help-39

[merry python]

after that, you have two factors of a number that's supposed to be prime, so contradiction!

[tidal otter]

Yeah that's very clever proof

[merry python]

I have to go now, good luck! (open a new channel if you have more questions)

[tidal otter]

Thanks

I got it

.close

[pearl pond]

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[frozen solar]

Am I doing this wrong?

[frozen solar]

It’s supposed to be like this.

[chrome niche]

ale co masz zrobić?

[frozen solar]

I have to graphically present the points

[frozen solar]

.

[chrome niche]

aren't you polish?

[pearl pond]

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[pearl pond]

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[vast lichen]

Help

[vestal sable]

at what value of x does -2x + 4 = 0?

[pearl pond]

@vast lichen Has your question been resolved?

[vast lichen]

Idk

😭

[vestal sable]

Could you rearrange -2x + 4 = 0 until you achieve x = something?
Try adding 2x to each side and then dividing through by 2.
The answer is below if you get stuck but give it a go first :))
||4 = 2x -> x = 2||

[vast lichen]

So x=2

[vestal sable]

yep, that's your c

[vast lichen]

Aah

[pearl pond]

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[unkempt yacht]

given this diagram (ignore blue writings), find the correct statement

find the correct statement out of the following

these are my calculations so far

the things i box are vital information

[glacial scroll]

If the graph is at scale, then d = 0 and (x-3)^2 and x factorize y ??

[unkempt yacht]

oh youre right

hmm, i think the first is correct

doesnt check out with the anskey, but its probably wrong anyway

thanks

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[glacial scroll]

what's the key

?

[unkempt yacht]

second

[pearl pond]

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[lilac trail]

hey, im completely stuck on a math problem. The problem is to determine if the function has an inverse. f(x) = x* tan((pi*x)/2) with domain (-1,1). with my knowledge, a function has an inverse if it is one-to-one in the domain area. i know you can derivate and find out if the derivated function > 0 or < 0 in the domain to find out if it is increasing or decreasing and then also if it has an inverse.

well, the derivative doesnt look nice and im just thinking if there is any other way i can see that this function doesnt have an inverse? without looking at the graph itself

[pseudo oxide]

i mean
swap vars and solve for y?

if u can, i guess it has an inverse

pretty cool once u graph it

[lilac trail]

yea, it isnt x/2 but pi*x is divided by 2

[pseudo oxide]

oh, nevermind

yea i saw that

$x\tan\left(0.5\pi x\right) \rightarrow x = y\tan\left(0.5\pi y\right)$

[lilac trail]

yea thats the one

[jolly parrot]

ren

[pseudo oxide]

the inverse is the one on the right.

hm.

[lilac trail]

idk how you would go from swapping vars and solve for y

[pseudo oxide]

that's how u calculate an inverse function

[lilac trail]

well from my ujnderstanding the function doesnt have an inverse in the domain (-1,1)

[pseudo oxide]

from intuition i agree

idt it's simplifiable

,w inverse function of x sin x

[jolly parrot]

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[pseudo oxide]

ok yea

that just confirms my hypothesis

hmm

[pseudo oxide]

wait what if we convert to polar

does that help? i got no idea

[lilac trail]

yea, i just dont know how i can actually calculate or use any theories to support that it doesnt have an inverse in that domain

[pseudo oxide]

cot (0.5piy) = tan theta
so on

then arctan cot 0.5pi * y = theta

etc

,w arctan (cot (x))

omfg

why can't WA just simplify

[jolly parrot]

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[pseudo oxide]

i know what it is

just might as well check

,w differentiate 0.5x(x + 2 arctan (cot x))

[jolly parrot]

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[pseudo oxide]

yea, idt it's very reducible, is it?

[lilac trail]

these looks like pretty ugly numbers so i cant imagine that my professor wants me to do it like this

[pseudo oxide]

ok hang on

googled to reconfirm MY answer

pi/2 - t

so basically, arctan (cot 0.5piy) = theta
arctan (cot t) = theta
pi/2 - t = theta
pi/2 - piy/2 = theta
(pi/2)(1 - y) = theta

oops, dc italics hang on

wait--no

not in the interval (-1, 1)

sighs

one sec.

[lilac trail]

it doesnt have an inverse in (-1,1)

[pseudo oxide]

ok

well that works too

isntead of solving the problem

*instead

[lilac trail]

how can i show it?

[pseudo oxide]

dk

[lilac trail]

yea, been stuck for like 2 days lol

i mean it looks impossible to calculate the derivate and solving for f'(x) = 0 using only simple calculator, so i imagine there must be some theorem or something that can be used to explain why it dont have an inverse in that domain

[pseudo oxide]

,w inverse function theorems

[jolly parrot]

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[pseudo oxide]

ok yea no

the best i could find

is smth in advanced calc

which even idk

[lilac trail]

yea idk

[pseudo oxide]

well
sorry

couldn't help much, could i

[lilac trail]

np, this is university type question though am i in the wrong channel

[pseudo oxide]

wait, no
I GOT SMTH

i think
wait--no

nvm

[lilac trail]

what

[pseudo oxide]

im an idiot

[pseudo oxide]

nothing

hm

[lilac trail]

rip

[pseudo oxide]

ok wait i have a small idea

$\cot 0.5\pi y = \frac yx \rightarrow \frac {\cos (0.5\pi y)}{\sin (0.5\pi y)} = \frac yx$

[jolly parrot]

ren

[pseudo oxide]

we need to isolate y, btw

wait--yea got it

[lilac trail]

impossible

[pseudo oxide]

there're gonna be 2-4 or smth values

with cos 0.5pi*y = y

that's an assumption we're gonna have to make

so that means y is constant

which means that x is constant

which means inverse doesn't exist

HOWEVER this assumes that cos 0.5pi*y = y

[lilac trail]

idk that logic, x can be whatever number between (-1,1) and limx->1 = infinite

[pseudo oxide]

ok ykw

idk

IF you figure out that cos (0.5pi*y) has to equal y for the inverse

then there is no antiderivative, basically

(by my logic)

[lilac trail]

hmm idk

not sure if i can use that

[pseudo oxide]

idk man

[lilac trail]

damn, thanks for your help though

[pearl pond]

@lilac trail Has your question been resolved?

