help-10

[graceful stirrup]

.close

[obtuse pebble]

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[obtuse pebble]

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[west delta]

<@&286206848099549185>

[leaden ginkgo]

!15m

[obtuse pebble]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[west delta]

o oops im new

[hexed zinc]

kill

hi

[hexed zinc]

it’s 1

[west delta]

tf

no

wait here is the solution

im not sure if ur trolling

[hexed zinc]

buddy ur arguing with someone in calculus 3

[west delta]

or not

[west delta]

bruh i did that too

also

[hexed zinc]

oh wait

what math is this

[west delta]

calc has nothing to do with

[west delta]

uhh number theory or alg

[west delta]

what is bro

waffling abt

[hexed zinc]

oh

ok nvm

[hexed zinc]

i love you

[west delta]

same'

❤️

💀

bro pulled

[hexed zinc]

no one saw that bro

[west delta]

the uno reverse

[limber quartz]

This ain't calculus

[west delta]

halp

[hexed zinc]

ok if this ain’t calculus then ur real analysis

[west delta]

👁️

[hexed zinc]

i’m math 55

[west delta]

im math 69

[hexed zinc]

the hardest class at harvard

[west delta]

hardest class at clown school

[limber quartz]

I don't remember this stuff. You need somebody who's recently done Discreet or maybe Combinatorics

[west delta]

wait heres the solution

i jut need someone

to decipher it

[limber quartz]

I can't do that, there's too much unspoken

[west delta]

oh ok

alr

[hexed zinc]

are you a discrete function or continuous

[west delta]

💀

bro is unironically using math in real life

[limber quartz]

But I recognize the series as being used to solve problems in Combinatorics

[west delta]

yeah its supposed to be a generating function

just no idea how he factored it

<@&286206848099549185>

[limber quartz]

There's the words

Generating functions

[west delta]

ye

its the only gf ill ever expand

[hexed zinc]

hi

how can i help

[west delta]

wow

[west delta]

halp

[wary trench]

hi

[glacial crag]

Hi

[west delta]

help

[obtuse pebble]

@west delta Has your question been resolved?

[obtuse pebble]

@west delta Has your question been resolved?

[obtuse pebble]

@west delta Has your question been resolved?

[west delta]

.close

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[obtuse pebble]

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[uncut parcel]

How would you get f(-3^(1/2))

[tardy epoch]

f is just made up of straight lines

Find h' with product rule first

[obtuse pebble]

@uncut parcel Has your question been resolved?

[uncut parcel]

h' is f(x) * 1/(1+(x/2)^2)^(1/2) + arcsin(x/2) * f'(x)

<@&286206848099549185> \

[uncut parcel]

<@&286206848099549185>

[hot harness]

Look at the graph

f(x) is a line from x <0

So if you find the slope you can find f(-sqrt3)

[hot harness]

Do you mean *1/2 not ^1/2

[uncut parcel]

ok

[uncut parcel]

yeah

thx

[uncut parcel]

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[frosty brook]

I need to prove this statemen

[frosty brook]

Here is my attempt with induction

Can somebody proofread it?

[frosty brook]

I tried to induct on number of edges. Denoted with n in the proof

[obtuse pebble]

@frosty brook Has your question been resolved?

[obtuse pebble]

@frosty brook Has your question been resolved?

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[devout yarrow]

I'm currently training for a math competition, and in several practise questions i found that it is required to be able to isolate a variable in a two variable quadratic equation (usually not factorable). One such example that I need help with understanding: Find all integer solutions to $x+y=x^2-xy+y^2$. The solution involved isolating y, and I do not know how to do so. Any help is appreciated.

[warm shale]

everything_addict

[drifting roost]

isn't the quadratic just:

x^2 - (y+1)x + (y^2-y) = 0

[tidal gazelle]

those questions are written with the intention that you should not be able to isolate any of the variables, you solve them with other ways such as finding inequalities or in your quadratic finding the roots and setting them equal to zero to break it into smaller equations

[obtuse pebble]

@devout yarrow Has your question been resolved?

[devout yarrow]

The solution in the book itself finds y in terms of x

[tidal gazelle]

could you show the equation the book has for y in terms of x

[devout yarrow]

oh

ok

$$y=\frac{x+1\pm\sqrt{-3(x-1)^2+4}}{2}$$

[warm shale]

everything_addict

[devout yarrow]

here

[tidal gazelle]

yeah thats the same as finding the roots of the quadratic, in this case the quadratic is $y^2 + y(-x-1) - x + x^2 = 0$. Apply the quadratic formula and you will get the same equation.

[warm shale]

nova

[devout yarrow]

ok

[tidal gazelle]

Now the term under the square root must be a perfect square in order for y to be an integer

[devout yarrow]

ill try manipulating to get that form

[devout yarrow]

yep

its easy from there

thanks for ur help

i managed to manupulate it to a typical quadratic

.close

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[plain sparrow]

Two mutually directed chords are drawn from one point of the circumference

[plain sparrow]

how to draw this

[obtuse pebble]

@plain sparrow Has your question been resolved?

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[dim breach]

i need help

[final thunder]

!da2a

[obtuse pebble]

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

[dim breach]

hi

[final thunder]

hi

[dim breach]

can u dm me

[final thunder]

no

[dim breach]

i need personl help

[final thunder]

Just post your question here.

