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#help-0
alpine sable
need help What must be the value of a for the graph of f(x) = ax2 to pass through the point C(-1; -1)?
vale wigeon
did you mean f(x) = ax**^**2?
alpine sable
What must be the value of a for the graph of f(x) = ax2 to pass through the point C(-1; -1)?
alpine sable
did you mean f(x) = ax**^**2?
no its f(x) = ax2
severe sluice
no its f(x) = ax2
is that equivalent to 2ax?
vale wigeon
$ax \cdot 2$?
ocean sealBOT
Kanga Gang Annihilator Ann
vale wigeon
so the 2 isn't an exponent?
alpine sable
i can show orginal photo
vale wigeon
k so you were given a poorly typeset problem...
vale wigeon
you need to have f(-1) = -1
regardless of what formula f is given by, it comes down to that and nothing else
alpine sable
ok thanks 👍
lone heartBOT
@alpine sable Has your question been resolved?
ivory zodiac
lone heartBOT
ivory zodiac
lim t approaches 0
can someone explain why the bottom was mutiplied by 8
shiuldnt it be only mytiplied by 2
to make the bottom t into 2
2t
ivory zodiac
ahh and it was just moved out of the parantehis!
i didnt see that
thank you!
.close
patent lotus
How do I solve these two?
severe forge
for the 3 do 2 integrations by parts
lone heartBOT
@patent lotus Has your question been resolved?
rancid narwhal
The line whose equation is y=x+5 is the perpendicular bisector of the line joining the points points P(3, 10) and Q(a, b). a] write down the midpoint of PQ in terms of a and b. b] show that b=a+3 and a+b=13?
how to do number b?
severe sluice
The line whose equation is y=x+5 is the perpendicular bisector of the line joini...
try using the definition of a perp bisector first
severe sluice
she's substituting y and x into the equation
i mean
y and x of the midpoint of the section PQ
shutup that might be a bit rude, sorry
@rancid narwhal the point where the line and the segment intersect would be the midpoint of PQ
according to the definition of a perpendicular bisector
rancid narwhal
ughh no fighting pls i have a very hard exam coming at 8:00
rancid narwhal
<@!742415866922139759> the point where the line and the segment intersect would ...
yes
,ti
ocean sealBOT
The current time for General Winfield Scott is 05:27 PM (+06) on Tue, 21/12/2021.
You haven't set your timezone! Set it using the interactive timezone picker with ,ti --set.
severe sluice
yes
and in the second step, your friend is trying to calculate the coordinates of that point
rancid narwhal
its physical exam so i have to go 30 mins before
severe sluice
its physical exam so i have to go 30 mins before
you got 2 hours left...
rancid narwhal
yea
severe sluice
and in the second step, your friend is trying to calculate the coordinates of th...
you understand this? @rancid narwhal
severe sluice
the midpoint of a segment AB where A has coordinates $(x_a,y_a)$ and B has coordinates $(x_b,y_b)$ is $(\frac{x_a+x_b}{2}, \frac{y_a+y_b}{2})$
rancid narwhal
1 min
ocean sealBOT
Kanga Gang ¬Sam
severe sluice
**Kanga Gang ¬Sam**
@rancid narwhal does this formula ring any bells?
rancid narwhal
**Kanga Gang ¬Sam**
yes i know it
severe sluice
yeah that's good
rancid narwhal
<@!742415866922139759> does this formula ring any bells?
yes i know it and i got it
pls stfu
pls go away
severe sluice
shut up
will you shut up
severe sluice
you know why she got that gradient?
severe sluice
yeah
rancid narwhal
🙂
severe sluice
the rest should be easy
severe sluice
(anyways i must go afk for around 15 minutes-)
rancid narwhal
(anyways i must go afk for around 15 minutes-)
ok bye
ok but it wasn't ur day
sorry for that
.close
lone heartBOT
lone heartBOT
Available help channel!
hollow pelican
guys i dont know how to solve this
lone heartBOT
guys i dont know how to solve this
hollow pelican
split cloud
can i interupt and ask what subject that is so i dont take it?
severe sluice
can i interupt and ask what subject that is so i dont take it?
integral calculus
you should take it @split cloud
split cloud
i might have too.. :/
hollow pelican
you messed up both of your integrals
wait how tho
severe sluice
wait how tho
for the first one, take a look at your u first
it's $\frac{x}{3}$, not $\frac{1}{3x}$
ocean sealBOT
Kanga Gang ¬Sam
hollow pelican
Ok
lone heartBOT
@hollow pelican Has your question been resolved?
lone heartBOT
@hollow pelican Has your question been resolved?
severe sluice
impudent, i think you should really retake algebra 1
lots of simple algebraic mistakes
@hollow pelican
raw shard
vague coral
pog
severe sluice
<:flonshed:629835071465062410>
uh sorry quantum but i dont really understand what this emoji means
raw shard
concern or something
severe sluice
i don't understand any of these...
