@steady basin Has your question been resolved?
#help-0
lone heartBOT
boreal vector
paper sage
$y'=(y^2-1)x$
lone heartBOT
$y'=(y^2-1)x$
ocean sealBOT
Yousssef
ocean sealBOT
Yousssef
rigid smelt
note that your equation is seperable and is in the form of dy/dx=f(y)*x, just use some algebra to seperate all of the functions in terms of y onto one side
paper sage
i got $ln\frac{y-1}{y+1}=x^2+c$
ocean sealBOT
Yousssef
paper sage
i dont know how to move on
rigid smelt
it should be x^2/2
paper sage
yes but we have 1/2 on the left side
rigid smelt
nvm, forgot about the 1/2 on the other side
paper sage
yes
rigid smelt
right, now just plug in the initial value and find C
paper sage
so $ln\frac{-1}{1}=x^2+c$
ocean sealBOT
Yousssef
rigid smelt
also, it's $\ln \left|\frac{y-1}{y+1}\right|$ for the lhs
ocean sealBOT
waler
rigid smelt
and dont forget to plug in x=0 for the rhs as well
paper sage
i dont know what ln-1 is
rigid smelt
it's undefined, but anyway, the antiderivative of 1/(y-1) is ln|y-1| not ln(y-1), this is because we dont know if y>1 or not
similarly for 1/(y+1)
hence, the final result would be ln|(y-1)/(y+1)|
paper sage
yes
rigid smelt
so instead, when plugging in the initial value, you would get $\ln |-1| = 0^2 + \mathrm{C}$
ocean sealBOT
waler
rigid smelt
not sure what you are trying to say there
but y=0 when x=0
i plugged in both x=0 and y=0
paper sage
oh yeh
rigid smelt
this still makes the equation true because this is the initial value that is given
rigid smelt
just evaluate everything and solve for C
paper sage
0
rigid smelt
and what is |-1|?
rigid smelt
the absolute value of -1 is undefined?
rigid smelt
yeah but again, we dont have that
once again $\int \frac{1}{x^2-1} \mathrm{dx} = \ln \left|\frac{x-1}{x+1}\right|$
ocean sealBOT
waler
rigid smelt
yes it should be
rigid smelt
just follow what im asking, what is |-1|?
paper sage
undefined
rigid smelt
the absolute value of -1 is undefined?
.
rigid smelt
oh dear
paper sage
yes
rigid smelt
which means C=0
paper sage
yes
rigid smelt
so now our solution is $\ln\left|\frac{y-1}{y+1}\right|=x^2$
ocean sealBOT
waler
rigid smelt
now can you find a way to get rid of the ln?
paper sage
yes by multiply in e on both sides
rigid smelt
well we dont say multiply, we say raise e to the power of both sides
paper sage
raise i mean
rigid smelt
but yes, we will get $\left|\frac{y-1}{y+1}\right|=e^{x^2}$
ocean sealBOT
waler
rigid smelt
now, technically here is what your answer got wrong, but technically, there is two cases to this, but I suppose they just wanted (y-1)/(y+1) to be positive
keen socket
now, technically here is what your answer got wrong, but technically, there is t...
its equivalent
due to the modulus
rigid smelt
anyway, yes, we would get (y-1)/(y+1)=e^(x^2) and just cross multiply
rigid smelt
its equivalent
no, not really
getting rid of the absolute value bar would need to address for two cases, where (y-1)/(y+1) can be positive or negative
here we have no information on y except for the fact that y(0)=0, and this at all does not help
paper sage
$y-1=e^x^2(y+1)$
ocean sealBOT
Yousssef
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
ocean sealBOT
Yousssef
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)
rigid smelt
just gather y to one side and x to one side
then solve for y, basically similar to what you did to the de
paper sage
divide by y+1?
paper sage
then solve for y, basically similar to what you did to the de
should i divide by y+1
rigid smelt
no, you multiplied both sides by y+1 to ultimately just divide it again
that doesnt quite make sense
rigid smelt
just gather y to one side and x to one side
.
gilded vessel
here we have no information on y except for the fact that y(0)=0, and this at al...
i mean if you were to choose (y-1)/(y+1) to be negative then that condition would not be met
rigid smelt
i mean if you were to choose (y-1)/(y+1) to be negative then that condition woul...
