a proposal submission for a conference

the actual deadline is a few weeks away but I told my supervisor I would send my draft to her last week for comments

[pearl star]

Dang goodluck bro

I will see tbh maybe its fast quesiton

[astral canyon]

yeah same thing, pythag is your relation this time

[pearl star]

is 0.6 km north and 0.8 km east the y and x axis?

for pythagorean

[astral canyon]

yeah

[pearl star]

okay I got for the hypotenuse 1

cuz sqrt 0.6^2+0.8^2

[pearl star]

I think the drawing below it is wrong

[astral canyon]

why would it be wrong?

[pearl star]

nvermind

[astral canyon]

Yeah the hypotenuse is 1

[pearl star]

i just saw

[astral canyon]

x^2 + y^2 = s^2

differentiate both sides wrt like last time

[pearl star]

wait ds/dt is 20 or 30

[astral canyon]

Typo in the question, use whichever you want i guess

[pearl star]

yes thats why i was confused a little about the drawing

[astral canyon]

Well actually the text says ds/dt = 30, dy/dt = -100

[pearl star]

yep

[astral canyon]

The drawing says ds/dt = 20, dy/dt = -60

[pearl star]

Yes

[astral canyon]

It's just a typo, use whichever pair you want

[pearl star]

I used the 30 an -100

and s is the hypotenus we found?

[astral canyon]

yes

[pearl star]

oh the s is not changing it remains consatnt?

[astral canyon]

make sure you differentiated this properly

[pearl star]

Yep

[astral canyon]

no its changing

[pearl star]

2x dx/dt + 2y dy/dt = 2s ds/dt

[astral canyon]

ds/dt represents the change of s. If you plugged in other numbers ds/dt would differ

[pearl star]

ohhhhhh ok yes thats true nvm

i was thinking something else

My teacher used the numbers in the drawing instead of the question 😂

its ok same process

[astral canyon]

yeah, numbers dont really matter as long as you understand how to do the problem

[pearl star]

yep

i got (2)(0.8)(dx/dt)+2(0.6)(-100) = 2(1)(30)

so dx/dt = 112.5

[astral canyon]

,w 2x * w + 2y * v = 2s * u, x=0.8, y =0.6, s = 1, v = -100, u = 30

[grand pond]

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[pearl star]

225/2

okkkkkkk

i think i got the right angles now i just have a last question

this one i have no idea about all i know is 1/3πr^2h

[astral canyon]

The trick here is that the view is always a similar triangle

[pearl star]

wait so the main tank

has x = 1.5 and y = 3 ?

and the actual water cone or wtv has a height of 1.8 so y = 1.8

but x is not given

[midnight plank]

@pearl star Has your question been resolved?

[small jasper]

yes sure

[small jasper]

The trick here is that the view is always a similar triangle

set up an equation based on this

it should be of the form ||x/y = ...||

[midnight plank]

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[limber plaza]

I'm a bit confused on this example application. He says "the limit exists, but it's divergent."

But a limit = L only if L is a real number. We explicitly write that the limit doesn't exist if the limit goes to infinity (unless there are some algebraic tricks, etc., which I don't think there are here).

[limber plaza]

Did he make a mistake or am I just being goofy?

[lavish venture]

where?

what are you referring to?

[limber plaza]

The lim_{n \to \infty} n^2 = \infty

[lavish venture]

whats the problem with that?

[limber plaza]

He says that the limit exists, and hence we can use the rule on the right. But the rule on the right states that we need the limit to exist ( = L)

[lavish venture]

the statement $\lim_{n \to \infty} a_n = \infty$ has an entirely different meaning

[grand pond]

knief

[limber plaza]

I don't think I'm clear on this.

[lavish venture]

i mean he is definitely abusing notation here

f(L) is nonsense when the limit is infinite

regardless this still works

[limber plaza]

I agree that it still works, just wanted to make sure that it was indeed an abuse of notation. Was concerned I was missing something.

Thanks!

[lavish venture]

you're welcome

[limber plaza]

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[wicked walrus]

I am confused on how to solve algebraiclly for the x v and a

[atomic jolt]

,rot

,rotate

[wicked walrus]

Thank you

[atomic jolt]

what are the equations

oh wait there are two pictures

I don't know how to rotate the other one lol

[wicked walrus]

The ones under question 2

C I didn’t really do it because I didn’t know how

I rotated it

.close

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[twin jungle]

how do i do this

[twin jungle]

i dont know where to start

[slate heath]

Use $ln(x)=0\iff x=1$ and $ln(e^{x})=x$

[nova yoke]

how do you "undo" log?

[grand pond]

Mathlympian / Gab

[twin jungle]

you would do like e^of something for the ln

[twin jungle]

i dont get this

[slate heath]

Properties of log

[twin jungle]

what would x be to get ln(x)=0

and wouldent ln(1)= 0 think

[slate heath]

1

because e⁰ = 1

[twin jungle]

yeah

ok

wait

ok nvm

ok i get the next part then

cause it would be xln(e)=x right?