raw shard
unpog
severe sluice
i didnt learn b4 algebra in my diploma
then take it
life will seriously be better if you to=ake algebra
take*
hollow pelican
life will seriously be better if you to=ake algebra
its still wrong damnn
severe sluice
at school or online
for a course like algebra 1 online is okay...
actually, you shouldn't be wasting your school time for that
maybe use khan academy: https://www.khanacademy.org/math/algebra
oak chasm
Yeah, Khan Academy is great for refreshers.
hollow pelican
Yeah, Khan Academy is great for refreshers.
ohh i see thanks
hollow pelican
Yeah, Khan Academy is great for refreshers.
i think i did smth wrong here but idk which part
oak chasm
You had (1 - x tan(u)) du/(2x). You then distributed the 1/(2x) to only one term.
Like if you have (a + b)c, you need ac + bc, but you have a + bc.
The c only got distributed to one term.
ocean sealBOT
Chai T. Rex
oak chasm
See how I distributed it to both terms?
hollow pelican
yea
ocean sealBOT
Chai T. Rex
hollow pelican
ohh ok
oak chasm
You're on the right track though.
Just use the sum rule.
Oh, I noticed you have a minus.
ocean sealBOT
Chai T. Rex
oak chasm
Use the sum rule to separate it into two integrals.
hollow pelican
i should be doing it this way?
oak chasm
Then you can use your substitution on the second one.
And the first one is easy to do.
hollow pelican
Ok
hollow pelican
Then you can use your substitution on the second one.
i need to times the 1/2x to x and the tanu also?
oak chasm
Not to both. That's only with addition in the parentheses: (a + b) c = ac + bc.
With (ab)c, you just have abc.
oak chasm
No problem.
hollow pelican
No problem.
for this question i have to make u cos theta?
oak chasm
You want to simplify the full square root argument.
So, 1 + cos(θ) would be better.
(for u)
And then when you get du, you get sin(θ) in it, which is the part outside the sqrt, which is nice.
severe sluice
impudent has a weird skillset :/
hollow pelican
And then when you get du, you get sin(θ) in it, which is the part outside the sq...
is it like this?
oak chasm
No, your reversal of the derivative power law is incorrect
uᵃ⁺¹ → (a + 1) uᵃ.
That's how the power rule works.
hollow pelican
oh yea
oak chasm
So, if you have an integral, you do the reverse.
hollow pelican
i forgot about the half on the denominator
oak chasm
(a + 1) uᵃ → uᵃ⁺¹
uᵃ → uᵃ⁺¹/(a + 1)
hollow pelican
okay
hollow pelican
(a + 1) uᵃ → uᵃ⁺¹
uᵃ → uᵃ⁺¹/(a + 1)
i dont know how to start for this question tho
oak chasm
Just pretend the constants are numbers.
ocean sealBOT
Chai T. Rex
oak chasm
Do the same thing except with the variables instead of numbers.
hollow pelican
Ok
oak chasm
Sorry, forgot the dx.
hollow pelican
Sorry, forgot the dx.
i need to change the limits for u also?
oak chasm
You can do that or you can do the indefinite integral separately and use that.
hollow pelican
You can do that or you can do the indefinite integral separately and use that.
ok im completely stuck im not sure i change the limits correctly anot
oak chasm
OK, so you had limits in terms of x.
0 and I₀.
What do you do to x to turn it into u?
@hollow pelican
hollow pelican
times u?
hollow pelican
yea
oak chasm
What gets done to the x to make it equal to u?
hollow pelican
divide by u?
oak chasm
No.
Look at the equation u = -T x.
Let's say you know x.
Let's say it's 5.
How do you get u?
hollow pelican
+T?
oak chasm
No, what does -T x mean to do with T and x?
hollow pelican
meaning -T times x?
oak chasm
Right!
So, to get u, you multiply x by -T.
u = -T x, so you multiply x by -T and you get u.
Does that make sense?
hollow pelican
yea
oak chasm
So, we found out how to change x to get u.
hollow pelican
yep
oak chasm
Now your original integration limits are xs. Change them to us using your tranforming method.
What do you do to x to change it to u?
hollow pelican
times -t?
hollow pelican
postive T*
oak chasm
What's 0 times -T?
hollow pelican
0
oak chasm
Right, so the new bottom limit is 0.
OK, now do that to I₀ to get the top u.
What do you get?
hollow pelican
-TIo
oak chasm
Good.
ocean sealBOT
Chai T. Rex
oak chasm
That way it has du with u limits instead of x limits.
hollow pelican
Ok
oak chasm
Oh, change the limits on your first line, too.
Because it's also du instead of dx, so it should use the u-based limits.
hollow pelican
okay
oak chasm
Now you know the limits for the bottom line.