[lilac trail]

<@&286206848099549185>

[pseudo oxide]

it's legit PINNED

[lilac trail]

its pinned yea

correct yea

its (pi*x)/2 not pi/2x

hmmm

but looking at the function x*tan(blah blah) it doesnt have an inverse in that domain or am i tripping?

[valid plover]

it doesnt have an inverse

or hold on

[lilac trail]

yea but at point 0 it goes from decreasing to increasing

[valid plover]

to disprove that it has an inverse just plug in -1 and 1

and f(-1) = f(1)

or going the derivative way

the function cant "turn around"

[lilac trail]

yea, but the function isnt defined at -1 or 1

[valid plover]

then look at the left, right hand limit

or literally plug in any point from (-1,1)

you get that its equal

1 sec

f(-x) = -x * tan(pi/2 * -x) = -1 * (-1) x * tan(pi/2 * x) = f(x)

its symmetrical around x = 0

[lilac trail]

hmm alright thanks, i guess i can use that

[valid plover]

or

2nd derivative has to be 0

but why would you even bother if you can disprove it with a counterexample

[lilac trail]

yea, but there are other functions to prove also, and just for example checking each number in the domain doesnt really work, so it feels like cheating kinda just doing f(-1) = f(1)

[valid plover]

nah just show that its symmetrical around x = 0

meaning 2 inputs share the same output

[lilac trail]

ye alr, thanks

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[surreal smelt]

Hello, how to show that this is a distribution for all φ in D(R) ?

[pearl pond]

@surreal smelt Has your question been resolved?

[surreal smelt]

<@&286206848099549185>

<@&286206848099549185> 😭

[marble sigil]

I think I'd need context for what phi is, isn't a distribution just integral = 1 and always nonnegative

[surreal smelt]

phi is a test function D(R)

I don't think distrubution is just integral = 1 !

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[silent bramble]

how do they know Xn+1/xn - c goes to 0

in the proof

[prime bramble]

pretty sure it's because x_(n+1)/x_n goes to c

[pearl pond]

@silent bramble Has your question been resolved?

[silent bramble]

oh okok

how do you know there exists such M_0 > n_0

is it because there exists such n_0 for

and by the triangle inequality you can say there exists such m_0

i think so

ok

[silent bramble]

Im guessing here, they imply for all y in B

but if someone can confirm thatd be appreciated

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[frosty badge]

if we divide a by a number by 11 we will get 8 remaining if we divide b by 11 we get 3 remaining what will we get if we divide a+b by 11

[frosty badge]

i dont get it

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[rigid trench]

a) 4800
b) 200

Idk how to solve it

[hoary birch]

what S/. ?

is it like $?

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[stoic siren]

convert to component form

,calc 12sin(-125)

[jolly parrot]

Result:

7.3924855102639

[stoic siren]

,calc 12cos(-125)

[jolly parrot]

Result:

9.4525741457308

[stoic siren]

what

my calculator gives me smthn else

[gentle hatch]

are you working in radian or degres on your calculator?

[stoic siren]

degrees

what did i do wrong

[light helm]

you shouldn't be evaluating sin or cos of -125° for exponential form

[stoic siren]

wym

so i should be keeping it in exact value?

[light helm]

you shouldn't really be touching sine or cos at all

[stoic siren]

how do i convert it to component form then

[light helm]

do you know what exponential form of complex number is?

or looks like

[stoic siren]

no

[light helm]

i'd recommend that you start by looking that up

[stoic siren]

either way

the question asks me to do this

through what we know

so there has to be a way

we havent gone over complex numbers

for (5e)

why is my answer wrong

[light helm]

oh sry,, misread the question

misread component as exponent

what answer were you getting

[stoic siren]

oh 😭

[stoic siren]

i got -6.88i and -9.83J

[stoic siren]

this gave me different answers

but

heres the thing

neither of these are correct

i do have an answer sheet

if you'd like to see

[light helm]

what does the answer sheet say

[stoic siren]

my rounding up was definitely not wrong when doing 12cos(-125)

and the J value is just so off

i dont know what i even did wrong

[light helm]

were there any issues with a,b,c,d?

[stoic siren]

nope

all correct

so it cant be my calculator mode

[light helm]

seems like an error in the book then

[stoic siren]

even if it is, i changed modes to check and i didn't get anything like the answer

[stoic siren]

alright sweet

thanks champ

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[stoic siren]

how do I solve b

[pearl pond]

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[stoic siren]

[pearl pond]

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[loud lintel]

Can someone please help me

[loud lintel]

Q15

I don’t at all understand the question

[runic zephyr]

do you know derivative and its concept ?

[loud lintel]

No, I don’t think so

[runic zephyr]

it is that

[loud lintel]

Ok

Should I just search up the derivative and its concept on YouTube and watch a tutorial?

[runic zephyr]

idk because it needs some basics.

[loud lintel]

Like?

[runic zephyr]

what is function

slope of a line

and other stuff

[loud lintel]

Ok

[runic zephyr]

like limit

[loud lintel]

I think I’ve covered everything except for functions

[runic zephyr]

and continuity

and alot more

you can try tho

[loud lintel]

I’ll try

Thanks a lot

[runic zephyr]

sorry i cant help further then this

np 👍

[loud lintel]

Do you know what the best way to revise for GCSE is?