oh right, nvm, so yeah, only one case will happen
e^{x^2} for nested exponentiation
paper sage
like that
rigid smelt
yeah, almost there, now expand the brackets
ocean sealBOT
Yousssef
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)
paper sage
$y-e^{x^2}\cdot y-e^{x^2}=1$
ocean sealBOT
Yousssef
paper sage
this seems wrong
paper sage
yeah, almost there, now expand the brackets
this feels wrong
paper sage
$y(1-e^{x^2})-e^{x^2}=1$
rigid smelt
that sadly is not correct
ocean sealBOT
Yousssef
paper sage
$y=\frac{1+e^{x^2}}{(1-e^{x^2}}$
ocean sealBOT
Yousssef
paper sage
I don’t understand how to ask a question
not here
whole haven
Oh sorry
paper sage
ocean sealBOT
paper sage
I don’t see anything wrong
rigid smelt
apparently the antiderivative is wrong
the general solution should have been $\ln\left|\frac{1-y}{y+1}\right|=x^2+\mathrm{C}$ instead
ocean sealBOT
waler
rigid smelt
oh right
paper sage
oh yes
rigid smelt
hmm so that means, we went wrong at the step where we tried to get rid of the absolute value
ocean sealBOT
Yousssef
gilded vessel
hmm so that means, we went wrong at the step where we tried to get rid of the ab...
in essence
rigid smelt
though that doesnt quite make sense, or my head is not thinking straight
gilded vessel
you just need to take the other case for the right solution
rigid smelt
since the positive case should work
paper sage
you just need to take the other case for the right solution
what do u mean
rigid smelt
the problem that i had addressed to you earlier
gilded vessel
where (y-1)/(y+1) is negative
rigid smelt
yes, the second line however is the problem
paper sage
oh okay
rigid smelt
im not sure if your teacher requires you to prove why the negative is the one to take
paper sage
so i should just flip it
rigid smelt
but apparently i dont get why also
gilded vessel
i dont know much on differential equations but i reckon you could solve both cases and then validate them
although you dont need to go that far
if you look back to when you had ln|-1|
you know taking the positive case will result in no longer having y(0) = 0 be possible
paper sage
or i just set my own rules when we have inte gral of 1/x^-1 its always 1/2 times ln 1-x/x+1
gilded vessel
because you have ln(-1)
rigid smelt
yeah, i did that on paper (though admittedly written badly), and the positive case is the one where it should work. Lemme just recheck again
gilded vessel
meaning you should take the negative case
rigid smelt
oh wait
gilded vessel
yeah, i did that on paper (though admittedly written badly), and the positive ca...
well positive for the derivative you found
positive for (1-y)/(y+1)
rigid smelt
right, actually should account for that
gilded vessel
yeah
gilded vessel
why i am the only one who feels dumb right now
nah
im not even 100% sure if my reasoning is correct
im guessing out here ngl
gilded vessel
i understand that we should take negative case since the we have y(0=0
well we take it out to allow this to be possible
paper sage
so if we take negative case should i flip my quota then?
gilded vessel
because in the positive case you would get ln(-1) meaning the point does not exist for any c value
rigid smelt
so if we take negative case should i flip my quota then?
yes
gilded vessel
yeah to get the solution in this case its just the reciprocal
thats not general though
paper sage
if we take the negative case isnt that still ln -1
gilded vessel
if we take the negative case isnt that still ln -1
no because you would have ln(-1*-1)
paper sage
or no its ln (1) right
gilded vessel
thats not general though
in reality you should recognise this at the step of handing the absolute value and then make the decision for which case to take
i believe thats right because at that step is just algebraic relations
paper sage
in reality you should recognise this at the step of handing the absolute value a...
yes
rigid smelt
yeah i think that should also be how we eliminate the cases
gilded vessel
i havent even done differential equations or anything past polynomial integrals but im guessing this is fine since its past that step
rigid smelt
and i dont usually have to deal with cases evaluation like this when solving des because all of the DE's i do in phys just have formulas and stuff
right, i guess the question is resolved, im going to bed then
gilded vessel
sweet dreams
rigid smelt
u too
gilded vessel
yeah i should probably sleep its 2am
rigid smelt
oh are you aus time?
gilded vessel
yeah
rigid smelt
your e^(x^2) just disappear into thin air
gilded vessel
no after that step
from the 2nd last to last step
im guessing you meant to move it to the RHS
foggy nest
Who really smart here?