[polar fossil]

eyebrow raised

[dim breach]

i need help with 5a

and 5b

[final thunder]

[dim breach]

can u solve 5a?

i have no ide

?

[final thunder]

What do you know about a rhombus.

[dim breach]

it has 4 side

☠️

[final thunder]

What do you know about the lengths of the four sides?

[dim breach]

ig i gotta make triangel ccongurent

[final thunder]

right

[dim breach]

.close

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[final thunder]

[obtuse pebble]

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[inner helm]

Does anyone know how to answer these questions?

[tender stratus]

assuming those are functions, g/f seems to represent g(x)/f(x) for x in the intersection of their domains

[mossy adder]

It does seem like you do have to intersect the domains, so do that first

I never encountered that being a thing, but I guess it is here

and then you can answer for each point x in the domain what f(x)/f(x) is

and that way you have a new function

[inner helm]

huh

[inner helm]

I didn't get it

[tender stratus]

are you familiar with the concept of functions?

[inner helm]

yes

for every input of x it has to have only 1 output

[tender stratus]

so if you think of f and g as functions, you can say stuff like f(3) = 2 and g(3) = 8

[inner helm]

yeah

[tender stratus]

and then we define g/f (a new function) as g(x)/f(x)

but we cant really allow x for which we have no outputs from g(x) or f(x)

so we restrict x to be in their common domain

[inner helm]

what do you mean by "allow x"

[inner helm]

how

how did they get this

same here

I don't get how to answer these questions

[tender stratus]

oh so you didn't understand the + one too

[inner helm]

yes

[tender stratus]

might help to read on "function composition"

wait thats not the right term i think

one min

[inner helm]

I know how to find the composite of functions if they ask for (f(gx)) or g(f(x)

[tender stratus]

sorry, composition was the wrong word

when they say f + g it's just the sum of their individual outputs

[inner helm]

Sum and difference and product and quotient of function you mean

[tender stratus]

yes

[inner helm]

but Ik how to solve general questions

but when they give an order just points

idk

[inner helm]

oh wait lemme try

Yah no Idk

[tender stratus]

suppose I want to find (f+g)(2)

this is f(2) + g(2)

by definition

can you find that in the second question?

[inner helm]

f(2)=-1
g(2)=-3
f(2)+g(2) = (-1)+(-3) = -4

[tender stratus]

thats correct

[inner helm]

ohhh

[tender stratus]

so now we add (2, -4) to our set

[inner helm]

I see

okok I'm starting to get it

what about the division

[tender stratus]

it's the same thing, except you do g(x)/f(x)

[inner helm]

f(2)= -1
g(2)= -3
g/f (2) = -3/-1 = 3

THANKSS

I get it

[tender stratus]

np

I hope you understood about the domain now, too

[inner helm]

yes yes

.close

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[daring cairn]

could anyone help me prove this

[daring cairn]

uhh could u explain i am kinda confused

[glossy basalt]

sure

[glossy basalt]

do you recall any theorem that relate the ? with theta?

[daring cairn]

uhh no

[glossy basalt]

oh ok,do you recall
angle at centre=2×angle at circumference

[daring cairn]

uhh yes

but i don't see a relation of that with this question tho-

[tender stratus]

it's an isoceles triangle

[glossy basalt]

now that you notice that theta/2 you can try and notice which tan(theta/2) related to

specifically which right-angled triangle

do tell if you want to get the triangle highlighted

[daring cairn]

that would be extremely helpful

oh i see it now

[glossy basalt]

see if you can get it now 🙂

[daring cairn]

i'll check once if i don't get it i'll ping is that okay

and

thank you

.close

[obtuse pebble]

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[inner helm]

How do I solve this?

[versed stratus]

can you express 27sqrt7 as the product of two terms?

[inner helm]

yes

log(base 3) 27 + log(base 3) sqrt7

it'll be 3 +

idk how to get x/2

ohh

I got it

[inner helm]

thanks

.close

[obtuse pebble]

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[final thunder]

<@&268886789983436800>

[upbeat plinth]

.close

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[daring cairn]

could someone help me solve it i got a quadratic equation 11x²-78x-120=0 which doesn't solve to whole numbers

[royal basin]

show your work

maybe you made some kind of arithmetic error somewhere

also what are your pronouns?

[daring cairn]

anything i don't mind

[daring cairn]

it's kinda messy u won't understand

[royal basin]

ok you can take off all pronoun roles except "any pronouns" for future ref

[royal basin]

then rewrite it to make it not messy

otherwise it is impossible to diagnose where your error came from

[daring cairn]

alright is okay if i close the req now and reopen it in 10mins with same question or will this channel stay like this ?

[royal basin]

you'll get pinged after some time, press ❌ on the bot prompt to keep the channel open.

then ping me once you've got your work so i can look at it.

[daring cairn]

oh alright thanks

[royal basin]

i reproduced the thing and got almost the same equation as you though, except my x coefficient is different. mine doesn't seem to have any integer solutions either though.

although nobody said i didn't fuck up either

hang on

nevermind, my equation does have integer solutions (hopefully).

[obtuse pebble]

@daring cairn Has your question been resolved?