[slate heath]

Yep

[twin jungle]

ok wait i think i get it

then it would be like ln(ln(lnx))=1

then you do it again right

then ln(ln))=e

[slate heath]

Yeah

[twin jungle]

ok

thx

.close

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[slate heath]

yw!

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[lethal wren]

A right
triangle is formed in the first quadrant by the x- and y-axes and
a line through the point (1, 2)

[lethal wren]

Find the vertices of the triangle such that its area is a
minimum.

[pearl hull]

Have you tried anything yet?

[lethal wren]

uhh

yes

but im not sure

so im trying to find the constraint and i know we're minimizing A = 1/2xy

so i found the slope of the entire line using (0-y)/(x-0) to get -y/x

then tried substituing point slope to try to get something in terms of x and y

y - 2 = (-y/x)(x-1)

but thats about it so far

[pearl hull]

Let’s call x = a; y = b

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

[grand pond]

e4 e5 Qh5 Nc6 Bc4 Nf6 Qxf7#

[lethal wren]

so far for vertices i got

(0,4) (0,0)

cause i solved for y and got 0 and 4

wait nvm this might not be right keep going

[pearl hull]

how did you get that?

[lethal wren]

well before you said to give x and y the a and b variables i just did it all in terms of x and y

and i got x = y/2y-4 then substituted it into the 1/2xy

then took the derivative and found the critical numbers

where y was 0 and y was 4

[pearl hull]

yeah, you can't really do that since x and y mentioned in the question are technically representing slopes

They should not be mixed with the variables in the line equation

[lethal wren]

okay

[pearl hull]

have you tried plugging one of the three points into this equation?

[lethal wren]

no i havent

[pearl hull]

ye, try it out

[lethal wren]

could we plug in (x,0)

[pearl hull]

no

[pearl hull]

^

[lethal wren]

but isnt the (1,2) already baked into that point slope formula

thats what confused me

[vital flume]

hi I just solved it using similar things as yoursbut im not quite sure if you made a mistake or not

[lethal wren]

nice

[pearl hull]

yep, it has.

[vital flume]

using the expression of a linear fonction going through a point

[pearl hull]

how about the other two?

[pearl hull]

We've already at here.

[vital flume]

y=(2/(1-x))*(x-2)+2 then y=2/(1-x)

have you found the same ?

Then Area = x/(1-x)

find min of this function

[pearl hull]

$y = \frac{2}{1-x}\times (x-2)+2$

[grand pond]

e4 e5 Qh5 Nc6 Bc4 Nf6 Qxf7#

[vital flume]

(u can add 1 and substract it from the numerator)

ye

[pearl hull]

I legit have no idea what is this.

mind elaborating?

[vital flume]

it's using the same thing as the expression of a line going throught 2 point

[lethal wren]

the key says the vertices are (0,0) (0,4), (2,0)

[vital flume]

the points being (1,2) and (x,0)

[vital flume]

wdym ?

[lethal wren]

its minimizing the vertices of the triangle that produce the minimal amount of area

so those three vertices of a triangle give the smallest possible area

[vital flume]

sry I miswrote this I fixed it

[lethal wren]

bro i dont even know at this point apparently what i did was wrong but i almost gotr the answer

maybe thats just pure luck

.close

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[vital flume]

oh wait shit I missed something

it's not the same x im dumb

sry for confusing you

just use the equality between the slopes

-y/x = 2/(1-x) for example

[pearl hull]

.reopen

[midnight plank]

✅ Original question: #help-49 message

[pearl hull]

I did mention that earlier, replacing x and y with a and b to avoid any confusion.

[vital flume]

then y = -2x/(1-x)

[vital flume]

ok ye perfect

so we just need to find the min of A(x) = (1/2)xy = (x^3)/(x-1) (the area) now

@lethal wren

[pearl hull]

He’s left

.solved

[midnight plank]

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[vital flume]

🥲

[midnight plank]

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[copper linden]

doubt in this!

the answer given is 3, but cant the team members finish at 9th,8th,and 7th position respectively to earn no points as no point is awarded at those positions?

[last slate]

oh yeah weird

bc if all a team's members placed either 9th, 8th, or 7th then that would mean that the team gets no points, not 3

unless there's extra nuance to the problem that was missed

[copper linden]

yeah, i cant see a restriction in the question which states that each team atleast scored 1 point

[last slate]

that also actually doesn't account for 6th place either because according to the problem 1 <= n <= 5

[copper linden]

hmm true

[modern sapphire]

well, total points possible to earn is 5,4,3,2,1 for the candidates in the race

so thats a total of 15 points

if one team earns 0, then you have to split 15 between the other two teams

but that would mean one of the teams gets more than 6 points

[copper linden]

that makes so much more sense

thank you!

i think that solves my doubt

.solved

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[crystal rivet]

Let $U$ and $V$ be open sets in $\mathbb{R}^n$ and let $f:U\rightarrow V$ be injective (so there is an inverse map $f^{-1}:V\rightarrow U$). Suppose that $f$ and $f^{-1}$ are both continuous. Show that for any set $S$ such that $\overline{S}\subset U$ and $\overline{f(S)}\subset V$ we have $f(\partial S)=\partial (f(S))$.