Two natural numbers are given. The larger of them is equal to the square of their
difference, and the smaller of them is 9 times greater than their greatest common
divisor. Find the smallest common multiple of these two numbers.
quiet vector
u?
foggy nest
Nah I am bad. Help me pls
quiet vector
and u need to assume one of them is larger say x
it says x=(x-y)^2
now make the second equation and solve
foggy nest
Dude, I've already done that, but I'd like to compare my answer and someone else's
tall topaz
Okay then post your solution
lone heartBOT
@foggy nest Has your question been resolved?
foggy nest
I think I had many mistakes, so I need normal answer I'm really going to have problems if I don't do these tasks
narrow meadow
how would I solve something like this? Is it something like minimizing the k(n-k) part using the inequality that is included? If not then how else do I turn this into something solvable like a quadratic or smthn
narrow meadow
I'm assuming it's a quadratic function in k
but idk I could be wrong
the answer should be (n-1)(n-2)/2 but idk how to get there
lone heartBOT
@narrow meadow Has your question been resolved?
wind hawk
Is this a sum?
narrow meadow
is what a sum?
wind hawk
sorry im confused what the problem is asking you to do with the expression
narrow meadow
sorry im confused what the problem is asking you to do with the expression
hmm ok how about this, show that the function equals (n-1)(n-2)/2 given the inequality
wind hawk
you might need to wait a bit after you first join
wind hawk
^
you can just get your own channel, ask in #discussion or smthn
okay go there and someone will help you
wind hawk
the answer should be (n-1)(n-2)/2 but idk how to get there
maybe I'm confused but I'm not sure it's true? If you set k to always be (n-2), for example,
$\frac{n(n-1)}{2} - (n-2)(n-(n-2)) = \frac{n(n-1)}{2} - 2(n-2) = \frac{(n^2-n -4n+4)}{2} \neq \frac{(n^2 - 3n + 2)}{2}$
ocean sealBOT
THEBIGTHREE
narrow meadow
hm
wind hawk
that assumes k depends on n, but in the case its constant I think you could still exhibit a counterexample in the same way
narrow meadow
ic, lemme think about that rq
narrow meadow
wait sry I'm not familiar with desmos, what's the purple line?
wind hawk
its shown on left, the purple is the original statement, the green is the simplified one
plotted as a function of n
the {} specify a domain restriction, so it only plots when 1 <= k <= n-1
narrow meadow
hmm icic
yeah that makes sense
well shit, that means I did something wrong earlier on then
lol
wind hawk
howd you simplify initially?
narrow meadow
are you familiar with graph theory?
wind hawk
it seems odd to me that there would even be such a simplification, but I havent done those types of problems much
ah, not really, thats probably why
I wonder if you could prove it with induction, though?
that would be my first thought
narrow meadow
ah, not really, thats probably why
I'm actually doing this but I asked about the simplification cus I wanted to try it with my own method but yeah
wind hawk
ohh, I see
yeah it could be your initial equation doesnt correctly model the situation?
narrow meadow
yeah well, feel free to stop me anytime but, what I basically did was
let n be number of vertices,
using handshake theorem the umber of edges for a graph is n(n+1)/2
since the question is asking about a disconnected graph
I can do n(n+1)/2 - number of edges missing = number of edges of disconnected graph
when graph is disconnected it's split into 2 parts basically
if number of vertices in one part is k
second part is n-k
and so number of vertices difference of disconnected and connected is k(n-k)
and so n(n+1)/2 - k(n-k) = (n-1)(n-2)/2
that's what I had in mind
but apparently it's wrong
wind hawk
im not very familiar with graph theorey so this may be off, but how about thinking about each disconnected part seaparately and saying the max in one is
(n-k)(n-k+1)/2
and the other is
k(k+1)/2
and then their sum is the max number for the entire graph?
that doesnt seem to work though
I would imagine you have to do something more fundamental than using the max for an undirected connected graph?
so maybe look at how that's derived and make an adjustment
narrow meadow
so maybe look at how that's derived and make an adjustment
yeah I'll give that a go
on a side note, how familiar are you with combinatorial proofs
wind hawk
on a side note, how familiar are you with combinatorial proofs
not at all 😅
not a fan of discrete personally
narrow meadow
neither am I
😂
all good 👍
thanks for the help man
prob gonna go to bed it's 2am xD
cosmic swallow
Hey, I am having trouble solving this problem. It is a pre-university polynomial problem.
minor needle
,rccw
ocean sealBOT
cosmic swallow
go to math help available
grim wyvern
where
lament glen
"the rest" as in remainder or result of division?