[daring cairn]

on redoing it i have noticed i made a mistake

is the equation 11x²-98x-120=0

[royal basin]

matches mine now

[daring cairn]

yes got the answer

x=10

thanks

.close

[obtuse pebble]

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[odd wraith]

guys

[odd wraith]

f(x) 9x^2-2x and g(x) = e^8x

for f(x) i did 3 derivatives

and for g(x) i did 3 integrals

in the final i got

this

but when i check an example, it wants an answer like this:

but i couldnt get the last part

[odd wraith]

this one is from a differnet question

differnet*

different*

[inland matrix]

Answer like that? Wdym?

[odd wraith]

I can't get it this way

[inland matrix]

The first term, ig in the brackets it should be (9x²-2x) instead

[odd wraith]

yes

okay i got it this way

but im not sure is it correct or no

thank you

.close

[obtuse pebble]

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[sinful eagle]

Lg100-lg0.1

[sinful eagle]

Can I do

Lg10^2

2*lg 10 - lg10^-2

[sweet gulch]

what is lg?

[sinful eagle]

2*1+2

4

[meager aspen]

Uhh hello?

[timid silo]

log most likely

[sinful eagle]

log

[sweet gulch]

ok

[timid silo]

the base though? couldnt tell u

[sweet gulch]

yeah, 10^2 = 100 and 0.1 = 10^-1

[sinful eagle]

base don’t matter

[sweet gulch]

not 10^-2

[timid silo]

Tbh just stick it in one log

log 100/0.1

[sinful eagle]

ah so 3?

[timid silo]

log 1000

[sinful eagle]

.close

[obtuse pebble]

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[timid silo]

3 log 10

[obtuse pebble]

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[deft torrent]

can someone help pls

isnt angular velocity radians/time

how do i find radians

[timid silo]

angular velocity is radians/time in seconds yes

[mossy adder]

go through the definitions and information over time

you know the length, so you can calculate the area of its circle

then you can calculate how frequently it goes through that circle

[timid silo]

Yeah u can calculate the amount of radians swept in the 8s using the area given and the area of the circle

then just divide by 8 to get rad/s

[deft torrent]

why do you need the area of the circle ?

113.10 is what i got for that

[timid silo]

because your 50 square meter area thats been swept out is the area multiplied by some amount of radians

[deft torrent]

ok i got it

.close

[obtuse pebble]

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[fierce vale]

what does this mean ??

[fierce vale]

ik it's Laplacian but idk what this is :

[past sand]

Just a vector

[kind hawk]

problems with old books: old notation

[fierce vale]

prof is still using it till this day : (

.close

[obtuse pebble]

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[wicked tree]

no that’s a logarithm

probably functioned in with a square notation

[obtuse pebble]

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[deft torrent]

im having trouble with the rad/min

[timid silo]

imperial system… yikes

Basically the 65 miles per hour is your linear velocity

12 inches is your radius

do the conversions needed

then plug them into the formula

[hexed agate]

*15

[timid silo]

mb im dyslexic

[deft torrent]

ok i got the rest of it

ty

.close

[obtuse pebble]

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[vernal void]

The generating cone is inclined to the plane of the base at an angle ү. The distance from the top of the cone to the center of the sphere inscribed in it is equal to d. Determine the area of ​​the lateral surface of the cone.

[vernal void]

I think it looks something like this

so S= piRl

I need to find CB

and radius

idk how do it

[obtuse pebble]

@vernal void Has your question been resolved?

[vernal void]

<@&286206848099549185>

[frigid goblet]

what DO you know ?

[vernal void]

.

this

[vernal void]

and this

[frigid goblet]

it doesn't say if you know the value of anything

[vernal void]

Well, how should I answer the question what do you know?

2+2=4

I know what's next

[frigid goblet]

are you trolling or do you actually need help?

[vernal void]

really need help

[frigid goblet]

ok

[vernal void]

I just gave an example that your question is not very precise

[frigid goblet]

do you know the value of mAB?

[celest halo]

this is the only information that the question gives?

[vernal void]

yes

[vernal void]

m?

[frigid goblet]

mesure of point A to point B = mAB

[vernal void]

this is the diameter of the base

[frigid goblet]

!original:

[obtuse pebble]

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.

[vernal void]

it's in Ukrainian, can you translate it?

[frigid goblet]

yes, mostly with my phone

[vernal void]

this is the task

[frigid goblet]

what grade are you in, cause i'm going to use university level stuff

[vernal void]

Grade 11

the last one in Ukraine before the university

[frigid goblet]

first of all

this is the formula for the area

[vernal void]

not this?

[frigid goblet]

you are right my formula add the base under witch is out of the question

[vernal void]

I think we need to use trigonometric formulas here

[frigid goblet]

source is machinery's handbook

[vernal void]

so we have a circle inscribed in a cone and not a cone in a circle

oh

sry i blind

but I don’t quite understand how this will help us

[frigid goblet]

i showed you how to put a circle inside a triangle, all you have to do is the process inversed

[vernal void]

.close

[obtuse pebble]

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[timid silo]

just wanted to ask for this question

is $\frac{x+6}{(x+3)(x-1)}$

[warm shale]

mj

[timid silo]

also a valid solution?

