help-43

[sudden carbon]

ok what abt

area of a circle with the same raius r

[subtle helm]

$2 \int_{-r}^{r} \sqrt{r^2-x^2} \dl x$

[winged lion]

remove the space

[boreal girder]

k

[kind viper]

$2 \int_{-r}^{r} \sqrt{r^2-x^2} \dd{x}$

[sudden carbon]

wait

[boreal girder]

Ann

[subtle helm]

Jinx

[sudden carbon]

shouldnt they have the same bound

i forgot to mention i want the circle inscribed in the square

so like they should hae the same boud right

[winged lion]

reminds like a mcdonalds order

[subtle helm]

!xy

[compact pewter]

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

[subtle helm]

Lemme guess it is optimization of the area of the square inscribed within a circle of radius R

[upper bane]

is this related to the can problem you have?

[sudden carbon]

what

[sudden carbon]

no questio

im just thinking of it mysef

[sudden carbon]

uh

not rlly

[sudden carbon]

yea

like

if i print circles onto squares

i needa find out how much material is wasted

[winged lion]

,, \int_{-r}^{r}\int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}}, dydx = \int_{0}^{2\pi}\int_{0}^{r} \rho, d\rho d\theta

[sudden carbon]

wait i need double inetegral....?

[winged lion]

nah just think the integral is nicer over double integral

[sudden carbon]

wdym

[winged lion]

This is way simpler to compute

[winged lion]

Than this

[sudden carbon]

really...

[winged lion]

Anyway, I didnt get quite your help request

[sudden carbon]

how do i do that

[sudden carbon]

i was just wondering

lets say i have a circle inscribed in a square

how do i use integration to find the area wassted?

thi was what i thought

[sudden carbon]

oop

[winged lion]

You would integrate the red constant line and subtract from that the integral of the circle

[sudden carbon]

so the bounds are -r to r?

[winged lion]

Yes

[sudden carbon]

so for the square then i have

?

[winged lion]

,, \int_{-r}^{r} r , dx - \int_{-r}^{r} \sqrt{r^2-x^2} , dx

[sudden carbon]

huh

why is it r/2

[winged lion]

And that times 2 because of symmetry

Oh wait

[sudden carbon]

okay yay

and can you explain now what you meant earlier

like the use of double integrals

do you think it's necessary

cuz if it is i wanna incorporate it

to add dcomplexity

[winged lion]

It's more convenient to use polar coordinates because the computations become childs play

[sudden carbon]

but i thpought double integral gives volume lol

[winged lion]

That was the only thing

[winged lion]

Yes they do

[sudden carbon]

whats polar coordinates '-'

[winged lion]

But if f(x,y)=1 you really get area or a volume with height 1

[sudden carbon]

is this vectors 😭

[winged lion]

Lets just say coordinate transformation

[sudden carbon]

oh...

[winged lion]

Say $y=f(x)$ then $\int_a^b f(x) , dx = \int_a^b \int_0^{y} 1 , dydx$

[sudden carbon]

i just realized that the right hand funciton is gonna be a pain evaluating

[winged lion]

We can describe a one dimensional integral (which gives area) as a double integral

[winged lion]

Yes

[sudden carbon]

ohhhhhhhhhhh

[sudden carbon]

is there anyway i can simplify it...

[winged lion]

But you can do this instead and get easily πr²

[winged lion]

Trig sub

But its really work

[sudden carbon]

uhhh

i tryna comprehend this...

[sudden carbon]

whats the p looking thing?

[winged lion]

,, 2\int_{-r}^{r} r , dx - 2\int_{-r}^{r} \sqrt{r^2-x^2} , dx = 2\int_{-r}^{r} r , dx - \int_{0}^{2\pi}\left(\int_{0}^{r} \rho, d\rho \right) d\theta

[winged lion]

Just another variable called rho

[sudden carbon]

wow

do u think watchig a vid on double integrals will help

lol

[winged lion]

The double integral is pretty easy to compute

[sudden carbon]

oh rlly

[winged lion]

If you know normal integration yes

[sudden carbon]

i just learned normal integration last week 😹

ok wait

before i try to understabd that

this is acc th easier part

i lso

have to find the area wasted

[sudden carbon]

if i were to inscribe a circle in a hexagon....

is that even doable using integrals..

[sudden carbon]

help why is there a 2 infront of both functions

[winged lion]

It is but just means more work

[sudden carbon]

T-T

is it worth using integrals

to solve

[winged lion]

With a hexagon you just get two additional lines

[winged lion]

No

Integration is cool and all but really use geometry for these kind of stuff

[sudden carbon]

wdym

[sudden carbon]

😭

is it worth using integrals for the square even

[strange ermine]

Not at all

[sudden carbon]

💀

[strange ermine]

Why would you do $\int_0^L L\dd x$ when you can simply say the area is L² ?

[sudden carbon]

😹

to prove to my prof that yeah i know integral

but i feel like

it seems like such an overkill

[boreal girder]

Alberto Z.

[strange ermine]

Indeed!