wait I guess the latter interpretation doesn't make sense
cosmic swallow
It means that if you put 1 in this polynomial it returns -8
lament glen
oh
cosmic swallow
lament glen
ok I think I got it
lament glen
least possible degree would be 3, since we can't get an x + 3 term from m(x)
so we multiply m(x) and n(x) which is x^3 + 2x^2 - 2x + 3
plugging in 1 here to check the remainder for division with s(x), we get 4
we wanted -8, and multiplying the polynomial by -2 gets us that result
fallow grove
Hey guys is the marked circle in image one the same as $\gamma y^{\frac{1}{\gamma}}$?
ocean sealBOT
fluffy snail
warm wind
no
fallow grove
ahh right because the gamme is negative
warm wind
its simply because ${-\gamma}-1 $ isn't equal to $\frac{1}{\gamma}$
ocean sealBOT
Lyu
delicate jacinth
Can you use derivative?
lone heartBOT
@gritty pond Has your question been resolved?
ocean sealBOT
sopinha
lone heartBOT
@gritty pond Has your question been resolved?
delicate jacinth
why $2\sqrt[3]{18}$
ocean sealBOT
sopinha
ocean sealBOT
cookie2
delicate jacinth
i was lazy to take the derivative, so i did it on geogebra
cold grail
That does look complicated, either do something like find a taylor series for it, or replace tan with sin and cos
But let me ask what calc is this?
It should be bat the definite integral part is wierd
But let integral of tan(1 / u^2 + 4) du, bounded by s and -8
be f(s) - f(-8)
d/ds would be tan(1/ s^2 + 4) - 0
since f(-8) is a constant
Kind of yeah, If you take the derivate of a integral f(x) you get f(x)
With some extra steps you get rid of the -8 while keeping the s
lone heartBOT
@gritty pond Has your question been resolved?
placid stirrup
i have 10 questions like these, if someone could dm me and walk me through how to work this out i would greatly appreciate it
desert tusk
i have 10 questions like these, if someone could dm me and walk me through how t...
Why couldn’t you just share all it here
placid stirrup
oh i thouhgt i did my b
desert tusk
DMs limit your chance to get more help
lone heartBOT
<@&286206848099549185> if i could get any help would greatly appreciate it
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.
@placid stirrup Has your question been resolved?
placid stirrup
<@&286206848099549185>
tacit arch
i have 10 questions like these, if someone could dm me and walk me through how t...
you should do your own work first before just asking for solutions
it's also against the rules to simply juts give answers
lone heartBOT
@placid stirrup Has your question been resolved?
last ether
It's pretty self explanatory
open roost
k thx man
open roost
That's why
btw man could u help me with this q too
how would u know when to use radiant or degree on ur calcualtor
when i used this i tried radiant mode on cal
but got a different number
open roost
Unless it gives you the degree symbol
how am i getting this but the answer for rad is 5.55
last ether
,w 2 Pi - Arctan[2/Sqrt[5]]
ocean sealBOT
last ether
That's why
open roost
That's why
wait how come we munus it with 2pi
last ether
Because the angle terminates in Q4
open roost
Because the angle terminates in Q4
wait so this would be the reference angle tan^-1(2/√5) = 0.729
open roost
but we do 2π -tan^-1(2/√5) = for the real angle
open roost
That's the ref angle
right?
last ether
Mmhm
lone heartBOT
@open roost Has your question been resolved?
placid stirrup
i dont have a particular problem, but does anyone know how to solve second order ODE with laplace?
im struggling
open roost
Mmhm
yes rightr?
wait but for radiant how come there is no negative angle for it?
like u know how there is -45
would it just be -0.72
lone heartBOT
open roost
and also we write angles (in radians) 0 to 2 pi
oh so that is like the default domain?
lone heartBOT
Available help channel!
open roost
but how come for degree it can be -45?
lone heartBOT
but how come for degree it can be -45?
tribal haven
0 to 2pi rad corresponds to 0 to 360 degrees
and if you can convince yourself that 1: an angle of x and an angle of x+360 degrees are the same and 2: negative angles can exist then you can define a valid value for -45 degrees as having the same value as 315 degrees
open roost
wait but how come for this it could be -42 or 318
tribal haven
because they represent the same angle
open roost
could u not find something that represents the same angles for radiants then?
tribal haven
you can, but usually you work with radians in [0, 2pi] like you would for degrees in [0,360]
see how -42 is not in that [0,360] range so you only have 1 radian value given
open roost
oh like u could but it is not nessesary unless it asks for it?