[polar fossil]

i think so yeah

,w graph {6/((x+3)(x-1)), (x+6)/((x+3)(x-1))}

[warm shale]

Results provided by WolframAlpha

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

Requested by kitten.in.a.teacup

[timid silo]

uh oh

I guess not

[polar fossil]

why not?

[timid silo]

actually no it should be

it's just that their graphs

are different

[polar fossil]

they're not completely identical but they have the same properties listed in the question

[timid silo]

alright then, thanks for double checking

.close

[obtuse pebble]

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[molten fractal]

The question is a+b+c+d=4. Prove abc+bcd+cda+dab<=4

But I worked out abc+bcd+cda+dab<4, can someone help pointing out what is wrong please, or tell me the correct way to solve this problem.

My approach is as followed:
4>=(a+d)(b+c)
16>=(a+d)(b+c)(a+b+c+d)
16>=(2×sqrt(ad)×(b+c)+2×sqrt(bc)×(a+d))(2(sqrt(ad)+sqrt(bc)))
4(abc+bcd+cda+dab+sqrt(abcd)×(a+b+c+d) 4-sqrt(abcd)×(a+b+c+d)>= abc+bcd+cda+dab
We can easily prove that 0<sqrt(abcd)<=1 so
4<abc+bcd+cda+dab
Which is wrong because 4 can equal abc+bcd+cda+dab.

[smoky vigil]

try using (a+b+c+d)^3

[molten fractal]

Can you explain more please?

[smoky vigil]

AM-GM may be helpful

(a+b+c+d)^{3}=(a+b)^3...

[warm shale]

🫎Mοοsey🫎

[molten fractal]

You wrote a^2+b^2+c^2+d^2>=4 before, isn't it correct?

[smoky vigil]

i was just overthinking, yeah its correct

assume all but one have magnitude less than one (only case where squaring gets smaller)

[molten fractal]

Yeah I mean I can prove it, you dont need to go further here, please continue with a^2+b^2+c^2+d^2>=4

[smoky vigil]

ah wait i was thinking wrong :I

[molten fractal]

So can you point out where my proof is wrong? Maybe we can work from there?

[obtuse pebble]

@molten fractal Has your question been resolved?

[molten fractal]

<@&286206848099549185>

[devout yarrow]

OK I might not be able to help you but first format it in latex

[molten fractal]

Do you know a program that can do it automatically?

[devout yarrow]

Nah

[molten fractal]

.close

[obtuse pebble]

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[molten fractal]

The question is a+b+c+d=4; a,b,c,d are positive real number. Prove abc+bcd+cda+dab<=4

But I worked out abc+bcd+cda+dab<4, can someone help pointing out what is wrong please, or tell me the correct way to solve this problem.

I worked out
4<abc+bcd+cda+dab
Which is wrong because 4 can equal abc+bcd+cda+dab.

[smoky vigil]

,rccw

[molten fractal]

Well there's a simple solution by changing the unknown
a=1+i+j+k
b=1-i
c=1-j
d=1-k
The inequality equals when i=j=k=0 but it's not the intended solution (the author explicitly told to use cauchy, cauchy might refers to the simple case, AM-GM, in my country)

[molten fractal]

thank you

[dim atlas]

non equal cauchy

[molten fractal]

Can you explain in details please?

[dim atlas]

i want to ask you about the question which used inequality cauchy

i can't understand cauchy

[molten fractal]

So do you mean my solution is too messy or is there a line you don't understand?

[dim atlas]

I can't understand 4 >= abc +bcd +cda+ dab

[molten fractal]

Well it's what I'm trying to prove

[dim atlas]

oh

sorry

[molten fractal]

Is there anythin else?

[dim atlas]

I'm afraid to bother you while you think

[molten fractal]

<@&286206848099549185>

[obtuse pebble]

@molten fractal Has your question been resolved?

[timid silo]

You should use AM - GM inequality

firstly , apply it to a , b , c , d

and you will get a relation

and then apply it to abc, bcd , cda, dab

[molten fractal]

Can you give more hints please, I can't think of anything.

@timid silo

[timid silo]

$$ \text{Given , a + b + c + d = 4} $$

[warm shale]

🎀 𝒫𝒶𝓃𝒾𝒸 🎀

[molten fractal]

Please continue sir

[timid silo]

Using AM - GM inequality
$$ \frac{a + b + c + d}{4} \geq \sqrt[4]{abcd} $$

[warm shale]

🎀 𝒫𝒶𝓃𝒾𝒸 🎀

[timid silo]

But we know that a + b + c + d = 4. So,
$$ 1 \geq \sqrt[4]{abcd} $$
$$ 1 \geq abcd $$

[warm shale]

🎀 𝒫𝒶𝓃𝒾𝒸 🎀

[molten fractal]

Oh I get it

Thank you so much

[timid silo]

Now , we use the AM GM inequality on abc , bcd, cda , dab
$$ \frac{abc + bcd + cda + dab}{4} \geq \sqrt[4]{abc \cdot bcd \cdot cda \cdot dab} $$

[warm shale]

🎀 𝒫𝒶𝓃𝒾𝒸 🎀

[timid silo]

Welcome

[molten fractal]

Have a good day, conscientious helper.