[winged lion]

Give him this

[sudden carbon]

wgat the duck is this 😭

LOL

is that triple integral

[winged lion]

yes

[upper bane]

this is the kind of stuff i see in my nightmares lmao

[sudden carbon]

too scary

ty @winged lion

.close

[compact pewter]

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[tepid marten]

MID point theorem question:

[tepid marten]

If D, E and F are mid points of sides of AB, BC, CA of an isosceles triangle ABC, prove that triangle DEF is also isosceles

!status

[compact pewter]

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

[upper bane]

sketch

[tepid marten]

im at 2

[spark folio]

from midpoint theorem you should be able to calculate the side lengths of DE EF and FD

in terms of AB, BC and CA

[tepid marten]

in Triangle ABC : DE=1/2BC ( MPT)

DF= 1/2AC (MPT)

EF=1/2AB (MPT)

what to do after this

[spark folio]

isosceles triangle ABC
use this

[tepid marten]

how

[spark folio]

If the triangle ABC is isosceles, what does that tell you?

[tepid marten]

it got 2 sides equal

[spark folio]

mhm

lets say, its the sides AB and BC that are equal in length

[spark folio]

so, can you use that in these three calculations?

[tepid marten]

let me try

AB=BC so those 2 are equal

last side is unequal so hence proved?

[spark folio]

what did this prove?

that ABC is isosceles?

you already knew that

you need to bring in DEF

[tepid marten]

💀

i cant do it. Please tell me how to do it

[spark folio]

AB=BC right? So, 0.5*AB = 0.5*BC

[tepid marten]

why 0.5

[spark folio]

or DE = EF (from mpt)

[spark folio]

coz you have mpt that uses the factor of 0.5

u gotta use the stuff you got, otherwise whats the point

[tepid marten]

ok i got another question

when do you apply theorem 2

[upper bane]

i assume you have not specified what theorem 2 is

[tepid marten]

when 1 point meet

ig

and parallel to 3rd side

[spark folio]

can you draw a diagram of this and write what you mean in a proper complete sentence? It would be difficult to piece together and make sense of 10 different messages

[tepid marten]

scrap the previous question

here is what im saying:

Show that a quadrilateral formed by joining the mid point of the adjacent sides are equal

<@&286206848099549185>

[upper bane]

sketch?

[tepid marten]

idk how should i draw it

[upper bane]

wait let me think

you have an outer quadrilateral

then you are joining the midpoints of each pair of adjacent sides to form another quad

am i correct?

if not, please show the original problem

[tepid marten]

this is it there is no figure

[upper bane]

i mean i don't expect a figure, but the original wording of the problem

like a screenshot of it maybe?

or a picture

[tepid marten]

i wrote this from the book but ill send a screenshot

wait 1 min

[upper bane]

....you left out crucial parts of the problem!!!

[tepid marten]

mb 🙏

[upper bane]

ok so

this seems like another midpoint theorem question

draw a square, then dot the four midpoints

then join them

then i'm guessing you understand MPT usage from the discussion above?

[tepid marten]

yes

do we have to draw a square inside a square?

[upper bane]

yes

[spark folio]

...

you have to prove its a square

[upper bane]

i mean you have to draw a quad inside a square

you don't know yet if it is one

it's your job to show it

[tepid marten]

ok

.close

[compact pewter]

Channel closed

Closed by @tepid marten

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[topaz zephyr]

Is the channel occupied

[stuck haven]

no

You can ask your help

[spark folio]

yes, by you. It has your name on it now

[topaz zephyr]

I've solved all the problems but 47 no. Is tickling my brain. Been trying to do it for like the past 1 hour

[opaque iron]

Yes,

[plush belfry]

sine cosine pythagorean identity

[stuck haven]

does taking sin^6theta common reminds you of anything?

[plush belfry]

sin^2(x)+cos^2(x)=1

[topaz zephyr]

I've tried it

[plush belfry]

oh

did it work?

[topaz zephyr]

No

[stuck haven]

@topaz zephyr

what ab this

[topaz zephyr]

That's the reason I was asking help

[spark folio]

you can use 46 to show that this is 0

[topaz zephyr]

I gtg somewhere, I'll be back in an hour, till then can you guys try to solve it?

[spark folio]

and then do what lichtbach said

[stuck haven]

Yes

[topaz zephyr]

Did try it, didn't work

[stuck haven]

man

[spark folio]

do it again then, this time with help from ppl

[stuck haven]

im telling you

[topaz zephyr]

I can show you that it simplified to an expression after I come back

[stuck haven]

You be doing anything but that

Youre done from here

[topaz zephyr]

After that it doesnt simply anymore

[spark folio]

skill issue 💀

[stuck haven]

well

[spark folio]

it does give the answer

[stuck haven]

Yes

but we are helpers

[upper bane]

huh, is this something special

ok probably a bad idea to ask it in another person's help channel, sorry

[stuck haven]

It'll give you (sin^2theta+1)^3

[upper bane]

ahhh

[stuck haven]

so we can see that its cos^3theta*(costheta+1)^3, now taking the terms common

[upper bane]

wait then this applies to all of them no? they all can be split into (sin^2(theta) + 1)^(some power)

[stuck haven]

we arrive directly at the answer

[stuck haven]

since we have 3

i only can think of cubic

[stuck haven]

@upper bane as fom here and the thing I said

it is solved

[upper bane]

i see. from here i thought of turning this into a polynomial in (sin^2(theta) + 1)

[stuck haven]

that'd be hard

[upper bane]

not sure what i'm gonna cook from there

yeah

[stuck haven]

as ive said, I see the 3

so a cubic

and the degree is sweet

[upper bane]

i see, appreciate the insight

[stuck haven]

👍

[compact pewter]

@topaz zephyr Has your question been resolved?