[crystal rivet]

At first I was confused if there was a typo and it meant bijective (since only bijectivity implies the existence of an inverse) but I think I can just ignore that. Either way I don't really know how to approach this

[grand pond]

za

[subtle blaze]

How do you define boundaries in your course

[crystal rivet]

As in $\partial S$?

[grand pond]

za

[subtle blaze]

Yeah

[crystal rivet]

$\mathbf{x}$ is a boundary point if $\forall r>0$, we have both $B(r,\mathbf{x})\cap S\neq \varnothing$ and $B(r,\mathbf{x})\cap S^c\neq \varnothing$. The set of all boundary points is $\partial S=\partial S^c$.

[grand pond]

za

[crystal rivet]

That is what I have in my notes

[twilit jetty]

you can try proving these theorems, they look like a good direction to start

[crystal rivet]

Quick question about notation, what does int_X B mean? I assume int B is the interior of B, but what does the subscript of X mean? Same with with cl_X A

[twilit jetty]

thats to tell you what topological space the interior is being taken in

theyre using subscripts for wider context, such as $d_X,d_Y$ or $[A]\alpha,[B]\beta$ or $101_2,19_{10}$

[grand pond]

mtt

[twilit jetty]

that way its easier to get your bearings since its between X and Y

[crystal rivet]

Oh ok thank you, I'll try that out

.close

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[slow granite]

Need to see if this is right

[gusty falcon]

,w x^2 + 3(25 - x)^2 minimize

[grand pond]

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[gusty falcon]

yes

also you need to write out the y value as well

"determine the two numbers"

[spiral rock]

Also, you don't need derivatives as this is just minimizing a quadratic

But that is not a big issue

[slow granite]

Thing is that it is supposed to be application of calculus,so i have to use it unfortnately

[slow granite]

Thano you

.close

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[tidal turret]

how does the lagrange remainder works for second order degree polynomials in two variables, I want to estimate the error

[tidal turret]

taylor polinomial i mean

[mighty oracle]

what year maths is this 😭

[midnight plank]

@tidal turret Has your question been resolved?

[tidal turret]

like first or second month of university, so first semester

[shrewd pagoda]

do you know how to do taylor series expansion with 2 variables?

out of raw curiosity, wouldn't it just be doing the very same thing for the lagrange remainder

[tidal turret]

i just need the first order partial derivatives and the second order partial derivatives (at a point)

[tidal turret]

yes and no

do you know what lagrange remainder is?

[shrewd pagoda]

well yeah

[tidal turret]

explain

in a few words

[shrewd pagoda]

if it's giving you a function, just elaborate the lagrange remainder under that function for the 2 variables, (sounds like the logical step from the one variable approach)

[tidal turret]

what?

what does the gradient has to do with this?

[tidal turret]

you are trying to use single variable intuition for the two variables case?

i see you edited

[shrewd pagoda]

welll I mean, if it kinda works that way for taylor series with 2 variables, why would it not for the remainder

*speaking out of my arse though *

[tidal turret]

well its complicated

say for example the second order taylor polynomial

compare it from the single variable case with the multivariable case

[shrewd pagoda]

might be intensive, but solving math is trivial, speaking from a raw how to do it approach

[supple pilot]

i think it works the same as the one-variable case, you just have to consider only vectors in the direction of the argument you're trying to bound the error for

[tidal turret]

care to elaborate?

[supple pilot]

since proof of the remainder is based on MVT it should work the same when we restrict to 1-dimensional subspace

[supple pilot]

so you apply taylor's theorem and get a polynomial in 2 variables

[supple pilot]

at a point let's call it "a"
which is a function R^2 -> R as well
then you want to analyze the possible error in point "x"
lagrange remainder (iirc) says that there exists a vector eta on the segment between a and x such that remiainder equals (D^3 f(eta))/6 * (x-a)^3

well D^3 f is a trilinear form here so it has (eta, eta, eta) as arguments but otherwise i think it is correct

[midnight plank]

@tidal turret Has your question been resolved?

[midnight plank]

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[slow granite]

This was a question made with love

I hated doing it

Do you guys not like probability questions?

<@&286206848099549185>

[near haven]

but there are five body styles four interior ccolours and six exterior colours?

with a maximum of 100 cars within the space

hmmm

I belive not though

perhaps

because maybe

jst maybe

there are a total of 120 car designs

which takes up 600 msquare

i didnt dig too deep though

its prob wrong

🙁

[slow granite]

Can you give me a way to get the right answer

[near haven]

im in grade 10 man my school sucks so we dont do gud maths :/

everythign u did is correct

I belive

but

u forgot there was also another

car exterior

u only did 5 🙁

[slow granite]

Ohhhh..