tribal haven
ye
lone heartBOT
@open roost Has your question been resolved?
knotty mountain
if $w^2 + w + 1 = 0$ and $w^3 = 1$ what would be the best way to approach evaluating the following?
ocean sealBOT
vertify
knotty mountain
$(1-w)(1-w^2)(1-w^4)(1-w^5)(1-w^7)(1-w^8)$
ocean sealBOT
vertify
knotty mountain
is there any simple way or do i just need to go through the process
the answer is 27
viral pagoda
Uhm
knotty mountain
i guess after you break it down it's not too bad
little drum
yes ofc. first of all.. you simplify w^3 = 1, so you're left with
(1 - w)(1 - w²)(1 - w)(1 - w²)(1 - w)(1 - w²) = [(1 - w)(1 - w²)]^3 = [1 - w - w² + w^3]^3
knotty mountain
$(1 - w)(1 - w²)(1 - w)(1 - w²)(1 - w)(1 - w²) = [(1 - w)(1 - w²)]^3 = [1 - w - w² + w^3]^3$
ocean sealBOT
vertify
little drum
ah if you want in latex:
$(1 - w)(1 - w²)(1 - w)(1 - w²)(1 - w)(1 - w²) \ = [(1 - w)(1 - w²)]^3 \ = [1 - w - w² + w^3]^3$
ocean sealBOT
Arya
little drum
$= [1 - (-1) + 1]^3 = 3^3 = 27$
ocean sealBOT
Arya
ocean sealBOT
vertify
ocean sealBOT
vertify
alpine sable
can i write $\pi+\cot^-1(x)=\cot^-1(x)$??
lone heartBOT
can i write $\pi+\cot^-1(x)=\cot^-1(x)$??
ocean sealBOT
M O L T R E S
alpine ore
No, because that implies that $\pi=0$??
ocean sealBOT
jafar/جعفر
tacit arch
,w plot arccot(x)
ocean sealBOT
tacit arch
,w plot arccot(x) + pi
ocean sealBOT
alpine sable
what about $\pi + \cot (\frac{\pi}{6})$??
ocean sealBOT
M O L T R E S
ocean sealBOT
♡LexQa♡
knotty mountain
if w is a non-real root, show that $w + w^2 + w^3 + w^4 + w^5 + w^6 = -1$
ocean sealBOT
vertify
knotty mountain
.close
lone heartBOT
lone heartBOT
Available help channel!
strong schooner
lone heartBOT
strong schooner
So I found f'(x) and placed it in Newton's Method $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$
ocean sealBOT
ṼØῳ
strong schooner
Now the problem is that f'(x) = 0, so basically this thing will be undefined?
So the question is how do I change this into a form that works and does not implode on itself?
rose sigil
wait what is the issue?
strong schooner
(Answer is D but I got A)
I'm guessing that it has something to do with the fact that f'(x) = 0?
rose sigil
i thot A was right T_T
strong schooner
ehhhhhhhhh
rose sigil
i think it's an error that d is "correct" nvm HAHA
strong schooner
we are screwed
I do remember that Newton's Method will "break" if I put in something like an maximum or minimum point.
I tried it once with an absolute point of a quadratic equation 
rose sigil
ya but this doesn't have anything to do with making the initial guess lol
strong schooner
Question is probably asking how do I make Newton's Method work with min/max points.
rose sigil
oh @strong schooner i see what the problem is
strong schooner
I recommend using a calculator
rose sigil
you're basically applying newton's method to f, but it should be being applied to f''
alpine sable
Its unethical to provide answers to an exam
strong schooner
Or asking in an open help channel
rose sigil
so it should be $x_{n+1}=x_n - \frac{f'(x_n)}{f''(x_n)}$
ocean sealBOT
💜𝓁𝒶𝓎𝓁𝒶💜
strong schooner
wait
ocean sealBOT
🅿🆁🅾🅿🅶
strong schooner
does Newton's method still work for f'(x)/f''(x)
rose sigil
<@&268886789983436800> asking for test answers or something and spamming
strong schooner
sigh
rose sigil
does Newton's method still work for f'(x)/f''(x)
yea?
rose sigil
the thing you were doing before would find x's such that f(x) = 0
but we want x's such that f'(x) = 0
by find i mean approximate
f' is just a function
rose sigil
loll
strong schooner
Wait here's how I understand it
Since we are putting in max/min points of f(x) into Newton's method
So basically they will be roots in f'(x)
So we can use Newton's method for f'(x)
Is that correct?