[topaz zephyr]

This is the farthest I have went

[spark folio]

your 2 cos^2(x) became 2, which should have become 2 sin(x) instead

[compact pewter]

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[topaz zephyr]

Here you can see it clearly, it's the farthest I could simplify it

[upper bane]

it's already reopened

[topaz zephyr]

My bad

It does not work

Cause I've done it in the same way

Sinx = cos²x

[subtle helm]

u can factor the last expression maybe??

[ (\sin^3 \theta)(\sin\theta + 1)^3??]

[boreal girder]

k

[subtle helm]

it leads to smth weird..

[ [(\sin \theta)(\sin \theta + 1)]^3 = [(\sin^2 \theta + \sin \theta)]^3 = 1^3??]

[boreal girder]

k

[spark folio]

yea, thats the starting condition

[spark folio]

#help-43 message this is the original q

[subtle helm]

oh

[topaz zephyr]

Wait I think I solved it

[subtle helm]

yes

[subtle helm]

the final expression simplies to this

[topaz zephyr]

Using the relation (a+b)³ = a³ + b³ + 3ab(a+b)

[subtle helm]

ye..

[topaz zephyr]

Thanks for the help @subtle helm

.close

[compact pewter]

Channel closed

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[compact pewter]

Available help channel!

Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[molten badger]

In a square ABCD , point M and N lie on BC and DC respectively. $\widehat{MAN}=45^{\circ}$. Find the ratio between maximum area and minimum area of triangle AMN

[boreal girder]

Alexis_Fx

[molten badger]

the book provide this solution but I don't understand the first part

How BM+CN=MN here

[quiet sentinel]

Can you come up with a formula for where N is depending on where M is

Whats the relevance of P btw

[molten badger]

you rotate triangle ABM about A clockwise so that B lie over D then M is P

[cerulean steeple]

am i dumb or am i not finding it

[molten badger]

I mean the whole point of doing this way is not to do that

[molten badger]

Me too, and it doesn't even seem like it

.close

[compact pewter]

Channel closed

Closed by @molten badger

Use .reopen if this was a mistake.

[molten badger]

that part is non-sense I think

The rest is fine without it

hold on

.reopen

[compact pewter]

✅

[molten badger]

$MN \leq MC+CN$ this's not true

[boreal girder]

Alexis_Fx

[molten badger]

isn't it?

How can MN=MC+NC

this maximum doesn't exist right?

nvm I'm dumb

.close

[compact pewter]

Channel closed

Closed by @molten badger

Use .reopen if this was a mistake.

[compact pewter]

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!

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[sudden carbon]

is there anyway to solve this by hand

[hushed magnet]

u sub

[sudden carbon]

?

wdym

[hushed magnet]

do you not know what a u-sub is?

[sudden carbon]

oh i thought you meant

you sub

LOL

[sudden carbon]

what should i let u =?

[thorny urchin]

are yo allowed geometric interpretation

[sudden carbon]

no what i sthat ;

[strange ermine]

Hint: trig identity

[sudden carbon]

2hta

[signal willow]

draw the graph

[sudden carbon]

what

[sudden carbon]

its a cirxle

[hushed magnet]

and an integral is the area under the graph

[signal willow]

can you use that information to calculate the integral?

[thorny urchin]

semicircle

[sudden carbon]

yea and

[sudden carbon]

..

thats

[hushed magnet]

well whats the area of a circle

[sudden carbon]

the equivalence of

[hushed magnet]

or semicircle

[strange ermine]

Or even IBP, if you want to avoid both substitution and geometric interpretation

[covert crater]

yep

[sudden carbon]

F(r)-F(-r)

but i want

antideirvative

[thorny urchin]

"solve by hand" is vague,

to me means without a computer

[strange ermine]

So you want the indefinite integral?

[sudden carbon]

yes

[strange ermine]

Hence, without bounds?

[sudden carbon]

@hushed magnet can u explain

the u sub

what do i set u as

[sudden carbon]

well yeah ig

[hushed magnet]

lets imagine r=1 for a second

[sudden carbon]

ok

[hushed magnet]

can you tell me a trig sub which is somewhat connected to 1-x^2

[strange ermine]

You guess or you are sure? 😅

[hushed magnet]

if I set x = some trig function

[sudden carbon]

tan^&2?

tan^2x

?

[signal willow]

you can still get an antiderivative, just need more geometry

[sudden carbon]

brh

[hushed magnet]

thats not an identity

[sudden carbon]

ngm oop

[sudden carbon]

oops

the

sin^2x+cos^2x

= 1

[hushed magnet]

yes

so what happens if you set x=sin(u)

[sudden carbon]

what

1-sin^2u

cos^2u

[hushed magnet]

and that works out nicely because we also have a square root

[sudden carbon]

so its just

cosu?

[hushed magnet]

dont forget the du

[sudden carbon]

huh

oh 2wait

do i needa

adjust bounds too

[hushed magnet]

also absolute value

I thought you wanted without bounds

[sudden carbon]

oh

uhh

sorry

[dawn wadi]

[sudden carbon]

@hushed magnet

can i do it with bounds

how

[hushed magnet]

by transforming the bounds

if x=-1 and x=sinu, then u=?

[sudden carbon]

whyre we letting x = 1

[hushed magnet]

the lower bound

I assumed r=1 for a second

[sudden carbon]

idt

i should do that

cuz

r is just a costant representing variableradius

[dawn wadi]

What about setting x=rsinu to get a general formula

[hushed magnet]

well I wanted to do the easier case first

with one less variable to worry about

[sudden carbon]

okay

[sudden carbon]

huh?

what is u

[strange ermine]

The one you used...