[near haven]

i could be wrong

dont take my advice tbh

Hanoka is very gud at these things tbh

she is jst on another lvl

TvT

[slow granite]

Well,it's because of the condition being that exterior colours should not match with interior colours

So...

[near haven]

then u shouldve done 4?

ik im wrong

[slow granite]

Who is she?

[near haven]

but

luk

ther eare red and blue inter and exter

so they cancel out maybe?

[near haven]

idk but she jst so friking gud at maths

like if i ping her right

she responds

bfre the bot

can

[slow granite]

Okay ,can you ping her now?

[near haven]

@scenic wyvern

wait

I think she driving

[scenic wyvern]

you know you're not supposed to ping a specific helper??

!noping

[midnight plank]

Please do not ping individual helpers unprompted.

[near haven]

mb mb

noah needs help hanoka

plz help hims

[scenic wyvern]

that doesn't mean you can ping.

[near haven]

im sry 🙁

i jst joined this server

[scenic wyvern]

if OP wants to ping, he can ping Helpers

[near haven]

wait like this

[scenic wyvern]

#rules and #info,

[near haven]

<@&286206848099549185>

[near haven]

gotcha

noah ask her what ever

she knows

[slow granite]

I did,no one except them responed

[scenic wyvern]

I also do not appreciate this patronizing

[near haven]

im sorry

[scenic wyvern]

but anyway, what seems to be the issue?

[slow granite]

Thank you

[scenic wyvern]

....to that question? so I assume it's done?

[near haven]

well

[near haven]

I think the area covered is 400m squred right?

[scenic wyvern]

I know what the original question was

I'm asking what your issue with it is

[near haven]

so all cars should be able to fit within the 500 msquare

[slow granite]

Needed to see if i waas correct

[scenic wyvern]

and what is your calculation and justification?

[scenic wyvern]

justification?

[slow granite]

Uhhh...

[near haven]

ur gonna get a bit less than 11 marks

[scenic wyvern]

please tell me you have one, because the question asks for it...

hold up

I need clarification on the interior and exterior colors

are the top two interior colors limited to the top three exterior colors, and same for the bottom?

or are all four interior colors available against all six exterior colors (minus the req. that interior \neq exterior)?

[slow granite]

This

[scenic wyvern]

ok

[scenic wyvern]

now let's get back to this.

[slow granite]

But it's via calculation

[scenic wyvern]

you didn't explain what the calculations are doing

also, don't ping on reply please.

[scenic wyvern]

what even is the 1 x 4 x 5?

all five of them even

is it one row per body style, the second column being a choice of interior color, and the third a choice of exterior color?

[slow granite]

1 body style,4 int colour,5 ext colour

[scenic wyvern]

why only 5?

there are 6

[slow granite]

Because of the limitation of no ext col and int col being the same

[scenic wyvern]

but only two of the interior colors overlap with an exterior color

yes, that was a premove. I knew you were gonna say that

I suggest that you don't do it per body style, but rather per interior color

[slow granite]

Okay that makes more sense ,thank you

.close

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[static rampart]

@ruby veldt son please help me with math

[silent dock]

What have you tried?

[real minnow]

Try plugging in values like (0, y) and (x,0)

P(0,y) yields f(f(0)) = 2f(0)

[static rampart]

smart

[real minnow]

p(x,0) yields = f(xf(0)+f(x)) = 2f(x)

and if you plug in p(1,y), it proves the function is surjective

[silent dock]

You can also show its injectivity quite easily

[silent dock]

What have you tried? Nothing?

[real minnow]

yeh

[static rampart]

so x +1 for all x∈R

[silent dock]

…sure I believe that is the only solution, but how did you just jump to that conclusion?

I'm starting to think we've got a troll on our hands.

[real minnow]

[midnight plank]

@static rampart Has your question been resolved?

[midnight plank]

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[vernal field]

help

[vernal field]

dk how to integrate this

ik 2^(2x+3) = ((ln2)2^(2x+4)).dx

help channel is mine buddy, please post it in #help-2

[strong storm]

∫ a^(kx+b) dx = a^(kx+b) / (k ln a) + C

[vague seal]

there's an addition so you can split them into two integrals that are easier to think about

[hybrid widget]

Then use the fact that integrating a^x is a^x/(ln a)

[vague seal]

Also you could change it to base e

@vernal field

do these tips help?

[midnight plank]

@vernal field Has your question been resolved?

[midnight plank]

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[autumn folio]

hi, can someone show me how to solve this using u sub?

[stoic grotto]

What substitution have you tried?

[autumn folio]

idk any other substitution

[silent dock]

You want to simplify the thing inside the root with a substitution.

[autumn folio]

.

Nvm

[silent dock]

Almost correct..