[runic zephyr]

i dont take these exams because our education system is different so no sorry.

[loud lintel]

Ok

Thanks anyways

[distant goblet]

the best way is to just keep doing practice exams

it might be boring but thats the best way to understand what the questions want from you

[loud lintel]

I’ll keep doing that

The thing is that because we’re doing advanced maths

We have a normal GCSE (edexcel)

And an advanced maths qualification (AQA)

So I have 2 different exam boards to study for

[distant goblet]

yeah i did the same, realistically the edexel one is the one you should properly focus on and hte AQA one is a nice bonus, but it is worth revising for both. but you still have loads of time before gcses start anyway

[loud lintel]

We’ve got our mocks in May

For both

And if I get under 90% on any of them, my dad isn’t gonna be happy

And if I do bad on the AQA one especially, I’ll get moved down to normal maths

[distant goblet]

i find just doing past papers the best form of revision, you still have a long time to revise for it to get your 90%, there is a lot of resources on youtube as well that will walk you through exam prep and how to answer questions the best way

[loud lintel]

Ok

Ill to that

Thanks a lot!

[loud lintel]

also, do you have a preferred video on how to answer exam questions the best way?

[distant goblet]

i dont know any good gcse videos right now, it has been about 5 years since i revised for my maths gcse. there was a channel called the GCSE tutor that i found quite good but i dont know if they are still doing good videos now

[loud lintel]

Ok

Ill have a look

Thanks a lot for your help!

I really appreciate it

🙏

[distant goblet]

no worries, good luck for your mocks

[loud lintel]

ty

[trail spindle]

hey, just to hop into this rq, im revising for FSMQ Level 3 Additional maths rn, along with the normal maths gcse. I've found one of the more effective methods is past papers. focus on questions about topics that you aren't very strong on. for a 90%, make sure that you can do most questions in a reasonable time frame, preferably quicker than the time allocated, so that you have more time for those questions you'll inevitably find harder

just as an example, if you have a 90 minute long paper, and its worth 90 marks, aim to get around 60-70 of those marks within 45-50 minutes, so that you'll have a little more time on the harder ones

[loud lintel]

Ok

Thanks

And, do you know where to find good past papers?

[trail spindle]

well for the edexcel one, theres a tool on the pearson edexcel website to find past papers

https://qualifications.pearson.com/en/support/support-topics/exams/past-papers.html?Qualification-Family=GCSE&Qualification-Subject=Mathematics (9-1) from 2015&Status=Pearson-UK:Status%2FLive&Specification-Code=Pearson-UK:Specification-Code%2Fgcse15-maths&Exam-Series=November-2019

[loud lintel]

Thanks a lot

[trail spindle]

and is the aqa one advanced maths or further maths

[loud lintel]

I wasn’t able to find many for the AQA one tho

Further maths iirc

Let me check again

It’s AQA level 2 further mathematics

[trail spindle]

https://www.aqa.org.uk/subjects/mathematics/aqa-certificate/further-mathematics-8365/assessment-resources?f.Resource+type|6=Question+papers
i believe this then

[loud lintel]

Oh

Great!

Don’t know why I wasn’t able to find them, sorry for the trouble

[trail spindle]

its alright lol

[loud lintel]

Thanks a lot!

[trail spindle]

dont worry about it :))

[loud lintel]

🙏

[trail spindle]

good luck for the mocks

[loud lintel]

Thanks

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[midnight haven]

\textbf{Question:} Two spheres, each of radius $R$ and carrying uniform volume charge densities, $+\rho$ and $-\rho$ respectively, are placed so that they partially overlap. Call the vector from the positive centre to the negative centre $\vj d$. Show that the field in the region of overlap is constant. 

\vs{5 mm}
\textbf{My attempt}: I basically reduced the problem from a physics problem to a vector problem; the electric field for each ball independently is: \[
\vj E_1 =\4{\rho R}{3\eps_0} \vc*r \qq \vj E_2 =-\4{\rho R}{3\eps_0} \vc*r 
\]
my question now becomes how do I exactly deal with the intersection region vectorially?

[pearl pond]

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[midnight haven]

.close

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[shell brook]

Why do we call it closed domain if they don't contain their boundary points

[shell brook]

What's the reasoning behind it

[tropic saddle]

context?

I have never heard closed domain used in that way

[shell brook]

Uhh in trigonometry

Like for example we are given sin domain which is [2π, -2π]

That's open domain

Why is it called open

What's the logic behind naming them

[tropic saddle]

who calls that open

[pearl pond]

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[drowsy halo]

.close

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[meager imp]

I hope you can explain it to me in detail, because my English is not good

[pearl pond]

@meager imp Has your question been resolved?

[meager imp]

<@&286206848099549185>

[pearl pond]

@meager imp Has your question been resolved?

[pearl pond]

@meager imp Has your question been resolved?

[pearl pond]

@meager imp Has your question been resolved?

[pearl pond]

@meager imp Has your question been resolved?

[meager imp]

<@&286206848099549185>

[white stag]

!status

[pearl pond]

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

[white stag]

if you've tried to solve it then

!show

[pearl pond]

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

[meager imp]

here

but I find it not feasible, I want another more concise way

[pearl pond]

@meager imp Has your question been resolved?