rose sigil
sure
strong schooner
imma try solving it brb
$$x_{n+1} = x_n - \frac{cosx - \frac{1}{x}}{-sinx + \frac{1}{x^2}}$$
ocean sealBOT
ṼØῳ
strong schooner
wow that's really ugly
white falcon
I know
strong schooner
ehhhh
rose sigil
the x's should be x_n's but i think you're just about done
strong schooner
$$x_{n+1} = x_n - \frac{\frac{xcosx-1}{x}}{\frac{-x^{2}sinx+1}{x^2}}$$
ocean sealBOT
ṼØῳ
strong schooner
so we got this
rose sigil
haha why
ocean sealBOT
ṼØῳ
rose sigil
T_T
strong schooner
I blame my professor
Either I made some arithmetic mistake, or the answer is incorrect
lone heartBOT
@strong schooner Has your question been resolved?
white falcon
The answer is definitely a, what's the problem?
strong schooner
Actual answer is D.
It isn't A because it is asking us to use Newton's Method to approx the "critical value", not "root".
So it should be something like f'(x)/f''(x)
Problem is that even after doing so it is still wrong
Looking for someone to verify the answer at this point
rose sigil
it's def not a haha
white falcon
Oh if that's is so i think the answer c
strong schooner
Same, but it's D for some reason
white falcon
Have you checked the algebra? Maybe there is some errors in simplification or signs
strong schooner
Did, looks the same.
white falcon
Wait is the newton's method for finding critical point is $x_{n+1}=x_{n}-\frac{f'(x)}{f''(x)}$
ocean sealBOT
Gabok
strong schooner
Should be?
white falcon
Or is it plus
strong schooner
white falcon
Or this one $x_{n+1}=x_{n}+\frac{f'(x)}{f''(x)}$
ocean sealBOT
Gabok
strong schooner
minus
white falcon
Then i could not find any reason that it should not be c
The only way to find which is right is to graph the function and calculate up to x3 and see which one approaches the critical points more
strong schooner
Hmm
Welp I assume it’s a mistake then, @white falcon @rose sigil thanks
.close
golden vortex
Isn’t it 6? If not then why?
lone heartBOT
Isn’t it 6? If not then why?
rose sigil
why 6?
surreal meadow
how did you get 6?
tacit arch
Isn’t it 6? If not then why?
How did you get 6
surreal meadow
lmaooo
rose sigil
haha
tacit arch
EVERY HELPER GET IN HERE
rose sigil
which ones do you think are rational then haha
tacit arch
Do you know what a rational number is?
golden vortex
sqrt2 sqrt3 sqrt5 sqrt6 sqrt7 sqrt8
gray isle
(and give reasoning for each one)
golden vortex
sqrt2 is like ~ 1.4
tacit arch
I see
gray isle
decimal approximations don't tell you if something is rational or not
tacit arch
The definition of rational number you should use is if the decimal expansion repeats
surreal meadow
sqrt2 sqrt3 sqrt5 sqrt6 sqrt7 sqrt8
i fear you are confusing rational with irrational
ocean sealBOT
Result:
4
spring vigil
so, it was pretty easy finding it using the indefinite integral, but im having some trouble with part b
surreal meadow
+C
surreal meadow
ah didn’t see that sorry
spring vigil
thanks though
all good!
anyways, how am i supposed to set this up using a definite integral?
surreal meadow
i kind of figured it would be with $\int_0^{20} P’(x),dx$
ocean sealBOT
maximo
surreal meadow
but there’s no real motivation for that lower bound ig
spring vigil
ohhhhhhhhhhhhhhhhhhhhhhhh wait i misread the problem
lmao
he says to find p(x) and then p(20)
my bad
thanks for your time though!
.close
alpine sable
lone heartBOT
alpine sable
ok simplier question than before
but I'm confused conceptually
what do I minus from what?