.close

[obtuse pebble]

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[obtuse pebble]

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[unreal cloud]

How many two-digit numbers can Ahmet write on the blackboard such that the difference of any two of them is not a number consisting of two equal digits?

[unreal cloud]

Ahmet is a name

[royal basin]

so in other words: how many 2-digit numbers can we write on the board so that no two of them differ by a multiple of 11

[unreal cloud]

yes

the answer is 11 but i do not know how to get it

[royal basin]

do you know modular arithmetic?

if you do then my rephrasing of your question trivializes it

[unreal cloud]

i know a little bit about it

[royal basin]

you want your numbers to not be congruent to each other modulo 11.

or in other words you want them all to have different remainders mod 11.

how many remainders are possible when dividing by 11?

[unreal cloud]

10

[obsidian isle]

Are you sure?

Try again

[royal basin]

list them

list the 10 remainders

[unreal cloud]

1,2,3,4,5,6,7,8,9,10

[royal basin]

you forgot one

[unreal cloud]

0 ?

[royal basin]

yes obviously

there are 11 possible remainders

[unreal cloud]

.close

[obtuse pebble]

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[fierce vale]

$\frac{\partial f}{\partial x} \cdot \frac{\dd x}{\dd t}$

[warm shale]

Adam Chebil

[fierce vale]

can i simplify with partial x and dx ??

[final thunder]

What do you think?

[fierce vale]

we can ? 👀

i want to derive f(x,y) with respect to t

[royal basin]

f_x dx/dt + f_y dy/dt can't really be simplified much further can it?

also differentiate* not derive

[final thunder]

Simplify? I mean you would just find dx/dt and times it by del f/ del d

[fierce vale]

they didn't use del operator here ?

[final thunder]

I meant partial

[fierce vale]

i don't understand why we multiply by dx/dt ?

also shouldn't be written as partial f / partial t ??

[brazen gorge]

x changes as t changes

same with y

so we account for that (chain rule essentially)

[brazen gorge]

no since f "depends" on x and y, and both x and y "depend" on t

also i see some other person was covering the exact same thing lol, maybe their notes will be helpful to you #help-7｜zen1thxyz message

[fierce vale]

i don't see how we applied chain rule here : (

[brazen gorge]

let's make it a bit simpler and consider f(x)

1 variable

and we are looking for $\frac{df}{dt}$

[warm shale]

artemetra

[brazen gorge]

where $x=x(t)$ - some function that outputs $x$ given $t$

[warm shale]

artemetra

[brazen gorge]

so

*$\frac{df}{dt} = \frac{df}{dx}$

[warm shale]

artemetra

[brazen gorge]

but!

x itself is a function of t

so $f(x)=f(x(t))$

[warm shale]

artemetra

[brazen gorge]

we have composition of function hence chain rule

therefore, we also need to multiply by the derivative of x w.r.t t

hence, $\frac{df}{dt} = \frac{df}{dx}\frac{dx}{dt}$

[warm shale]

artemetra

[brazen gorge]

you might also want to take a look at related rates, a very similar topic

@fierce vale does that make some sense?

[obtuse pebble]

@fierce vale Has your question been resolved?

[fierce vale]

why df/dt = df/dx here ?

[brazen gorge]

it's not

i put a * there in front

it's more of like the process yk

i think that image was extra, sorry

[brazen gorge]

start with this

[fierce vale]

here we differentiate the inside function which x(t) with respect to t

but why did we differentiate the outside function f with respect to x and not t ??

[brazen gorge]

that's what we are trying to find

df/dt is what we are looking for

[fierce vale]

shouldn't this be equal to : $x'(t)\cdot f'(x(t))$

[warm shale]

Adam Chebil

[brazen gorge]

it is!

that's what i wrote

$x'(t) = \frac{dx}{dt}$

[warm shale]

artemetra

[brazen gorge]

$f'(x(t)) = \frac{df}{dx}$

[warm shale]

artemetra

[fierce vale]

yeah this is what i don't understand

f'(x(t)) = df/ dt ? no?

[brazen gorge]

no

actually i am not sure how to express it with lagrange's notation

[fierce vale]

how do i know which variable i put ?

[brazen gorge]

but $\frac{df}{dx} = f'(x(t))$ because it's the derivative of f as x changes, not t

[warm shale]

artemetra

[fierce vale]

wait f here is a single variable function ??

[brazen gorge]

yes

[fierce vale]

oh ok

[brazen gorge]

^

[fierce vale]

my problem is with multivariable fcts

[brazen gorge]

multivariable is the same just done twice essentially

[fierce vale]

do i write df/dx or df/dy ??

[brazen gorge]

$f(x,y) = f(x(t), y(t))$

[warm shale]

artemetra

[brazen gorge]

wdym?

[fierce vale]

$\frac{\dd f(x,y)}{\dd t}=x'(t)\frac{\dd f(x,y)}{\dd x}+y'(t)\frac{\dd f(x,y)}{\dd y}$ ?

how do i do this ?

[brazen gorge]

df/dt

[warm shale]

Adam Chebil

[fierce vale]

i think i understand it now
can u explain when do should i write partial f or df ??