[slate ravine]

That is the scariest thing I have ever seen

[cursive wolf]

Throwing my hat in the ring: Change of coordinates $x \mapsto u := x/R, y \mapsot v := y/R$, then we integrate over the domain $u^2 + v^2 = 1$ and get a new integral $$ R^2 \cdot \int \int_{u^2 + v^2 \leq 1} \frac{1 + 2 v^2/R^2}{1 + (...)/R^4} - \frac{1 + v^2/R^2}{2 + (...)/R^4} du dv$. Note that the factor $R^2$ in the beginning comes from the change of variables

[jolly parrot]

Lartomato
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[cursive wolf]

The integrand itself (without the R^2 in front) converges to 1/2 (first fraction to 1, second fraction to 1/2) as R goes to infinity, and I am pretty certain that it does so uniformly. I think for fractions of polynomial functions, this isn't terribly hard to try and check

Because of the uniform convergence, the integral itself converges to $\int \int_{u^2 + v^2 \leq 1} \frac{1}{2} du dv = \frac{\pi}{2}$, we're just integrating over a circle. But then, clearly, $R^2 \cdot \int\int(...)$ must diverge

[jolly parrot]

Lartomato

[cursive wolf]

(uniform convergence implies we can exchange integral and limit)

@meager imp what say you

FUCK I got the change of variables wrong

that's what it should be

$ R^2 \cdot \int \int_{u^2 + v^2 \leq 1} \frac{1 + 2 R^2 v^2 }{1 + R^4 (...)} - \frac{1 + R^2 v^2}{2 + R^4(...)} du dv$

thanks for nothing bot

[cursive wolf]

what demon has asked you to solve this integral

[pearl pond]

@meager imp Has your question been resolved?

[fallen tartan]

[pearl pond]

@meager imp Has your question been resolved?

[cursive wolf]

@meager imp thought: heuristically, you could rewrite the integral into an integral over the unit disk like in this screenshot

then, if you rewrite it in polar coordinates, you can write it as an infinite sum over annuli, first one from r = 1/2 to r = 1, next one from r = 1/4 to r = 1/2, next one from r = 1/8 to r = 1/4 etc etc

so $\int_{0}^1 (...) dr = \sum_{k = 0}^\infty \int_{1/2^{k+1}}^{1/2^k} (...) dr$

[jolly parrot]

Lartomato

[cursive wolf]

heuristically (???) maybe you can pull the limit $R \to \infty$ into the infinite sum (that one you'd have to think about) and into the integrals (that should be fine, since we are in each individual integral not close to any singularities, the convergence under the integral should be unfirom)

[jolly parrot]

Lartomato

[cursive wolf]

so then you can carry out the limit $\lim_{R \to \infty} R^2 \cdot \frac{1 + 2 R^2 v^2 }{1 + R^4 (u^4 + 6u^2v^2 + v^4)} - \frac{1 + R^2 v^2}{2 + R^4(u^4 +v^4)} = \frac{2v^2}{u^4 + 6u^2v^2 + v^4} - \frac{v^2}{u^4+v^4}$

[jolly parrot]

Lartomato

[cursive wolf]

any by god, that doesn't look so bad anymore in polar coordinates: https://www.wolframalpha.com/input?i=simplify+v^2(u^4+%2B+v^4+-+6u^2v^2)%2F(u^4%2B6u^2v^2%2Bv^4)%2F(u^4%2Bv^4)+with+u+%3D+(r+cos+phi)+and+v+%3D+(r+sin+phi)

and by god, the integral over phi is zero: https://www.wolframalpha.com/input?i=integral+-(16+cos(4+ϕ)+sin^2(ϕ))%2F((-17+%2B+cos(8+ϕ)))+dphi+from+0+to+2+pi

so my guess: the integral will converge to zero

[fallen tartan]

you can close this channel if your question has been resolved

[cursive wolf]

i don't think this question has ever been resolved though

i'm also guessing my solution won't work b/c there is no reason for the limit and the infinite sum to be interchangable unfortunately

dominated convergence will likely fail b/c the function is approaching a singularity for R to zero

[meager imp]

.close

[pearl pond]

Channel closed

Closed by @meager imp

Use .reopen if this was a mistake.

[cursive wolf]

did you do it or did you give up 😦

[glad portal]

seeing the amount of messages here I'd give up too 😭

[spiral pivot]

@meager imp there might be an approach that revolves around the residue theorem; however, for multivariate functions of complex variables it's a little more complicated, and not something I'm completely familiar with

[spiral pivot]

https://math.stackexchange.com/a/482060/25445

[cursive wolf]

@meager imp i have now bruteforced both integrals so much

and i am getting increasingly sure that the answer is 0

[cursive wolf]

what i have done is: rewrite in polar coordinates, x = r cos(theta), y = r sin(theta), substitute u := sin theta; this then gives """relatively""" tame integrands which wolframalpha can indefinitely integrate to horrible artanh expressions which simplify relatively well on the bounds u = 0 and u = 1

ah fuck but then i still haven't done the integral over r

omg

suffering never ends

[meager imp]

yeah, actually, at first I thought the answer was 0, I just had trouble proving it

[meager imp]

No, I didn't give up, I just saw someone reminding me to close the channel when I finished solving the problem so I thought this was necessary. 😦

But somehow miraculously, this channel is still here and we are still struggling with the problem

It's true that suffering never ends

[cursive wolf]

but honestly i think this strategy works

i wasn't very careful but the polar coordiantes -> u = sin(theta) substituion -> integrate for u -> integrate for r will proably give you a result

@meager imp can i ask if there is any context at all for this problem?

if this was for a class where you're currenlty learning about green's theorem for example, that would be hopeful

[meager imp]

From the beginning we approached the problem using polar coordinates and some people have also said this is the right direction

[meager imp]

oh okay

This is not a problem that must be solved, in fact we are only 11th graders, this problem was given to us by a 12th grader.