[dawn wadi]

Are you more familiar with theta

[dawn wadi]

Oh that's true 😂

[sudden carbon]

yea

[sudden carbon]

..

can i do that...

r = some radius not 1 tho.....

[hushed magnet]

ok lets do the more general situation

we can write $\int_{-r}^r \sqrt{r^2-x^2} dx = r\int_{-r}^r \sqrt{1-\left(\frac{x}{r}\right)^2} dx$

[boreal girder]

Denascite

[hushed magnet]

and now just like before, we can set x/r = sin(u)

aka x = r sin(u)

[sudden carbon]

so its

sqrt (u-sin^2u)?

[hushed magnet]

where are you getting that first u from

[sudden carbon]

i thouht you get u =1

[hushed magnet]

what

where

[sudden carbon]

nvm]

bruh

what is u

[strange ermine]

........

[timid dune]

Or you could start with x=sqrt(r)sinu

[strange ermine]

Do you know what u-substitution is?

[hushed magnet]

do you know what a u sub is

[sudden carbon]

bro

[hushed magnet]

well you arent acting like it

[strange ermine]

Seems not at all

[sudden carbon]

but what did you let u =??

[timid dune]

It’s more so letting x=something

[hushed magnet]

you dont have to always write u = something explicit in terms of x

if you want you can solve x = r sin(u) for u

[timid dune]

With the objective of manipulation s^2+c^2=1 in order to remove the square root

[hushed magnet]

but that makes it harder to see whats going on

[sudden carbon]

my internet

i didnt rlly learn the sub method that wel

[sudden carbon]

but okay this makes sense

the way i learned it was express u in terms of x

mmb

[timid dune]

Nope you can do x=something too which would be essentially equivalent to u=something since

For example u=x^2+9

we could just rearrange that to x=something too

[sudden carbon]

okok

makes ense

[timid dune]

So using this, a common sub would be x/r=sin(u)

or cos(u), they produce the same result

The idea behind the sub is that when you sub that in, the thing inside the square root becomes simply sin^2u, which we can easily square root

[sudden carbon]

ty

i see it now

and we should also adjust bounds right

as well as dx

[timid dune]

Yep

[sudden carbon]

did i adjust the bounds correctly

i got pi/2 and 3pi/2

[timid dune]

using x/r=sinu?

It would be sinu=1 and -1 so u= pi/2 and -pi/2

[sudden carbon]

wait what

[timid dune]

Yeah so I see that u=3pi/2 also gives sinu=-1

But I believe that the values you take have to be in the same period

[sudden carbon]

i thought you sub x = -r and r into u = sin^-1(x/r)

[sudden carbon]

oh

oop

[timid dune]

Note that arcsin has a range of -pi/2 to pi/2

[sudden carbon]

oh ok

so i cant use 3pi/2

[timid dune]

I believe not

But you should double check

[sudden carbon]

is there an easier way outta this T_T

the u sub method is a bit annoying hkjskd

[timid dune]

In solving this integral?

[sudden carbon]

ye

[timid dune]

I mean this is simply just a semicircle with radius r

[sudden carbon]

why a semicircle?

[timid dune]

Geometrically, it’s just half the area of a circle with radius r

Because the semicircle function is that

y=sqrt(r^2-x^2), x^2+y^2=r^2 !

[sudden carbon]

nahhhhhhh

i made a huge

mistake

oops

i actually wanted to integrate a circle

(x - r)² + (y -r)² = r²

[timid dune]

That’s quite ambiguous

Integrate the area under a circle?

[sudden carbon]

yea

LOL

sounds funny

cuz thats just pir^2

but im tryna use itegrals

[timid dune]

I don’t quite get it

Trying to use integration to find A=pir^2?

[sudden carbon]

ya..

ok heres whbat im doing

i have a circle with radius r

inscribed in a square

and

im tryna find area wasted

using integrals 😹

[timid dune]

Ah

Well again you don’t need to use integration

But if you did, it wouldn’t be too difficult

[sudden carbon]

really

[timid dune]

But unfortunately I gtg

[sudden carbon]

nw

tysm

[compact pewter]

@sudden carbon Has your question been resolved?

[compact pewter]

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[fair thunder]

i dont quite get this solution, especially that part where Q(x) = 1 at most k times and after

[kind viper]

the equation Q(x)=1 i.e. Q(x)-1=0 has at most k solutions

do you agree or disagree w this

[fair thunder]

oh right, yes

same for the R(x) then, i got that

now after that im confused

and why are we choosing Q(x) = 1 or -1 and same for R(x)

[hushed magnet]

if P(x) is a prime, then Q(x)*R(x) is that prime, so that product has to be of the form (+-1)*(+- the prime)

[fair thunder]

aahhh

okay

i got it

thank you so much

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[kind glen]

$\int\frac{dt}{(t^2 +b^2)^2}$

[boreal girder]

Prathmesh

[covert crater]

I assume b is a constant?