.close

[obtuse pebble]

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[timid silo]

am i missing something? question asked about the stationary points, why are they doing the 2nd derivative? as far as im concerned f'(x) = 0 are the stationary points

[timid silo]

part b btw

i got x = 3 as a stationary point but i dont get why they did the 2nd derivative and found the minimum, someone explain if im missing something pelase

[median yacht]

It's another method to classify stationary points

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

how

also pls ping when u reply cuz i didnt see ur first message

anyways whats another method of classifying them

the 2nd derivative?

id understand if it was another method but its saying that the 2nd derivative has points yk

meaning i have to do both first derivative and second derivative to get the full grade

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

no

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

Please don't occupy multiple help channels.

[mint arrow]

I would like to clarify my logic.

Zero Velocity = The object is stationary.

In this position vs. time graph, "B" is the stationary object, while A moves away from point of origin and C returns to origin.

Therefore, the letter on the chat that corresponds to zero velocity would be B.

If I am not on the right track, please redirect..

[timid silo]

Yeah with 0 velocity, the objects position doesnt change as it doesnt move

Objects stationary

B is at 0 velocity

[mint arrow]

Okay,t hanks.

that would mean that

C is moving toward the starting point right

[obtuse pebble]

@mint arrow Has your question been resolved?

[timid silo]

yes C has a negative velocity

as its a straight line, velocity is constant

so C is where the velocity is constant towards the starting point

[mint arrow]

This indicates position

as opposed to time

or no

.close

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[noble scroll]

hello

[noble scroll]

I made a mistake by clicking on the left-side in spyder, i can't go back to my codes lines

I'm stuck, anyone can help me please

[lethal breach]

Hey umm, this is a math server. Please keep your questions limited to that subject.

There are many other servers that'll help you with your coding and ide stuffs.

P.S. I would've still helped if I would've known this, but I don't.

Try going to the server that would actually help you out

[obtuse pebble]

@noble scroll Has your question been resolved?

[obtuse pebble]

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[wanton kraken]

my lecturer said that this is true

[wanton kraken]

but I think I lost the explanation of such property

was there such property?

I'm new to integrals

[versed stratus]

the derivative of ln(x) is 1/x, here d(ln(x)) simply acts as the element/variable wrt which you're integrating

[wanton kraken]

wrt?

[versed stratus]

with respect to

[buoyant zealot]

have you heard of integrating by change of variable?

[wanton kraken]

uhh I think no

let me think this through

so

integral is basically f(x) is derivative of what

right?

[integral sign]1/x(dx)=ln(x)

ohhh

I see the tendency

[wanton kraken]

Ohhh so it's because

or these have to be the same

yes?

[wanton kraken]

does that mean that these have to be the same?

(1/x)(dx)

[versed stratus]

no, I can integrate for instance dx/tan(x)

[buoyant zealot]

this is basically just integration by substitution (or change of variable)
you basically define u=lnx
substitute it in the integral
and that eliminates 1/x

[versed stratus]

but yes the variable has to be the same

[wanton kraken]

ohh

[wanton kraken]

ohh

wait

so

[buoyant zealot]

if you understand integ by substitution itll be quite obvious what they did here

[wanton kraken]

let me write this using paint

excuse my bad writing

[buoyant zealot]

but instead of actually replacing lnx by u
they just left it as it is

[wanton kraken]

if we have it like that, we can use substitution

such so t=ln(x)

[buoyant zealot]

i mean you could replace ln x by t here mentally. but at this stage it isnt needed

[wanton kraken]

but then why did this happen?

[buoyant zealot]

because the boundaries are originally in x1 and x2

so when you change the variable you have to adjust the boundaries accordingly

[wanton kraken]

🤔

[buoyant zealot]

basically here you deifned t=lnx. and changed everything inside the integral to the variable t

but the boundaries also have to be changed since they were in x

[wanton kraken]

oh, right

[buoyant zealot]

t1=lnx1
and

t2=lnx2

[wanton kraken]

right

[buoyant zealot]

at this point tho it isnt really necessary to use substitution

you use substitution when whatever you are integrating is in the form of a function multiplied by its derivative

you dont need to try to figure out how integ by substitution works from scratch.
there is a lot of content on it online.
its the type of idea that really clicks when you solve a couple examples

[wanton kraken]

I get it yeah

tysm @buoyant zealot

.close

[obtuse pebble]

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[willow dust]

Someone help me solve this regularly without using the L’hopital’s rule

[spice rune]

1 - 2sin^2(x) = 2cos^2(x) - 1 = (2cos^2(x) - sin(2x)) + (sin(2x) - 1) = (2cos^2(x) - sin(2x)) - (cos(x) - sin(x))^2
can you continue from here?

[willow dust]

yes

I got 2 answers from doing it

first one is 1 and the second one is 0

Idk which one is correct

[spice rune]

how did you get them?

[willow dust]

And the other one is photomath

[spice rune]

is that in any way related to what i wrote above?

[willow dust]

Yes

[spice rune]

isn't it just the first step of the transformations i provided?

[willow dust]

Yea it is

is it correct

Or do i have to transform in all the way

[spice rune]

yes

[willow dust]

so the answer is 1?

0 i mean

[spice rune]

how did you derive that?