[autumn folio]

Yeah i forgot the 2

Thanks

.close

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[lucid glacier]

Hi i would like to ask about the logic behind problems such as these how does two balls simultaneously graze each other tangentially while being on the same plane my brain cant brain it

[quiet hinge]

They don't graze each other(and even if they do it's possible)

[lucid glacier]

then whats goin on

o_o

i hate mechanics

[quiet hinge]

The vector whose direction is the direction of motion of the ball and placed at the center of the ball spans a line that is tangential to B

Im just re stating what is written

[lucid glacier]

im so buns at vectors lul

why would a ball have a direction of motion

[quiet hinge]

Cause it's moving

[lucid glacier]

but isnt it on a plane

hows it goin upwards

[quiet hinge]

That's a top down view of the plane

[lucid glacier]

oh

OHHH

omg

that makes sense now

[quiet hinge]

I mean you shouldn't even be looking at the picture in this case because the problem is very clearly stated and you should draw your own picture

Then there won't be any room for misinterpretation

[lucid glacier]

god forbid

dk how to

i should

but i dont

[quiet hinge]

Well anyways if your question is resolved you can type .close

[lucid glacier]

can i keep it for a lil bit

if i dont get it i might come back

[midnight plank]

@lucid glacier Has your question been resolved?

[lucid glacier]

.close

[midnight plank]

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[winter merlin]

hello there

[pearl hull]

hi

[winter merlin]

any recommendations of exponential e functions learning resource ?

I have this test coming up yet I still do not understand the whole idea of e functions

[spiral rock]

do you know exponents?

2^n?

[fallow scarab]

https://en.khanacademy.org/math/algebra/x2f8bb11595b61c86:exponential-growth-decay/x2f8bb11595b61c86:exponential-vs-linear-growth/v/exponential-growth-functions

[winter merlin]

kinda

[spiral rock]

so its the same thing,
e^n is repeated multiplication

e^2 is e * e

e^3 is e * e * e

etc

[winter merlin]

huh that makes sense

[spiral rock]

@lyric charm 😩

[winter merlin]

tbh I thought it would be some kind of magic going on with this types of functions

[fallow scarab]

why are you pinging ann

[lyric charm]

cause i opencried

[winter merlin]

lmao

[pearl hull]

[lyric charm]

i mean uhhhh like how are you gonna explain how e even happens

beyond "oh it's 2.718ish"

[winter merlin]

true

[woven tangle]

e = mc^2

[surreal moon]

Better question: what have you not been understanding? Are there exactly problems you are struggling with

[surreal moon]

!redir

[midnight plank]

This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.

[winter merlin]

yeah just a sec ill pull it up

I understand most of it

but when it comes to e^-x*-2 it's a bit rough

[surreal moon]

$e^{-x-2}$

[grand pond]

SWR

[surreal moon]

What do you find rough about it? What is the question even asking you to do?

[charred thorn]

i think it should be e^(-x-2)

[winter merlin]

it asks firstly (sorry i might get it wrong it's in a different language)
what's the defined range?
what's the max and the lowest points?
and I also need to find what's the values when I do this function = 0 and also x = 0

[surreal moon]

Ah i see. So basically, you need to learn a lot of properties of e^x

[winter merlin]

yeah

[surreal moon]

I think the easiest question to start with here is what do you get when x=0? Asked another way, what is f(0)?

[winter merlin]

well it's all zero?

wait

e^0 is 1 is it?

[surreal moon]

Why is it all zero?

[winter merlin]

sending it again for it to be easier

firstly
the (0+1) = 1
1 * e^-x-2 = e^-x-2

e^-0-2

e^-2 = 0?

[charred thorn]

Do you understand what negative exponent means?

[surreal moon]

Why would this equal zero?

[winter merlin]

sorry mb thought about f(x) = 0 insted of f(0)

[winter merlin]

I don't think so

[surreal moon]

You have correctly calculated that it is $e^{-2}$, but yeah you're not setting it equal to zero

[grand pond]

SWR

[surreal moon]

If $a\ne0$, then $a^{-b}=\frac1{a^b}$

[grand pond]

SWR

[winter merlin]

oh that's very useful

so is it f(0) = 1/e^2?

[surreal moon]

It's practically the definition for negative exponent

[surreal moon]

Very good. That's one question done

I think the next question for you to try is to now solve f(x)=0

[winter merlin]

okay ill try gimme a sec

remembered I connected my drawing tablet so ill just draw it

can I do that?

[surreal moon]

Yes, but do you know why?

Also, it's not x=1/2

[winter merlin]

well if anything is multiplied by zero it's zero

oh right -1/2

[charred thorn]

The $e^(-x-2)$ could be zero too

[grand pond]

Hacaric

[charred thorn]

or could it? I dont know

[winter merlin]

is it not e^-2 -> 1/e^2?

is that not final?

[surreal moon]

Why did the x disappear?

[winter merlin]

oh right I again did it but thought it's the f(0)

so

[charred thorn]

Oh, my bad, looks like exponetion can't result in zero.