[kind glen]

yea

[covert crater]

I'm thinking of a trig substitution like t = btan(u)

[kind glen]

ohh

[covert crater]

you essentially end up with this $\int\frac{b^{2}\sec(u)}{(b^{2}\tan^{2}(u)+b^{2})^{2}}du$

[rotund sphinx]

@royal terrace

[boreal girder]

឵឵MxRgD

[covert crater]

you can factor out the b^2 of the denominator to get (b^2(tan^2(u) + 1))^2 and then apply 1 + tan^2(u) = sec^2(u)

[compact pewter]

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[near bronze]

I'm trying to find an equation in rectangular coordinates for the curve by eliminating the parameter. And I did so by solving for (t) which, is t=(x-3) and substituting for t in the y equation

[near bronze]

and then cancling the (x-3) to get y= 1/3

but it just feels super wrong

like I can't do that for some reason but I have no idea why

[glacial prairie]

You can't eliminate because you dont have a product in the denominator

[near bronze]

so

if it were like this

[vivid breach]

,,y=\frac{x-3}{(x-3)+3}=\frac{x-3}x

[near bronze]

I could eliminate?

[boreal girder]

mtt

[vivid breach]

yea

your eliminating is really just dividing the top and bottom by a factor

if you tried to eliminate youd get $\frac1{1+\frac3{x-3}}$

[boreal girder]

mtt

[near bronze]

oh dang

yeah thats not pretty

ok so this is how we'd do it instead

and thats fully simplified

[vivid breach]

its y = (x-3)/x

[near bronze]

oh

whoops

yeah thats what I meant

thank you

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[vivid breach]

np

[compact pewter]

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[quartz yoke]

v = d/t, I tried t = 1/2(v1 + v2)d

[quartz yoke]

i dont think its corret tho

correct

[quiet sentinel]

v1t1=d/2

[short silo]

V1=d/t, v2=d/t right, distance in both instances are equal

[quiet sentinel]

v2t2=v1t1

[quartz yoke]

yes

[short silo]

So if we rearrange both formuals you in terms of d

[short silo]

It comes to this

[quiet sentinel]

$\frac{d}{t_1+t_2}=v_b$

[boreal girder]

ImOakley

[short silo]

This u can use logic ngk

Ngl it says deduce

[quartz yoke]

time is diff tho

so d = v1t1 + v2t2?

[short silo]

If i travel 20km/hr, for 1hr I travelled 20km right

[quartz yoke]

yes

[short silo]

That's the case travelling same two speed for equal times

So 30mims 30mins

Makes it ahr

[quartz yoke]

so t is the same?

[short silo]

Yes as in the question

[quartz yoke]

ok thank you

[short silo]

Hold om a sec

Sorry I'm weogn with my approach, it's more of a prove

Then deduce

[short silo]

This is correct

Use this

[quiet sentinel]

k ive found it

you get it in terms of v1 v2 and t1 and then the answer will become clear after simplifying

[quartz yoke]

so there are different times for the distances?

[quiet sentinel]

yes

if you travel the same distance at a higher speed you get there quicker right?

[quartz yoke]

yea

so d = v1t1 + v2t2

[quiet sentinel]

d/2 = v1t1

and so does v2t2

[quartz yoke]

oh ye

and then v1t1 = v2t2

[quiet sentinel]

so if you make it like $t_2=\frac{v_1t_1}{v_2}$

[boreal girder]

ImOakley

[quiet sentinel]

and take that $d=2v_1t_1$

[boreal girder]

ImOakley

[quiet sentinel]

plug those into this

[quartz yoke]

what about t1

do i put in t1 = v2t2/v1

[quiet sentinel]

it cancels out

$\frac{2v_1t_1}{t_1+\frac{v_1t_1}{v_2}}=v_b$

[boreal girder]

ImOakley

[quiet sentinel]

$\frac{2v_1t_1}{\frac{v_2t_1+v_1t_1}{v_2}}=v_b$

[quartz yoke]

2/t1 + 1/v2

[boreal girder]

ImOakley

[quiet sentinel]

$\frac{2v_1v_2t_1}{t_1(v_1+v_2)}=v_b$

[boreal girder]

ImOakley

[quartz yoke]

thats the correct awnser yea

[quiet sentinel]

$\frac{2v_1v_2}{v_1+v_2}=v_b$

[boreal girder]

ImOakley

[quiet sentinel]

yeah

[quartz yoke]

thanks this one was tricky

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[humble mortar]

Is there a standard way to construct a transversal set?

[humble mortar]

Specifically, I have a relation ~ on the real numbers, and I want to construct an arbitrary set of representatives using the axiom of choice.

I can't find any widely-used notation. Everything is wordy.

[lime juniper]

transversal meaning, containing one element from each equivalence class?

just call it a choice set

[wind tapir]

you could use "$R=\bigl{f(x):x\in X/{\sim}\bigr}$, where $f$ is given by AC".

but what's the problem with being wordy if it's clear; something like "let R be a transversal of X"?

[boreal girder]

Desync

[humble mortar]

Thanks. Those will work.

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[high shadow]

hi, what is the formula for this

i have 10000 and interests of 1% in year scale with it increasing by 0,5% each year.

[eternal pulsar]

Try to write it out, then put it in sum notation

[compact pewter]

@high shadow Has your question been resolved?

[glacial prairie]

And then tell us what your result is

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[fossil reef]

do you get zero when you differentiate a vector with constant magnitude?

[fossil reef]

but also changing direction

[narrow prairie]

No. If you imagine a car going in a circle at a constant velocity, the time derivative of its velocity vector would be an acceleration vector in the direction the car is turning

[keen granite]

I mean if r(t)=(cost,sint) then |r(t)|=1 but r'(t)!=0 but r'(t)=(-sint,cost)

[fossil reef]

okay thanks

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[merry lily]

so confused, its 9 or g?

[merry lily]

AM-GM Inequality

[upper bane]

the context of the multiplication tells me it's a 9, but the question says it's a g

typo maybe?