[willow dust]

this

[spice rune]

the denominator is incorrect

there you only multiply the first term by 2

but it should be both

[willow dust]

So its 2-2/2-1

[spice rune]

no

[willow dust]

can you solve it for me i will try to understand

[spice rune]

it's a 0/0 indeterminate form

[spice rune]

the tricky part is transforming the numerator
that I've done above: #help-10 message

then just split it into two fractions and reduce each separately

[willow dust]

uhh

can u tell me the answer i will try to find solution to the answer

i dont really understand english math so its pretty difficult for me

[spice rune]

there's no such thing as "english math"

math is math

[willow dust]

yea but im not that good at english

idk whats the thing u try to point out to me

[spice rune]

did you understand these transformations?

each step involves some formulas

[willow dust]

i understand the 2cos^2x - 1

but the others no

[spice rune]

the second step is just adding and immediately subtracting sin(2x)

[willow dust]

I see

[spice rune]

we do this so that a part of the numerator exactly matches the denomirator

[willow dust]

What about the next step

[spice rune]

so we'll be able to reduce this part to 1

the last step is to transform sin(2x) - 1 into -(cos(x) - sin(x))^2

[willow dust]

Is it trigonometry formula

[spice rune]

you can derive it this way:
1 - sin(2x) = (sin^2(x) + cos^2(x)) - 2 sin(x) cos(x) = cos^2(x) - 2 cos(x) sin(x) + sin^2(x) = (cos(x) - sin(x))^2

we use the pythagorean identity and the double-angle formula for sine

[willow dust]

i have no idea what that thing is

idk what it called in my language

[spice rune]

which one?

the pythagorean identity is sin^2(x) + cos^2(x) = 1

[willow dust]

phytagorean idenity

And double angle formula

do u memories all the formula or what

[spice rune]

there are formulas for sin(2x), cos(2x) etc

e.g. here we use that sin(2x) = 2sin(x)cos(x)

[willow dust]

in my country try we just memorize all of them

[spice rune]

you seem to have used it already judging by your denominator transformations

[willow dust]

but there are too many so i only memorize a few of them

do i dont really understand the formula

[spice rune]

jut the frequently used ones (like these)

[willow dust]

so if u transform it further its pretty hard to understand

[spice rune]

well, from this point everything is reduced pretty fast

[spice rune]

so did you understand this part?

[willow dust]

What are the frequently used formula for cos/sin 2x and cos/sin x^2

[willow dust]

I understand it now

[spice rune]

there are actually just 3 main formulas that are used very often:

1. sin^2(x) + cos(^2x) = 1
2. sin(2x) = 2 sin(x) cos(x)
3. cos(2x) = sin^2(x) - cos^2(x)
   by combining (1) and (3) you also get cos(2x) = 1 - 2sin^2(x) = 2cos^2(x) - 1

there also half-angle formulas (for sin(x/2), cos(x/2)) and formulas for things like sin(x + y), cos(x - y) etc

and many others but you don't see them too often

anyway, if you understand where they come from, you can always derive any of them even if you forget it

[lilac sun]

I keep forgetting the sin(nx) formula so I just derive it using complex numbers and de moivre's theorem

[spice rune]

lmao

don't mean that one lol

[willow dust]

Alright tysm

[spice rune]

do you understand what to do next?

[willow dust]

I did it and got

Its still 0/ 1/2

If i substitute pi/4 to x

[spice rune]

good, i assume you mean 1 - lim ...

also it should be sin(2x) in the denominator

[willow dust]

What is 1 - lim ...

[spice rune]

not sin(x)

[spice rune]

1 minus what's on your picture

[willow dust]

why 1

ohh

i see

[spice rune]

because of the other part of the numerator

[willow dust]

Yess

i forgot

So its 1-0

The final answer is 1

[spice rune]

not so fast

[spice rune]

you have an error here

[willow dust]

yes

i wrote it correctly in my book

omg

its my homework

[spice rune]

well, 2cos^2(x) - sin(2x) for x = pi/4 is still 0

so in this form it's still 0/0

there's one additional step missing

maybe you've already done it and just not mentioning it?

[willow dust]

yea but there’s 1 minus the 0

[spice rune]

1 minus 0/0

0/0 is not 0

[willow dust]

wait

[spice rune]

0/0 is not even a number

[willow dust]

thats right

why is it 0/0

I think its 0/ 1/2

2cos^x

[spice rune]

because both numerator and denominator are 0 for x = pi/4

[willow dust]

is 2x1/2

[spice rune]

2cos^2(x) is 2 * 1/2

yes

but sin(2x) is also 1

[willow dust]

wait no

omg

i got it now

its 0/1-1

0/0

What do i do next

[spice rune]

can you see how you can reduce the fraction?

and why such reduction would be legal (i.e. why you don't divide by 0)

[willow dust]

i dont really understand what u’re saying

can u write it down

thats what i meant “ english math “

[spice rune]

do you what reduction is?

[willow dust]

yes i know what reduction is

[spice rune]

can you figure out how that can help you simplify the fraction?

[willow dust]

idk

its 2am my half of my braincell is gone

also im new to calculus

[spice rune]

it's not calculus, just basic algebra

do you see any common factors in numerator and denominator?

[willow dust]

In this?