[spiral pivot]

💀

[cursive wolf]

they will be hunted down

[meager imp]

Usually we will encounter calculus problems that are less complicated than this, the transformation process of this problem is quite complicated for me.

[meager imp]

😆

[meager imp]

Yesterday you told me I could approach it through the residue theorem, but actually it's quite complicated for me to deal with multivariable functions.

But something keeps this channel from disappearing

Will it be here forever?

[spiral pivot]

Yes, the residue theorem for multivariate functions is an area of active research and deals with high level constructions

I didn't know you were a high schooler

[cursive wolf]

we are making horrible progress

the last steps of either equation chain is bruteforcing with wolframalpha

i fear for our purity

[meager imp]

:))) really can't imagine a scenario without wolframalpha

[cursive wolf]

hm

i mean, i guess, could make sense, this whole thing goes like 1/r which notoriously does not converge

[cursive wolf]

yeah this is the 1D-integral that this whole chain ended up with

i'm just curious whether one could have foreseen this from the integrand itself

[acoustic path]

suspicious

[sleek cloud]

hmmmm

[finite jolt]

The previous oldest thread: “why is sqrt(a+b) not sqrt(a)+b”
Oldest thread now:

[meager imp]

haha XD

[cursive wolf]

Owo whats this

[cursive wolf]

lastly, subtracting the polar coordinate expressions and just checking the highest-degree terms, we get

$ r \cdot \frac{1 + 2r^2 u^2}{1 + r^4(1 + 4(u^2(1-u^2)))} - r \cdot \frac{1 + r^2 u^2}{2 + r^4(u^4 + (1-u^2)^2)} = \frac{3 r^7 u^6 + p(r,u)}{r^8 (1 + 4 (u^2(1-u^2)))(u^4 + (1-u^2)^2) + q(r,u)}$

where $p,q$ are polynomials in $u$ and $r$ and of degree $< 7$ and $<8$ in the variable $r$ respectively. So the large scale behaviour of the polynomial is like $1/r$, whose integral diverges for $r \to \infty$

[jolly parrot]

Lartomato

[cursive wolf]

I guess that's the general rule here, never should've assumed this converges to anything nice -- would've saved me some calculations

@meager imp the answer is confidently $\lim_{R \to \infty} f(R) = + \infty$, by the above reasoning and since we see that the coefficients of the highest degree terms in denominator and enumerator have positive coefficients

[jolly parrot]

Lartomato

[cursive wolf]

and if you want to make the large scale behaviour argument more precise, factor out $r^7$ in the enumerator and find that, because the degree of $p$ in $r$ is less than 7, the function $r \mapsto p(r,u)/r^7$ is bounded away from zero (so bounded on $[c,\infty)$ for any $c > 0$). Same argument for $q$. So as $r$ becomes large, we have that the expression is larger than $\frac{r^7 \cdot const}{r^8 const} = \frac{1}{r} const$ for some constants that only depend on $u$.

[jolly parrot]

Lartomato

[cursive wolf]

hence, for large enough r, the integral can be estimated from below with the integral over some constant times 1/r, and that's divergent

i'm done with this exercise now omg

[meager imp]

[meager imp]

I really admire your efforts

even though this problem isn't yours, you're awesome

From the beginning, I tried to approach it using polar coordinates

The appearance of the arctanh function made me pretty sure it was going to 0

[pearl pond]

Available help channel!

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[meager imp]

Will this be our end?

[meager imp]

[pearl pond]

@meager imp Has your question been resolved?

Channel closed

Closed by @meager imp

Use .reopen if this was a mistake.

[pearl pond]

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>.
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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[dawn tartan]

1+tan^2Angle/1+cot^2angle

[dawn tartan]

Help me solve this problem using identities pleasee

I dont get it

[broken spear]

use tan = sin/cos and cot = cos/sin, and write 1 as a fraction

[dawn tartan]

Huhhh, i don't get it😭

[robust oar]

there is no left side or right side

also do you mean the "angle" is theta?

[broken spear]

well I guess they're trying to get a simpler expression

[dawn tartan]

Yess

[robust oar]

okok

have you tried anything yet

[dawn tartan]

Umm i didnt since i dont get anything

What is the first steppp?

[robust oar]

pretty they said it here

much*

[dawn tartan]

Do i just have to substitute or stufff

[broken spear]

when you've got trigonometric expressions like that, you'd always want to express everything in terms of sines and cosines

[dawn tartan]

Can you give an example please

[broken spear]

so you go "What's tan(angle) ? -> It's sin(angle)/cos(angle) ! "

then " what's cotan(angle) ?"

you rewrite the expression step by step with what you just found

[dawn tartan]

I have to rewrite this?

[broken spear]

yeah, I mean I guess that's what you want

so you start a chain of equations

step by step rewriting the same thing in different terms

then things will cancel out, and you'll get a simpler expression

[dawn tartan]

I still dont get it😭

Wait imma do youtube tutorial stuff

I dont even understand youtube

[broken spear]

you know that tan = sin/cos right ?

[dawn tartan]

Yes

[dawn tartan]

It's basically Soh cah toa

Wait

Nvm

Oh i just know it

Now what is next step

[broken spear]

well tan = sin/cos so you can rewrite the tan²(angle) as something else now

[dawn tartan]

What is the angle hereee? 2?

Ohhh i get ittt

So basically i can just cancell it

[broken spear]

well the angle is just "angle" or "theta" if you like, its value is not important

[dawn tartan]

Do i cancell?

[broken spear]

for any angle, it can be 2° or 145° we don't know, but for any angle tan(angle) = sin(angle)/cos(angle)

so you just continue the calculations with that "angle" remaining there

[dawn tartan]

So answer is sin⁴/cos⁴?