[native shard]

multiple typos

in the red box the second alpha beta gamma should be raised to the power of 2/3

and it should be = alpha beta gamma

[compact pewter]

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[merry lily]

Im not really understand, can anyone explain futher please

[upper bane]

why does this just seem like creating a GP by inserting three more terms in between 1 and 16

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[jagged scarab]

@quartz yoke

[quartz yoke]

Hello.

Wait.

Can I send attachments

[jagged scarab]

I’m gonna close it but u stay here

yes

[quartz yoke]

Yes I can

[jagged scarab]

.close

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[twin idol]

Lmao what even happened here

[jagged scarab]

He was having trouble 😭

[old mica]

eh?

[twin idol]

lmao

[quartz yoke]

I am kinda stuck on these two, don’t solve, I just want hints.

[old mica]

he needs to open his own channel if he wants to get help

.reopen

[compact pewter]

✅

[quartz yoke]

Just give me subtle hints

[jagged scarab]

I don’t think he could even find it 😭

I just @ed him

[old mica]

sure, but don't close it for him

it'll close the channel otherwise

[jagged scarab]

Ohhh I thought it would just become his channel my bad

[old mica]

now you know

[old mica]

^^ for future helpers, here's the question

[quartz yoke]

This btw

...

My internet

[compact pewter]

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[molten badger]

Simon owes the bank $5,000 each year for a duration of 4 years to support his college education. The annual interest rate is set at 3%, and during this 4-year period, Simon did not make any payments to the bank. After completing his college education, exactly 4 years after the initial loan, he did not make any payment in the first month. Nevertheless, in the second month, he decided to start making payments in installments of $x every two months, with each payment made at the end of the month and spaced precisely two months apart. At this point, the bank's interest rate is calculated at 0.3% per month. It is understood that Simon plans to completely pay off the debt within a period of 5 years. Calculate the value of x.

[molten badger]

Holy shi that's long

[icy nymph]

yuck

[wet lichen]

Maybe finding the present value of the debt first

Or maybe its accumulated value at the end of year 4. (This should be more convenient in this case.)

[molten badger]

that's the easy part

[wet lichen]

Ok

[tranquil prawn]

should we assume as "within 5 years" as a total of 30 payments?

[molten badger]

uhh

yes

[tranquil prawn]

the payments being discrete is the biggest problem ngl

[molten badger]

wait are you Dexter

[tranquil prawn]

If this serves any purpose, this is basically the behaviour of the amount of debt

[wet lichen]

Use formula of geometric series to get the PV at year 4 of the payments in terms of x. (Note there are 30 payments (payment per 2 months) and the interest between each payment is (1+0.3%)^2-1)

[molten badger]

year 4?

[tranquil prawn]

on this, yeah, you can probably manage to simply ball the value by "repeating" this process 30 times

basically
D -> D-x -> D(1.003^2)

[molten badger]

$T(1+0.3%)^{59}-x\frac{1-((1+0.3%)^2)^{29}}{-(0.3%)^2}$

I got this earlier not sure if it's true

T is the debt

[tranquil prawn]

Thing is that, the more you pay, the less the interest really affects the current debt
so calculating 1.003* ^ 59 is basically ignoring the factor of you paying

unless im not seeing something

[molten badger]

idk

[boreal girder]

Alexis_Fx

[tranquil prawn]

ill give it a lil bit of thought, ill try to come back with something

[compact pewter]

@molten badger Has your question been resolved?

[compact pewter]

@molten badger Has your question been resolved?

[wet lichen]

Hint: The formula for pv of annuity immediate with payment of 1 at the end of each period is $\frac{1-v^n}{i}$ where i is interest rate per period, $v=(1+i)^{-1}$ and n is the number of payments.

[boreal girder]

T

[slender kraken]

for 9) use 7-th roots of unity

[molten badger]

-1? Why -1

I know this's GP right?

sum of GP

[wet lichen]

And my apologies: I forgot that the loan is still accumulating after year 4. So it may be more convenient for you to think at the end of all payments (year 9).

[wet lichen]

v is the discounting factor per period

[molten badger]

Okay but how do we get this formula

[wet lichen]

With sum of gp

[molten badger]

I mean it look like GP to me but v=(1+i)^-1

Why -1?

No sum of GP is -1 here

Or I'm missing smth

[wet lichen]

You can take each payment as a new bank account

Because it is present value so the present value is like $(1+i)^{-1} + (1+i)^{-2} + \dot + (1+i)^{-n}$

[molten badger]

$S=\frac{1-(\frac{1}{1+i})^n}{1-(1+i)}$

[boreal girder]

T

Alexis_Fx

[molten badger]

No sum of GP look like this

[wet lichen]

The denominator is $1-1/(1+i)$ and first term is $1/(1+i)$

[boreal girder]

T

[molten badger]

It suppose to be $S=\frac{1-q^n}{1-q}$

[boreal girder]

Alexis_Fx

[molten badger]

You wrote it 1-v^n/i

Also I don't think the exponent is negative here

[wet lichen]

$v + v^2 + \dots + v^n = v \frac{1-v^n}{1-v}= \frac{1-v^n}{v^{-1} - 1}= \frac{1-v^n}{i}$

[boreal girder]

T

[wet lichen]

Note $v = (1+i)^{-1}$ as defined here

[boreal girder]

T

[wet lichen]

And the accumulated value is (1+i)^n times of this formula

[molten badger]

wut

So v^-1-1 is not i isn't it

How $v^{-1}-1=i$

[boreal girder]

Alexis_Fx

[molten badger]

Oh iee

[wet lichen]

The pv formula is actually the same as yours except for the ± sign

[molten badger]

I'm lowkey don't understand why can you just write

Like

Sum of GP with a1=1

I know the first 4 years is $5000\frac{1-(1+3%)^4}{3%}$

[boreal girder]

Alexis_Fx

[wet lichen]

This is correct if you take loan as negative

[molten badger]

Yeah I said I can do this part

But the second part is confusing

Atleast for me

[wet lichen]

Now the problem is the interests of loan and payment are different. If I understand correctly, the person will put his money into his bank account and pay the loan at once after 5 years.