[spice rune]

yes, in the fraction

(cos(x) - sin(x))^2 / (2 cos^2(x) - sin(2x))

[willow dust]

‘Cosx-sinx’ ?

[spice rune]

yes

do you think it would be legal to divide both parts by it?

[willow dust]

idk actually

[spice rune]

well, it would only be illegal if it could be equal to 0

the thing is though, it's only 0 when sin(x) = cos(x)

and as x approaches pi/4 it never actually gets to exactly pi/4

so sin(x) and cos(x) are always different

[willow dust]

by a small bit?

[spice rune]

yes

[willow dust]

so

what do i do next

[spice rune]

what have you got so far?

did you do the reduction step?

[willow dust]

lemme write it all down together

This is what i got so far

[spice rune]

what about reducing by that?

[willow dust]

idk how

do i break it down first

[spice rune]

wdym

[willow dust]

using a^2 - b^2

[spice rune]

i don't see a difference of squares anywhere here

why did you sugget cos x - sin x?

when i asked about common factors

[willow dust]

cuz i see it both has it

[spice rune]

yes, here you go

divide both numerator and denomator by that expression

[willow dust]

numberator is the upper part and denomator is the bottom part right

[spice rune]

yes

numerator, not numberator

and denominator, not denomator

but otherwise you got it

[willow dust]

Is this correct

[spice rune]

no

you just removed 2

[willow dust]

so its equals to 1-1

[spice rune]

you have to divide both numerator and denominator by that

not just the numerator

[willow dust]

i see

So (cosx-sinx)^2 is equals to 2cos^2(x) - sin2x

[spice rune]

no

they only have a common factor of (cos x - sin x)

like 10 and 4 have a common factor of 2 but still not equal

[willow dust]

the thing is idk how to devide the denominator by that expression

[spice rune]

well you said it yourself that it is a common factor

to deduce that you'd need to factor the denominator (at least in your head) in the first place

[willow dust]

Is this correct

[spice rune]

still no

can you write the denominator for me such that it contains (cos x - sin x) as a factor?

[willow dust]

2cosx ( cosx-sinx )

[spice rune]

oh wait you are right, sorry

my bad

[remote skiff]

I know you've spent a long time on this problem going down this path for the last 2 hours (which works), but the clever trick for this kind of problem is to notice that a limit as x tends to pi/4 doesn't allow you to use a small angle approximation, nor does a substitution y = x-pi/4 make it easy to do so. So you use the substitution y = 2x-pi/2 which has y tend to zero as x tends to pi/4 and that immediately lets you simplify everything down into polynomials and calculate from there.

[spice rune]

other than that you forgot about the 1 at the beginning

so now the problem is almost solved

you just substitute

there's no 0/0 anymore

[willow dust]

i didnt forget it

i just keep it aside

[willow dust]

I’ll try this method after i solved it this way

So this is the final aswer

Answer

[spice rune]

yes

[willow dust]

yesssssss

its 2:35AM in my country

also Tysm for your time tutoring me

[spice rune]

indonesia i guess?

[willow dust]

Cambodia

[spice rune]

nice

you're welcome

[willow dust]

Imma try this method

Is this how you do it

i use “t” instead of “y”

[obtuse pebble]

@willow dust Has your question been resolved?

[willow dust]

.close

[obtuse pebble]

Channel closed

Closed by @willow dust

Use .reopen if this was a mistake.

[obtuse pebble]

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[silent furnace]

how’s this supposed to be done

[prisma hatch]

!status

[obtuse pebble]

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

[silent furnace]

im at the lcm part

[prisma hatch]

Can you send your work?

[silent furnace]

(1+sin)^2+(cos^2y)/cosy+siny

what to do after this

[prisma hatch]

Your denominator doesn't seem to be right

[timid silo]

expand (1+sin)^2

[silent furnace]

1+2sin+sin2

[timid silo]

^^

[silent furnace]

cosy+sinycosy

[timid silo]

Yep

wait

yes

[silent furnace]

ok how do i get 2sec

[prisma hatch]

Write cos2 in terms of sin2

[timid silo]

1+2sin+sin^2+cos^2/cos+sincos

u got that right?

[silent furnace]

2+2sin is?

[timid silo]

2(1+sin)/cos(1+sin)

what can u do now

[silent furnace]

2/cos

um

YEAH

SEC

[timid silo]

2* 1/cos is what?

[silent furnace]

2SEC

[timid silo]

yes

[silent furnace]

THANK U SO MUCH

[timid silo]

np

[silent furnace]

.close

[obtuse pebble]

Channel closed

Closed by @silent furnace

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• 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.

[distant wasp]

I've been learning a lot about the Discrete Fourier Transform lately, and though I'd try to recreate something in Desmos. What I came up with ends up flagging the correct frequencies, but also flags a bunch of incorrect frequencies as well. I'd be really grateful if someone could take a look at my Desmos thing, and I'd appreciate some pointers on how to fix it.
https://www.desmos.com/calculator/iiyzjtws9r

[distant wasp]

Also, sorry this is a duplicate, but there was a misunderstanding with the original question that caused it to be buried

[obtuse pebble]

@distant wasp Has your question been resolved?

[obtuse pebble]

@distant wasp Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

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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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[lament atlas]

i did the integrals over dxdydz