Like I thought one of the two regions that lies between the curves be 0.5 * integral from 0 to pi/3 of (2sin(2x))^2 - (2sin(x))^2 and therefore the answer would be that multiplied by 2
(which is wrong)
surreal meadow
if i’m interpreting the question correctly, the area inside both of the curves on the first quadrant is the area inside 2sin(theta) up to their intersection and the area inside 2sin(2theta) from their intersection up to pi/2
surreal meadow
pi/2 because i specified inside the first quadrant
alpine sable
,w 2sin(2x) = 2sin(x)
ocean sealBOT
alpine sable
isnt it pi/3?
surreal meadow
yes
alpine sable
oh
surreal meadow
that’s their intersection
alpine sable
Like I thought one of the two regions that lies between the curves be 0.5 * inte...
thats what I did here though
I subtracted
alpine sable
cant I just do 0.5 * integral from 0 to pi/3 of (2sin(2x))^2 - (2sin(x))^2
and multiply by two
because symmetry
alpine sable
yeah
surreal meadow
but they want the area inside both curves
alpine sable
OHHHHHHHHHHHH
thank you
im gonna see if I can do it now
,w integral from 0 to pi/3 of 0.5 * (2sin(x))^2 + integral from pi/3 to pi/2 of 0.5 * (2sin(2x))^2
ocean sealBOT
surreal meadow
*2
alpine sable
if i’m interpreting the question correctly, the area inside both of the curves o...
thank you sorry for being dumb
surreal meadow
np, you weren’t being dumb
alpine sable
.close
strong schooner
lone heartBOT
strong schooner
So this is basically asking me $$2\int_0^1{sin^{-1}{x} dx}$$
ocean sealBOT
ṼØῳ
alpine sable
yeah
strong schooner
But I do not know how to integrate arcsin(x)
So I assume it's trying to ask me to find the area using Riemann Sum?
alpine sable
do you know IBP
strong schooner
Nope
alpine sable
what integration methods do u know
strong schooner
Graph looks suspiciously like many rectangles
So my initial guess is using Riemann Sum
So it would look something like
alpine sable
id just assume it wants you to use IBP to integrate arcsin
riemann sums seems like alot of work
strong schooner
$$2\frac{1}{n}({arcsin{\frac{1}{n}} + arcsin{\frac{2}{n}}... + arcsin{\frac{n}{n}}})$$
strong schooner
riemann sums seems like alot of work
Welp they didn't teach IBP yet so
alpine sable
🤷♂️
ocean sealBOT
ṼØῳ
surreal meadow
i don’t think doing riemann sums will take you there nicely though, unless you can somehow evaluate those arcsins
surreal meadow
you could try the taylor expansion for this but i doubt it would get you anywhere near those pi answers.
ibp is the easiest way to go
alpine sable
is a calculator allowed
strong schooner
~~Or is this out of the curriculum ~~
strong schooner
is a calculator allowed
Nope
alpine sable
yeah no clue how you would solve with ur current techniques
surreal meadow
ah, perhaps just integrate over y?
strong schooner
let arcsin = u?
surreal meadow
so take sin(2y) = x
surreal meadow
then just take $\pi -\int_0^{\pi} \sin (2y),dy$
ocean sealBOT
maximo
alpine sable
🧠
surreal meadow
hopefully that should do it (i may have screwed up some details but the idea is there)
strong schooner
$\pi -\int_{TBA}^{TBA} \arcsin(sin(2y)),dy$
ocean sealBOT
ṼØῳ
surreal meadow
nah
the curve y=2arcsin(x) from x=0 to 1 is the same as x = sin(y/2)
from y = 0 to pi
so instead of finding the integral of arcsin, we just find the area above the curve, that is, $\int_0^{\pi} \sin (\frac{y}{2}), dy$
strong schooner
ocean sealBOT
maximo
surreal meadow
then to find the area under the curve we take the whole area of the rectangle with corners (0,0) and (1,pi), so pi, and subtract what we found
so $\pi - (\text{the integral from above})$
ocean sealBOT
maximo
ocean sealBOT
surreal meadow
we just take the area under this curve (or 1-area under the curve if you’d like)
strong schooner
Processing
surreal meadow
perhaps it’ll make more sense with something more familiar. for example, graph y=x^2, and notice that it’s the same thing as saying x = sqrt(y), for x>=0
surreal meadow
what do you mean transform
ocean sealBOT
wraith sun
lone heartBOT
wraith sun
Why this equation is not linear for y???🥲
surreal meadow
do you know what it means to be linear
notice you have a y^3 and you’re dividing by it
modern wyvern
@jade hollow is this almost the same as the previous one where i had to balance the newtons?
desert warren
Trying to see whether the series converges
lone heartBOT
Trying to see whether the series converges
desert warren
but apparently this one is actually divergent?