The PV of the installments is $\frac{1-(1+0.003)^{-60}}{(1+ 0.003)^2 - 1}$ at year 4.

[upper bane]

your \frac is starting with a [ instead of a {

not sure if you intended that

wait nvm

[wet lichen]

Oh

[upper bane]

sorry

thought i fixed it

[wet lichen]

Maybe % sign.

[molten badger]

No he pays each month

[boreal girder]

T

[molten badger]

You have to write the % with a slashback

[wet lichen]

Thanks for telling me that

[merry lily]

what GP?

[upper bane]

sorry, this is not your channel atm

if you need to ask, please forward your question to another open channel

[molten badger]

So he pay $x then the debt drop and next month it's increasing by 0.3%

Next month he pay $x and continue

Every 2 months

Sorry

Every 2 months not each month

[wet lichen]

If that's the case, you can find x by equating the accumulated value of the loan and the pv of payments at year 4

[molten badger]

Okay how would you do that

[wet lichen]

At interest rate 0.3%

[wet lichen]

x times this and equate it with the accumulated value you found for the first 4 years

[molten badger]

Wait why? The payment also affects the debt

Why can we still use that formula?

[molten badger]

We can't just use this can we?

[wet lichen]

This gives the present value of all payments when each payment is 1 so the pv of payments is x times this formula

But at year 4 we already have a loan, which is the accumulated value of the first 4 years

To repay the loan, we want something with the same PV

[molten badger]

Okay

[molten badger]

So this time x is the payment you say

And what about the debt

[wet lichen]

We can take the reference time as year 4, i.e., exactly when no more debt occurs, so the value of the debt is the accumulated value you calculated here.

[molten badger]

Uh...

The debt also has an interest rate after 4 years

It's 0.3% per month

The first 4 years is 3%/year and the next 5 years is 0.3%/month

[wet lichen]

You can imagine it is a lump sum loan occurring exactly at the end of year 4, with value same as the accumulated value. In this case, we can forget the 3% p.a. interest

[molten badger]

Yes, but how can we calculate the debt in the next 5 years?

[wet lichen]

Instead of accumulating the debt to year 9, we discount the level payments back to year 4

If you are not comfortable with this, you can also accumulate everything to year 9

[molten badger]

Uhhh I genuinely doesn't understand

Do you have like a formula to calculate that 5 years debt

I may know what's going on when I see the formula

[wet lichen]

You can imagine you are not repaying the debt and accumulate it for 5 years, i.e., multiply its value at year 4 by (1+0.3%)^60

[molten badger]

What

[wet lichen]

We can separate the debt and the payments (Note (Debt - payment)(1+i) = Debt(1+i) - payment*(1+i))

[molten badger]

It can't work that way, can it?

The payment affect how the debt increasing

That make no sense to me

[wet lichen]

Like $((Debt(1+i) - x)(1+i) -x)(1+i) -x = Debt (1+i)^3 - (x(1+i)^2 + x(1+i) +x)$

[molten badger]

How do you get this?

[boreal girder]

T

[wet lichen]

I iterate the interest accumulation process

[molten badger]

How exactly?

[wet lichen]

Debt = Accumulated value of the loan at year 4
x = payment
1 + i = accumulation factor for 2 months, i.e., (1+0.3%)^2
Year 4 + 2 mon: Debt accumulated to Debt(1+i) and repay x, so the value of loan is Debt(1+i) - x
Year 4 + 4 mon: Debt(1+i) - x accumulated to (Debt(1+i) - x)(1+i) and repay x, so the value of loan is (Debt(1+i) - x)(1+i) - x.
And so on

[wet lichen]

And if you look into the resultant formula, you will see it can be separated as Accumulated value of Initial Debt - Accumulated value of level payments at interest rate at which the debt is accumulating.

[molten badger]

So, in year 9 is?

Uh

[molten badger]

I found closed form of this sequence

[molten badger]

This one

T is the debt

[molten badger]

Your work seems like what I did except this part

[molten badger]

I don't understand how you get this equation

a1=T(1+i)
a2=T(1+i)^2-x
a3=T(1+i)^3-x(1+i)
a4=T(1+i)^4-x(1+i)^2-x
.....

The payment make a sum of a geometric

[wet lichen]

Oh. I took 2 months as a period

So my 1+i is your (1+i)^2

For this equation

[molten badger]

Okay but I got a sum of GP

So it's like

$a_n=T(1+i)^n-x\frac{1-((1+i)^2)^{\floor{\frac{n}{2}-1}}}{1-(1+i)^2}$

[boreal girder]

Alexis_Fx

[molten badger]

I don't understand how you get this

[wet lichen]

It was an example showing why we can separate the debts and payments

[molten badger]

You mean like this?