Anyway thx

.close

[pearl pond]

Channel closed

Closed by @dawn tartan

Use .reopen if this was a mistake.

[pearl pond]

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>.
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• 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.

[tulip cradle]

For this

[tulip cradle]

I don’t really understand how it relates to AB=/ BA

[versed remnant]

ab is not equal to ba so we can not cancel those out , if thats what u are asking about

[tulip cradle]

Ohhhhhh

Ohhhhh

Thanksss

.close

[pearl pond]

Channel closed

Closed by @tulip cradle

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

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

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[keen thicket]

Hello, I need help with this certain problem my teacher gave us. I was able to solve for the derivative of the following but i dont know how to proceed further with the conclusion that symbolab gave me

[keen thicket]

implicit differentiation involving transcendental functions

[versed remnant]

btw if u are differentiating with respect to x

then derivative of cos y would be -sin(y)* (dy/dx)

[keen thicket]

ah yeah mybad forgot to put

[versed remnant]

u should now try get all the dy/dx terms to 1 side

[keen thicket]

ok wait

[hybrid agate]

I got up to here 👇

[keen thicket]

wait

[versed remnant]

changed acc?

[keen thicket]

i had not clarified whether

nope

i had not clarified whether its

lnx(y) or ln(xy)

[hybrid agate]

what is it then

yln(x)

or the latter

[keen thicket]

ylnx

[versed remnant]

i think its lnx(y) cuz they wrote the derivative like that

[hybrid agate]

burhhhhhh

[keen thicket]

most of my classmates r writing ylnx

[hybrid agate]

i thought it was ln xy

[versed remnant]

also

!nosols

[pearl pond]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[keen thicket]

my bad

[hybrid agate]

always write the multiplying factor behind the function

not after it

in this case the factor is a variable y

[keen thicket]

wait

i will have 2 clarify

is this read as ln(xy) or lnx y

[hybrid agate]

that has to be ln(xy)

[keen thicket]

okok wait ill try to

[hybrid agate]

if the y was not in the brackets they wouldve totally writen it as yln(x)

[keen thicket]

thank you for bearing with me

[hybrid agate]

❤️

[keen thicket]

is it right that i

left side only

[versed remnant]

ln(xy) = lnx + lny = 1/x + 1/y(dy/dx) it would be impossible to get this in ur answer

[keen thicket]

using chain rule

i cant trust symbolab anymore

[hybrid agate]

nah left side u cant do that

you must split the logs

[keen thicket]

wym

oh

like

lnx + lny

from there find their derivatives separately?

[versed remnant]

btw

i solved the question and sorry but u wrote it wrong

the way u wrote it first was correct

[keen thicket]

its ok

[versed remnant]

im getting ur answer

[keen thicket]

i genuinely need help understanding it haha

[versed remnant]

its y lnx = sinx/cosy

[versed remnant]

this was correct

[keen thicket]

why couldnt it have been ln(xy)

tho

[versed remnant]

the book should have used brackets but they didng

[versed remnant]

cuz with that u dont get the answer u posted

with y lnx i am getting the answer u posted

i solved it both ways

[keen thicket]

💀

i just messaged my teacher

ln(xy)

[versed remnant]

eh

did i solve it wrong ? i swear im getting ur answer with
y lnx = sinx/cosy

[keen thicket]

its not my answer and

symbolab might have interpreted it as

ylnx

[versed remnant]

oh u used a online calculator

[keen thicket]

ye

[versed remnant]

well if ur teachers says its ln(xy) then its gotta be that

[keen thicket]

it only gives the ans but not the soln

[versed remnant]

now try to seperate the y'

[keen thicket]

okay wait

[versed remnant]

wait what happened

how is there a y' in the denominator

[keen thicket]

i moved y'sinxsiny from the right

[versed remnant]

oh

what i wanted u to do was something like this

wait

[keen thicket]

im very sorry and thank you so much 🙏

[versed remnant]

$$\frac{1}{x} + \frac{y'}{y} = \frac{cosy\cdot cosx + sinx \cdot siny (y')}{cos^2y}$$
multiplying both sides with $cos^2y$

[jolly parrot]

JustToPro

[versed remnant]

no worries

[keen thicket]

oh ok so no denominator

[versed remnant]

so u get this
$$cos^2y(\frac{1}{x} + \frac{y'}{y}) = cosy\cdot cosx + sinx \cdot siny (y')$$

[jolly parrot]

JustToPro

[versed remnant]

yeah no denominator , it becomes hard kinda

[keen thicket]

ok wait ill try

and go on from there

and try to go on from there*

so i multiply cos^2y to both fractions

[versed remnant]

yes

and then u can get the sinx siny y' thing to the left side

$$cos^2y\frac{1}{x} + cos^2y\frac{y'}{y} - sinx \cdot siny (y') = cosy\cdot cosx $$

[jolly parrot]

JustToPro

[versed remnant]

now u can take y' common / factor out y'

[keen thicket]

i moved the first part to the right side tho

the one with no y prime

[versed remnant]

its good

[keen thicket]

okok wait

[versed remnant]

i was going to do that next step

[keen thicket]

n then move right side the inside of parentheses

?

[versed remnant]

u can just divide the entire left side parenthesis on right side

[keen thicket]

ahye

[versed remnant]

and then with a bit of simplification , combine them into 1 fraction and u got ur answer

[keen thicket]

my worst enemy

ill try

[versed remnant]

just know that $\frac{\frac{1}{x}}{y}
= \frac{1}{xy}$

[jolly parrot]

JustToPro

[keen thicket]

ohh

ok wai

t

ok so

from the bottom

y

the denominator of the bottom cos2y

i can put that on the numerator

[versed remnant]

yes

[keen thicket]

as (cosycosx cos2)y

but im not too sure as to how i can do the same for the x denominator

[versed remnant]

(cosycosx - cos2)y

[versed remnant]

u can do the same

[keen thicket]

n bring it down?