but i think it converges by alternating series test
bn is definitely positive, the derivative is negative and its limit is 0
rose sigil
molten pivot
Lol
desert warren
molten pivot
Now try taking the sum from 2 to infinity
molten pivot
,w Σ_(n=2)^inf (-1)^n/(n*ln(n))
ocean sealBOT
desert warren
https://socratic.org/questions/how-do-you-test-the-series-sigma-1-nlnn-from-n-is...
is this wrong then?
desert warren
ohh i see
ty for ur help
kinda weird why u cant just say the series that starts at n=1 doesnt also converge
desert warren
🤦♂️
molten pivot
😂
desert warren
https://cdn.discordapp.com/attachments/1021869096289189899/1048880188311744542/image.png one last thing
why is symbol lab saying it’s diverging
molten pivot
Haha all good, I understand
lone heartBOT
molten pivot
Reading closely will save your ass. Lol
lone heartBOT
Available help channel!
uncut pelican
I genuinly feel like im being dumb and can't solve such a simple question because im missing something. 
I dont know how to even start, its not as if its a right angle triangle, the angle's aren't given and the sides are all in x. Question is Find Perimeter of the Triangle, you have been given the base and the height as visible in the image
prime badge
what's the question though
uncut pelican
what's the question though
op my bad, find perimeter
find perimeter of the triangle
<@&286206848099549185>
prime badge
i'm a noob but looks impossible
lone heartBOT
uncut pelican
i closed since it genuinly looks impossible even to me 🙏
lone heartBOT
Available help channel!
alpine sable
hi can someone explain me how to do this
lone heartBOT
hi can someone explain me how to do this
alpine sable
determine a such that the straight line:
i) steps through the point A = (2, 0)
ii) is parallel to the straight line y = 2x − 1
iii) is perpendicular to the straight line 3x − y + 1 = 0
iv) is parallel to the bisector of quadrant I and III
v) make an acute angle with the x-axis
rose sigil
for the line to go through (2,0), the given equation needs to be true when x = 2 and y = 0
so if you substitute those in, you’ll have a one variable equation in a to solve to find what a must be 🙂
alpine sable
i will try thx
remote sand
$fortnite battle pass$
lone heartBOT
$fortnite battle pass$
ocean sealBOT
Abstraction
vague coral
worn fox
.close
quasi trail
lone heartBOT
alpine sable
ACD = 1/2 AOB
lone heartBOT
@quasi trail Has your question been resolved?
zenith shale
i need help
analog falcon
zenith shale
i can’t find the help 0
lone heartBOT
lone heartBOT
Available help channel!
pastel turret
How to find area of trapezium using length of four sides and without height
echo socket
Show how the problem is stated
Cuz lengths of the sides isn't always enough to determine the area
obsidian totem
How to find area of trapezium using length of four sides and without height
Can you send a screenshot?
pastel turret
obsidian totem
Ok one minute
lone heartBOT
@pastel turret Has your question been resolved?
pastel turret
.close
alpine sable
How do we associate a geometric sequence with a function
lone heartBOT
How do we associate a geometric sequence with a function
alpine sable
How do we associate a geometric sequence with a function
for exemple for 1;2;4;8;16;...
delicate jacinth
$f(x) = a_1\cdot q^{n-1}$
ocean sealBOT
sopinha
lime sigil
What are we doing here?
alpine sable
$f(x) = a_1\cdot q^{n-1}$
then i replace the a1 ?
alpine sable
but its like An=A1*q^n-1
delicate jacinth
yes
alpine sable
but why did u replace the An by f(x)
delicate jacinth
because you wanted a function
lone heartBOT
alpine sable
@delicate jacinthbut wait
for f(x)=a1*q^n-1
we replace the N by what then
the N is the X ?
.reopen
lone heartBOT
✅
alpine sable
@delicate jacinth hey help pls
raven dune
f(n) = a1*q^n-1
lone quartz
help
lone quartz
i dont know how to start
vale wigeon
lone quartz
kinda
oh
i just need to sub in half angle formula
then how about the sqaure
hi
probably?
breh
frozen tree
ask him to go to a non occupied channel
hes been bugging me and not willing to read the #❓how-to-get-help
ebon otter
then how about the sqaure
it's the same as (cos(β/2) )^2
lone quartz
it's the same as (cos(β/2) )^2
i cant get answer tho
frozen tree
it is
frozen tree
lone quartz
answer is 49/50
frozen tree
sorry guys for interupting..
ebon otter
But not pretty sure...
lone heartBOT
Available help channel!