[surreal moon]

It can't, but that's an important thing that omer should realize

[winter merlin]

Isn’t it a rule that e^x cannot be zero?

Or are there exceptions

Or did you two mean it cannot be zero in this specific situation

[charred thorn]

$e^-x = 1/e^-(-x)) = 1/e^x that means that e^(-x-2) = 1/e^-(-x-2) = 1/e^(x+2)$

@surreal moon is that correct?

[winter merlin]

This looks chaotic

[charred thorn]

XD

[surreal moon]

Correct. And this is worth knowing. You use the fact to say that you only need to solve $2x+1=0$ and do not need to consider $e^{-x-2}$

[grand pond]

SWR

[winter merlin]

Ah I understand now

So e functions are just like some regular functions with some specific useful key points?

[charred thorn]

I'm sorry for misleading

[surreal moon]

Kinda. There are a number of a important properties to be aware of

[winter merlin]

Okay I think I know how to learn for the exam now

Thank you all very much

I found this really helpful

.close

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[thorny tusk]

hello

[thorny tusk]

water is poured into a cone with the speed 5 ml/s

the angle of the cone is 60 degrees

how fast does the radius of the cone increase after 20 seconds

dv/dt = 5 cm/s

dr/dt = dr/dv * dv/dt

but unsure on how to proceed from there

[wet pollen]

What is the equation for the volume of a cone?

[wet pollen]

.

[thorny tusk]

v = pi r^2 * h

[wet pollen]

That's a cylinder.

[thorny tusk]

/3

[grand pond]

Kookiemon

[wet pollen]

You are looking for the radius, r.

You are not looking for the height, h.

Can you eliminate or redefine h somehow?

[thorny tusk]

eliminate h from the formula

guessing it has to do with the angle being 60 degrees

[wet pollen]

Yes.

[thorny tusk]

no I

I'm stuck

the area of a circular sector can be calculated

v/360 degrees * pi * r^2

but don't think that'll help us

[wet pollen]

Does this help?

Can you redefine h in terms of r?

[thorny tusk]

tan(30) = r/h

h = r/tan(30 degrees)

[wet pollen]

And what is tan(30 deg)?

[thorny tusk]

1/(root 3)

[wet pollen]

So h = sqrt(3) * r

[thorny tusk]

yes

[wet pollen]

Now you can rewrite the volume formula.

[thorny tusk]

v = (pi * r^2 * sqrt(3) * r)/3

v = pi * r^3 * sqrt(3)/3

[wet pollen]

What do you think you should do next?

[thorny tusk]

let's

dv/dr = pi * r^2 * sqrt(3)

[wet pollen]

So this is a bit of an annoying part of this type of problem because you aren't specifically taught why it's done this way (that will come later), but it should be

dV/dt = pi * r^2 * sqrt(3) dr/dt

The volume and radius are changing with respect to the change in time.

[thorny tusk]

but there's no t

in the formula

[wet pollen]

So this is why these types of problems are annoying to me because this is a parametric equation which is not taught until a later time.

[thorny tusk]

oh

[wet pollen]

The volume is a function of time, V(t), and the radius is also a function of time, r(t).

[thorny tusk]

yes

that makes sense

they change with time

[wet pollen]

So that's why it looks like

$\frac{dV}{dt} = \sqrt{3}\pi r^2 \frac{dr}{dt}$

[grand pond]

Kookiemon

[thorny tusk]

the dr/dt at the end comes from the chain rule?

[wet pollen]

Yes.

$V(t) = \sqrt{3} \pi r(t)^2$

[thorny tusk]

ye that makes sense

[grand pond]

Kookiemon

[thorny tusk]

v'(t) = sqrt(3) * pi * 2r * r'(t)

[wet pollen]

That works as well.

[thorny tusk]

dr/dt = dr/dv * dv/dt

we have

dv/dt

[wet pollen]

I just realized you need to solve for r.

[thorny tusk]

hmm

but we know what dv/dr is

and we know what dv/dt

we can invert dv/dr

(dv/dr)^-1

[wet pollen]

You know dV/dt is 5 ml/s. You are given a time to solve for, dr/dt, which is 20 seconds. At t=20 seconds, you know what the volume is equal to.

You have the formula V(t) = sqrt(3) pi r(t)^2.

At t=20, you can find r.

V(20) = 5 ml/s * 20 s

[thorny tusk]

100 ml

[wet pollen]

Plug that into the volume formula and solve for r(20).

[thorny tusk]

100 = sqrt(3) * pi * r(20)^2

r(20) = about 4.3

[wet pollen]

Now use that in the derivative formula from earlier.

[thorny tusk]

dv/dt = sqrt3 * pi * 4.3^2 * dr/dt

what do I do with the dr/dt

[wet pollen]

You solve for it, you know dV/dt.