[versed remnant]

x multiply with the entire term

yes

[keen thicket]

a h

a h

[versed remnant]

wait wait i abit wrong sorry my bad

u need to make sure the fraction have common denominator

like the numerator fraction is 1 single fraction before u can do that

[keen thicket]

such as the cos^2y

?

ah

[versed remnant]

(it would be to hard to type this out in the image thing i did before )

now u can bring down x and take y up

[keen thicket]

ohhh

h

[versed remnant]

what i said u didnt do D:

i told to make the denominators common before doing this

[keen thicket]

wait :'))

[versed remnant]

btw i gtg but this should be ur answer

ur final answer

[keen thicket]

okay

OH

oh

so

oh so ycosycosx/y is basically just cosycosx?

man.

[keen thicket]

just an example but ye

here

how to

disable

when done

[vestal mural]

type '.close'

[keen thicket]

.close

[pearl pond]

Channel closed

Closed by @keen thicket

Use .reopen if this was a mistake.

[keen thicket]

thank you

[pearl pond]

Available help channel!

Send your question here to claim the channel.

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

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[grand bear]

hello

[grand bear]

why is this incorrect? I just used the y=mx+b formula

oh

am i supposed to look for the line tangent to the derivative? or... how do i go about this

what i just said makes no sense

[snow sail]

youre probably missing the translation

so you have the slope

[grand bear]

whats a translation?

[snow sail]

use point slope

[grand bear]

yes

[snow sail]

like, the tangent line probably doesnt go through the origin

but youve written a line of form y = mx

[grand bear]

got it got it

[snow sail]

lines like this go through the origin

$(y-y_1) = m(x-x_1)$

[jolly parrot]

jan Niku

[snow sail]

line with slope m that goes through point (x1, y1)

[grand bear]

wait

[grand bear]

so i need a different y intercept?

and a new slope

[snow sail]

you dont control the y intercept

[snow sail]

no

we can write it more explicitly

you know the derivative gives you the slope of the tangent line at some point

[grand bear]

yes

[snow sail]

so, we can replace that for m here

[grand bear]

ohh that makes sense

ill try that out

[snow sail]

$(y-y_1) = \qty( \frac{14-x_1}{y_1} ) (x-x_1)$

[jolly parrot]

jan Niku

[snow sail]

only a little algebra remains on your end

this is the equation of the line tho

[grand bear]

does -(322 sqare root 51)/255 look right for the y intercept?

[snow sail]

I can't really say, I'd suggest plotting it

[grand bear]

ill do that

[snow sail]

i would sanity check on something more readily intuitive too

like, the slope you can maybe eyeball, at least as far as the sign and whether its bigger than or less than 1 or 2

and what happens at x=x1

im goin around ping if u reply

[grand bear]

man

my tangent is not tangenting

[grand bear]

i can tell that it should be negative and less than 1

oh i see my mistake

4.2 is my x value

[snow sail]

https://www.desmos.com/calculator/xycvkr3xie

if its helpful

this is a little more functionalized

[grand bear]

man what am i doing wrong

that helps

but im sort of confused with that notation

the most i learned was y=mx+b

[snow sail]

this is just one way to write a line

usually, its best to use the form of the line that corresponds most closely with the information you have

y=mx+b is slope intercept

but with a tangent line, point-slope is much more convenient

[grand bear]

ah got it

what does (x-a) mean?

[snow sail]

(a, b) is some point you want the line to go through

[grand bear]

so x is a?

[snow sail]

so given that the line has slope $m$ and goes through $(a, b)$, you can write the line as $$(y-b) = m(x-a)$$

[jolly parrot]

jan Niku

[snow sail]

you have to be a little careful here

x is a variable, its value can be anything in the domain

but a is some fixed value, its the x coordinate of the tangent point

[snow sail]

like, when you write this equation of the line, you assume that a and b are given, and are fixed. Then, we let x and y vary together.

[grand bear]

so a would be 4.2, and x can vary?

[snow sail]

a would be 21/5

oh, yea, 4.2

and x can be anything

well, i guess, we dont pick an x here, that x can be anything is what gives us a line

[grand bear]

🤯

[snow sail]

were getting a little into the weeds with parameterization i worry its making this process seem more complicated than it is

[grand bear]

sorry

my brain feels a little twisted but im learning 👍

you mean we can ignore x here right?

[snow sail]

its just like

when you have an equation of a line

you know its gonna have an x, and a y

and then some numbers

[grand bear]

right right

[snow sail]

so you dont want to plug in a number for x or y, because you want them in the final equation

but the rest, the a, the b, that all becomes numbers

so we assume you plug stuff in there

thats all im really saying

[grand bear]

got it!

and

b in this equation is different from the y intercept?

[snow sail]

yea, it is

https://www.desmos.com/calculator/s28iuubfda

heres a version without the b

if its less confusing

either way the name is a placeholder, remember, it ultimately gets replaced by a number

so we can call it more or less whatever we want

[grand bear]

got it

sorry im still very lost

how do i find the slope of this tangent line

when there the slope itself contains an x variable

[snow sail]

you gotta ping reply lol

[snow sail]

this equation you have for the slope

as an input, it takes in a point on the circle

and it spits out just a number

the number is the slope of the tangent line at the point on the circle

so you just use the formula $m(x,y) = \frac{14-x}{y}$

[jolly parrot]

jan Niku

[snow sail]

where x,y is some point on the circle

[snow sail]

youre maybe overthinking this

There are less moving pieces than it seems like at first

but im not certain exactly where you are worried

If youre concerned about the variable issue, I could suggest going back and reviewing finding the equation of a line using given information

with the entire calculus context removed, just algebra

[pearl pond]

@grand bear Has your question been resolved?