Yeah I clearly didn't understand what you said sorry

I misunderstanding it

[molten badger]

So is this formula correct

T is the debt after the first 4 years

[wet lichen]

I think we are paying at the end of each 2 months so the exponent there is simply floor n/2 without -1

[molten badger]

Alright I will fix it

Alright thanks you:D

@wet lichen

.close

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[open lotus]

i need help on 18 but don’t know where to start

[open lotus]

limits is a little difficult for me

do i plug in 0 to h?

[covert crater]

you can't do that unfortunately

[open lotus]

[covert crater]

you'd get the denominator as being 0 and you can't really divide by 0

[stark marsh]

do it and see what you'll get if you really want ....

[open lotus]

i wouldn’t since the denominator is 0

i see a 4 so i think that’s a special number

wait no

ln4 is really important here

i js don’t know why

[covert spear]

ln(a) - ln(b) = ln(a/b)

[stark marsh]

then you answered your own doubt
anyways try log properties

[open lotus]

ln(4+h/4)/h

[covert spear]

simplify

[stark marsh]

go on

[cold widget]

go w log properties , and also if you have done derivative defination as limits do you recognise something ( judging by the 19 question i felt you have done that )

[open lotus]

OH THE 4 CANCEL OUT

i think

[covert spear]

yes

[stark marsh]

there is still the h ...

[covert spear]

1 + h/4

[open lotus]

so we’re trying to get rid of the h?

[tulip spear]

try by expansions?

[stark marsh]

in someway yes

[stark marsh]

at least the h in the denom.

[open lotus]

separate the numerator and divide each of them by h and subtract them?

[tulip spear]

mb its l-hospital form

[verbal rover]

Set h= x-4 and work with hopital easier way

[tulip spear]

differentiate numerator and denominator differently and take the ratio of them

[open lotus]

why should we make h=x-4? did we set 4+h=x?

[stark marsh]

it's a limit definition of the derivative of ln(x) at x=4 yk ...

[tulip spear]

be careful this works for 0/0 and infinity by infinity form

[open lotus]

[verbal rover]

Simble variable change and yes

[tulip spear]

wont it work?

[verbal rover]

Simple*

[stark marsh]

it works but I don't think op learned derivatives rules yet ...

[tulip spear]

hahaa

[cold widget]

and its cyclic

[tulip spear]

that isnt my deal mate

i told an approach

now nothing more

[cold widget]

? you arnt here to solve problems mate , you are here to help the helpee

[open lotus]

so now i plug my new h term into the purple equation right?

[tulip spear]

its done.

[covert crater]

?

[steel ingot]

I wonder if OP has some identities in their arsenal, e.g. $$\lim\limits_{x\to 0}\frac{e^x-1}{x}=\cancel{0}{1}$$

[open lotus]

WHAT 😭 😭

[verbal rover]

Yes and just turn the -ln4 back

[kind viper]

that's 1 not 0

[cold widget]

khudka padhle jee ke liye

[steel ingot]

Yeah I got rusty, standing corrected...

[tulip spear]

thanks sir

[boreal girder]

Rafain

[open lotus]

ok so i change my equation back?

to the og?

[verbal rover]

Yep and set h = x - 4

Youll have lim x>4 (lnx - ln4 ) / x-4

Right?

[kind viper]

are we not allowed to recognize the limit directly as $\ln'(4)$ by the way

[boreal girder]

Ann

[kind viper]

i feel like there's been some beating around the bush

[open lotus]

[verbal rover]

Well this is the definition of differentiation

[kind viper]

yes that's what i am saying

[open lotus]

i hate that term

[kind viper]

do you know derivatives or not @open lotus

[verbal rover]

Do you know differentiation?

[open lotus]

derivatives are slopes of a function at a given point

[open lotus]

i know abt power rule

[kind viper]

1. do you know the limit defn of the derivative, Y/N
2. do you know what the derivative of ln(x) is, Y/N

i'd like a pair of straight answers this time instead of trying to say more words than i ask for

[verbal rover]

Yes and the definition or derivative in one point a is
lim x>a (f(x) - f(a) ) /(x-a) = f’(a)

[open lotus]

d/dx lnx = 1/x

[open lotus]

1. no
2. yes

[verbal rover]

[open lotus]

wait that’s wrong?

[verbal rover]

Here is the def

[covert spear]

in your case x is 4

[verbal rover]

Nothing im just crying because im happy

[strange ermine]

Mmh are you sure you haven't been taught the limit definition of the derivative?

[open lotus]

[strange ermine]

Check your notes/textbook

[open lotus]

i have i js have trouble grasping it

[strange ermine]

So you should know it but you don't remember it atm, right?

[open lotus]

lemme write this down and put a star next to it

[covert crater]

just to be clearer it's this. $\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$

[boreal girder]

឵឵MxRgD

[verbal rover]

Alr

[strange ermine]

Look at this @open lotus

[verbal rover]

The same thing

[strange ermine]

Yeah but OP's question resembles this other format

[covert crater]

Yeah I know. I typed in latex so it's clearer for them to see

[verbal rover]

Ye i need to learn latex

[open lotus]

same

[open lotus]

got it written down

[strange ermine]

Now apply it to your exercise 😉

[open lotus]

now i js have algebra work right?

my denominator is still 0

[covert spear]

don't just plug in 4

[open lotus]

oh…

[covert spear]

you can do it with the definition of the derivative but it's time consuming

[open lotus]

test is in 6 minutes btw

[verbal rover]

ln x - ln 4 and observe if there is any similarities with the definition