[thorny tusk]

oh right

5 = sqrt3 * pi * 4.3^2 * dr/dt

dr/dt =

0.05

ml/second

[wet pollen]

,calc 5/(sqrt(3) * pi * (4.3)^2)

[grand pond]

Result:

0.049696132632215

[wet pollen]

Do the steps make sense?

[thorny tusk]

the steps make sense

[wet pollen]

I would try rewriting what you need to do and explain it to yourself.

[thorny tusk]

the difficult part is

knowing which formulas and equations to deal with

[wet pollen]

Yeah, formulas are always tricky to remember.

[thorny tusk]

thank you for the help 😄

.close

[wet pollen]

yw

[midnight plank]

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[mental talon]

How do I determine if u and v span W? Question in first picture. In second picture I found general solution space for the equation Ax = 0.
We haven't learned linear dependence, basis, dimm, or nullspace yet.

[mental talon]

Quesiton in first image

I rref'ed the augmented matrix and got this

not sure where to go from here

[stoic grotto]

Well, what does it mean for a set of vectors to span a space?

The idea here is that you've figured out a space W that satisfies Ax = 0. Now, for any arbitrary vector in that space, can you write it as some au + bv

[mental talon]

that for any vector in the spanned space, it can be represented as a lin comb

W = { w_hat | w_hat = ju + kb} in this case

rght?

wait

[stoic grotto]

Depending on your choice of lettering, sure. Looking at your handwriting it might be better to say
W = {w_hat | w_hat = r <-1, 0, 1, 0> + s <0, -1, 0, 1>}

[mental talon]

uh

oh so the

[stoic grotto]

(I haven't verified this, but it comes from your RREF form in the image sent at 16:33)

[mental talon]

so parametric form makes up the condition on the space

[stoic grotto]

Exactly!
IF some vector x satisfies Ax = 0
THEN x can be written as x = r <-1, 0, 1, 0> + s <0, -1, 0, 1>

The space W is just every possible x that could be, so it includes r=0 s=0, r=1, s=0, r=15 s=20, and so on

[mental talon]

wait hold on..

[stoic grotto]

Step 1 of the problem is figuring out how to describe the space W -- let me know if you're satisfied with that part or if you have more questions about that. Gotta take this one step at a time

[mental talon]

im stuck on that still

[stoic grotto]

Alright. Why don't you tell me what you know about W and some of the properties you think it should have

[stoic grotto]

This is very abstract -- a good answer is gonna be only slightly more detailed than things like "W is a set" and "W has vectors in it"

[mental talon]

So I'm checking to see if W = span({u, v})

Right

[stoic grotto]

That's several steps ahead. How are you going to check if W = span({u, v}) if you don't know what W is?

[mental talon]

Because W = r[-1; 0; 1; 0] + s[0; -1; 0; 1]

[stoic grotto]

If we're willing to jump to W = r[-1; 0; 1; 0] + s[0; -1; 0; 1] then we've completed step 1: Figure out how to describe W

[mental talon]

I don't know if that's right though

All I know is that solution set = solution space, so I assumed that W equals that

[stoic grotto]

Yeah same thing. In English: If x satisfies Ax = 0, then x is in W

[mental talon]

x is in W? or x IS W?

[stoic grotto]

x is a vector
W is a vector space

Does that make sense?

[mental talon]

I guess?

[stoic grotto]

It's probably worth going back through the textbook and trying to really understand the difference between a vector and a vector space and how those two ideas are related

[stoic grotto]

a subspace is a vector space

[mental talon]

nonono I know what a vector space is. It's a set of vectors with internally consistent addition and scalar multiplication rules where all 10 axioms are satisfied. A vector is part of a vector subspace

I'm just confused what a solution space is vs a solution

If a solution is a singular value (if it's unique), then how can it be a space

[stoic grotto]

Now the gears are churning
A Solution Space is a Vector Space
A Solution is a Vector

Exercise: Is x = [0;0;0;0] a Solution to the equation Ax = 0?
Is x = [1; 0; 0; 0] a solution to the equatino Ax = 0?

And why not a third example: Is x = [-2; 0; 2; 0] a solution to Ax = 0?

[mental talon]

im not sure. But how do I go from a solution in the form of two free variables to a solution space? And how would I then check if that is spanned by u and v

[stoic grotto]

You need to be able to understand how to the 3 exercises I've asked as a prerequisite to understanding how to solve this problem. It's a very simple task if you know how to set it up right.

Lets step back to 10th grade algebra and ask a much simpler, but very simple question:
Exercise: Is x = 3 a solution to the single-variable equation 2x = 6?
Is x = 5 a solution to the single variable equation 2x = 6?

You should arrive at Yes and No -- you took a single value for x, plugged it in, and figured out if 2x = 6 was a true statement or a false statement

Turning it up a notch:

• Is x = -2 a solution to x^2 = 4
• Is x = 0 a solution to x^2 = 4
• Is x=2 a solution to x^2 = 4

[mental talon]

yes no yes but im failing to see how this is relevant 😅

[stoic grotto]

Well, if I ask
Is x = [0;0;0;0] a solution to Ax = 0, you want to follow a similar process!