[grand bear]

@snow sail thank you for the help! i got the answer

.close

[pearl pond]

Channel closed

Closed by @grand bear

Use .reopen if this was a mistake.

[snow sail]

np!

[pearl pond]

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[outer echo]

how do i solve a problem like

1.2 ^ x = 2

[young mural]

Logs

Have you done those yet?

[outer echo]

mm a little bit but i am not sure how to use them properly

[young mural]

Ok well logs basically tell you what power you have to raise "a" to, to get "b", so log to the base a, of b, would give you the power of a that makes b

So in your case, x = log base 1.2 of 2

[outer echo]

ohh

[young mural]

And you'd basically just put that in your calculator

[outer echo]

,calc log_1.2(2)

[jolly parrot]

The following error occured while calculating:
Error: Undefined symbol log_1

[outer echo]

oo

[young mural]

I'm not sure how you'd represent a base in plain text

,calc log(1.2)(2)

[jolly parrot]

Result:

0.36464311358791

[young mural]

,calc 1.2^0.36464

[jolly parrot]

Result:

1.0687414406298

[young mural]

Erm

Oh it probably did multiply

,calc log1.2(2)

[jolly parrot]

The following error occured while calculating:
Error: Undefined symbol log1

[young mural]

hm

,calc log^1.2(2)

[jolly parrot]

The following error occured while calculating:
Error: Unexpected type of argument in function pow (expected: number or Complex or BigNumber or Fraction or Unit or Array or Matrix or string or boolean, actual: function, index: 0)

[worthy lance]

,w log(6/5, 2)

[jolly parrot]

Results provided by WolframAlpha

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

Requested by _arkimond_

[young mural]

Thank you 🙏

[outer echo]

ohh log_base(x)

equals log(x) / log(base)

,calc log(2) / log(1.2)

[jolly parrot]

Result:

3.8017840169239

[young mural]

Yeah so if you write log without a base next to it, so for example log(100), this normally means log base 10

And you may also see ln, this is called the natural logarithm

And it is log to the base e

,calc log(100)

[jolly parrot]

Result:

4.6051701859881

[outer echo]

ohh i thought log was assumed to be base10

[young mural]

It is yeah

[young mural]

I don't know what this is

[worthy lance]

In programs like wolfram log means ln

[young mural]

Right but e^4 isn't 100 either?

[vestal tapir]

it is

[outer echo]

it says 4.6051701859881 for me

,calc e^4.6051701859881

[jolly parrot]

Result:

100

[young mural]

Huh

Am i stupid

YeH I am sorry

[outer echo]

ty for teaching me how to use logs... this was like the only basic algebra rule i was missing

i learned how to take the derivative and integral of logarithm before learning how to use logs now 😭

[pearl pond]

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[sand torrent]

can someone help me with this

[sand torrent]

f(x)= e^1-x

[crisp veldt]

what are you struggling with on this question? @sand torrent

[sand torrent]

idk how to find the area

[minor falcon]

Integrate the function for the given interval

[sand torrent]

wouldn't that be

e^1 - e^-1

[crisp veldt]

yes that's correct

[minor falcon]

Yes

[sand torrent]

so the answer is

e- 1/e

[minor falcon]

Should be correct

[versed mica]

why

it’s from 0 to 2

so -e^-1 -(-e^1)=e-1/e

[minor falcon]

Yes -e^(1)+e^(-1)

[versed mica]

you do 2 first

[minor falcon]

Woops you're right

My bad

[sand torrent]

yeah i forgot to integrate it, but i still managed to get the correct answer

[versed mica]

integral of f(x) from a to b is F(b)-F(a)

[minor falcon]

Yes

Sorry lol my bad

[sand torrent]

so for final answer

is this correct

[minor falcon]

Yes

[sand torrent]

ok great, thank you so much for the help!

.close

[pearl pond]

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[pearl pond]

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[timber elbow]

I don't understand how to do this using multivariable calculus

[timber elbow]

i know we have to treat S as a constant

[pearl pond]

@timber elbow Has your question been resolved?

[pearl pond]

@timber elbow Has your question been resolved?

[pearl pond]

@timber elbow Has your question been resolved?

[pearl pond]

@timber elbow Has your question been resolved?

[glacial scroll]

statouswwpr

!~statustautsuatuataifw ovf-0=2q3o-vg kp=-w[erbk yp[

ok

!status

[pearl pond]

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

[pearl pond]

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[midnight haven]

Hey doing test corrections, wondering if anyone can help me out its simple algebra 2

[midnight haven]

<@&286206848099549185>

[grim fractal]

8% is crazy 😭

[midnight haven]

hey man dont judge me bro

[grim fractal]

it doesn't look like you really tried to solve any of them...

[midnight haven]

i cant even becuase i dont even know the formula

[wanton latch]

Do you understand the concept of fractional exponents?

[midnight haven]

yes

my teacher teaches one thing

tho

its like different from online

so every hard

[wanton latch]

Well it really isn’t

It seems to me that your teacher is attempting u-substitution in a very weird way

[midnight haven]

for number 5 i got pl;us minus 36x^4(y-5)^3

[wanton latch]

First let’s focus on the top half then we will work on the bottom half

[midnight haven]

ok

first i got 256^-1/4

-64?

because 1/4 of 254 is 64, and cant forget about the negative

[wanton latch]

For fractional exponents, the denominator is the power of which the radical is (like square root or cube root) while the numerator power the number that is being radicalized is held to

[midnight haven]

so is it -64?

[wanton latch]

No