Take the matrix A, multiply it by the vector [0, 0, 0, 0] and see if that's equal to the zero vector!

[stoic grotto]

Eh, I'm trying to build you up to the solution so you can understand how this actually works. What I say won't make snese if you don't understand this smaller piece

Try and figure out if x = [0;0;0;0] solves Ax = 0.
Then, do the same for x = [1;0;0;0] and x=[-2;0;2;0]
Let me know what you get for those

[mental talon]

I don't know what you want for me to do?

I don't know what A is

so I can't say anything about this lol

[stoic grotto]

🤣 Take a look at the original problem statement

It defines A for you

[midnight plank]

@mental talon Has your question been resolved?

[mental talon]

.close

[midnight plank]

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[midnight plank]

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[coral belfry]

,tex
Hello. Suppose we have that: $ a_n > M>0 , \forall n\in\mathbb{N} $ \
And $ b_n > b , \ b< 0 $. \
Can we say that $a_n b_n < M\cdot b $ ? It seems wrong to me

[grand pond]

fijokazż

[fallow scarab]

why is it wrong to you

[coral belfry]

because usually we change the > symbol when we multiply both sides by the same negative number

b_n could be positive for some n for all we know

[fallow scarab]

great you disproved it

[coral belfry]

(but we do have that it converges to b)

alright lol i think i found a way

thanks

.solved

[midnight plank]

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[coral belfry]

.reopen

[midnight plank]

✅ Original question: #help-49 message

[coral belfry]

,tex
Hello. Suppose we have that: $ a_n > M>0 , \forall n\in\mathbb{N} $ \
And $ b_n > b , \ b< 0 $. \
Can we say that $ \exists n_2 \in \mathbb{N} $ \
\
s.t. : $ \forall n>n_2 : \ b<b_n < 0 $ ? \
\
If yes, then can we say that \
$\forall n>n_2 : \ a_n b_n < M\cdot b $ ?

[grand pond]

fijokazż

[coral belfry]

thought it up a bit more. now this looks more likely but still not sure

oh yeah sorry, we also have that b_n converges to b

[visual tiger]

well the first result is true, take epsilon = -b/2 in the limit definition

However, the second inequality is not true

[coral belfry]

hmm okeoke

lol thought you had more to say

thanks

.solved

[midnight plank]

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[silk otter]

I need help with solving the exponential equations below:
4ˣ⁺¹ + 2ˣ⁺² = 96
3²ˣ⁻¹ - 3ˣ⁺¹ = 54

[fallow scarab]

4 = 2^2

and then

,tex .exp rules

[grand pond]

riemann

[crude zealot]

hi i need help with a calclus question

[lament knoll]

!occupied

[midnight plank]

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).

[honest rain]

https://tenor.com/view/hello-hi-hy-hey-gif-8520159980767013609

[midnight plank]

@silk otter Has your question been resolved?

[honest rain]

if this is math in the future im cooked

https://tenor.com/view/dead-my-honest-reaction-damn-skull-skeleton-gif-15885937969029652188

[slate heath]

it looks like typical advanced Olympiad math tbh

-# nonetheless this is unrelated to the help sorry

[midnight plank]

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[tidal turret]

Si se sabe que cada unidad de un cierto producto A cuesta $39$ pesos y que cada unidad de un cierto
producto B cuesta $48$ pesos, ¿cuantas unidades de cada producto se pueden comprar gastando exactamente $135$ pesos? \

If its known that each unit of a certain product A costs 39 pounds and that each unit of a certain product B costs 48 pounds how many units of each product can we buy having exactly 135 pounds?

[grand pond]

Renato

[atomic magnet]

what have you tried thus far

[tidal turret]

i have done the preliminary work

it is a diophantine equation

39x + 48y = 135

with x,y in Z

[atomic magnet]

have you been formally taught how to solve linear diophantine equations?

[tidal turret]

yes

[atomic magnet]

okay, so whats the first step normally to begin

[sacred folio]

the best way would be to try possible integral solns by substituting the variable with the biggest constant

so y

[calm comet]

Can't we just use trial and error to find integral values

[sacred folio]

ye

[tidal turret]

we divide by gcd(39,48) and use bezouts identity

[atomic magnet]

In this specific question, since units of a product cannot be negative, the easiest way is to just use trial and error. However, if you wanna practice solving lin diophantine equations more generally, then yes begin by dividing by gcd and continue, finding a particular soln using extended euclidean algorithm, etc

[tidal turret]

is not that i want to practice X

is just that this whole exercises in this section is about diophantines

.close

[midnight plank]

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[midnight plank]

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[tidal turret]

I am learning about the Lagrange remainder for multivariable function
in this case, the lagrange remainder of a second order taylor polynomial
where did the limit comes from? the multivariable limit? it seems out of the blue