The bottom line is also a bit simpler if you leave the constant in front of the integral also in front of the [...] part.
hollow pelican
is it like this?
ocean sealBOT
Chai T. Rex
oak chasm
If you leave the constant in front of the integral outside the []s, then you get something nicer:
ocean sealBOT
Chai T. Rex
hollow pelican
but how do i simplify it to become like this?
hollow pelican
1
ocean sealBOT
Chai T. Rex
hollow pelican
yea
ocean sealBOT
Chai T. Rex
hollow pelican
yea
oak chasm
OK, so we used to have the -1 multiplied by the fraction.
Now we're going to move it so that it's multiplied by the parentheses instead.
ocean sealBOT
Chai T. Rex
oak chasm
I distributed the -1 to all the terms inside, so they became the opposite signs.
Does that make sense?
Like if you multiply a number by -1, it changes the sign.
-1(a - b) = (-a + b)
hollow pelican
yea
hollow pelican
okay
ocean sealBOT
Chai T. Rex
oak chasm
And that's how they got it.
oak chasm
You're welcome. Don't forget the algebra Khan Academy course.
It'll teach things like that.
hollow pelican
okay
hollow pelican
You're welcome. Don't forget the algebra Khan Academy course.
i am not sure how to solve this question too
oak chasm
OK, so we need to get y = something with t in it.
hollow pelican
Ohh
ocean sealBOT
Chai T. Rex
ocean sealBOT
Chai T. Rex
oak chasm
And that way, you can get what y equals.
hollow pelican
I see
hollow pelican
And that way, you can get what y equals.
i have to make u sub , u=ln t ?
oak chasm
Well, what will du be?
hollow pelican
1/t
oak chasm
OK, so what will you have after the substitution?
hollow pelican
OK, so what will you have after the substitution?
so basically i have to choose the correct u to cancel either the numerator or denominator?
oak chasm
Well, you want to have the new integral be easy to solve.
If you can change one part into u² and the other part into 1, then the integral of u² is easy.
So, the idea is think of a substitution you want to try, do the substitution, see if the integral looks easy.
If it doesn't work or it looks hard, try a different substitution.
hollow pelican
Ohhh okay
oak chasm
One thing in addition.
You will have a function.
Like squaring there.
You'll make a substitution to simplify what's inside that function.
severe sluice
Well, you want to have the new integral be easy to solve.
more specifically, you want the derivative of u wrt respect to whatever variable you're integrating with to be present in your integrand (or easy to get)
oak chasm
And the substitution will also get rid of the old variable in all the rest of the expression.
So, if you pick ln(t) as u, that simplifies the part inside the squaring.
It also gets rid of the ts outside the squaring.
severe sluice
@oak chasm offtopic: wow that's a lot of boosting
buoyant kayak
more specifically, you want the derivative of u wrt respect to whatever variable...
wrtrt
oak chasm
@hollow pelican In an earlier problem, you had sqrt(1 - cos(θ)) in the bottom.
sqrt is the function.
You simplify the inside of sqrt with u = 1 - cos(θ).
hollow pelican
yea
oak chasm
And then du is -sin(θ) dθ.
And that gets rid of the sin(θ) in the numerator.
So, simplifying the stuff inside the function gets rid of the original variable everywhere else.
Here, simplifying the inside of the squaring with u = ln(t) gives you du = 1/t dt.
That gets rid of the original variable in the rest of the expresssion outside the squaring.
That's important.
If the original variable doesn't go away, that substitution won't work.
So that's the first rule.
hollow pelican
Okay
oak chasm
Then, if there are a few substitutions that work, you want the one that gives you the simplest integral to solve.
So, you picked u = cos(θ) on your original one, which worked.
But u = 1 + cos(θ) gave a simpler integral to solve.
hollow pelican
I see
oak chasm
Because it made it sqrt(u) instead of sqrt(1 + u).
So, that's the general idea.
Substitute to simplify inside a function.
Make sure it gets rid of the original variable outside the function.
Of the ones that get that far, pick the one that gives you the easiest integral to solve.
hollow pelican
Ok
hollow pelican
Of the ones that get that far, pick the one that gives you the easiest integral ...
is it like this?
oak chasm
Good, now combine the constants into one on the right side.
hollow pelican
I didnt know it can be any number
oak chasm
Yeah, when you do integration, the k or C or whatever you add in can be any number.
Because x² + 2 gives you the same derivative as x² + 1003049304.
So, integration does the derivative in reverse.
hollow pelican
I see
oak chasm
So, it should be able to give us both x² + 2 and x² + 1003049304.
So, that's what the k does.
It means "I can be any number".
That means that the k on the left and the k on the right can be different numbers.
So the way to handle it is this.
You don't know what k is on the left, you don't know what k is on the right.
When you do k - k, you don't know what number that'll be.
Does that make sense so far?
oak chasm
OK, do you have any questions?
hollow pelican
so is there a way to solve for the constant?
oak chasm
Yes, but you have to understand what I've said so far.
Does it make sense that if you don't know what k on the left or right are, then you don't know what number you get when you subtract them?
hollow pelican
Does it make sense that if you don't know what k on the left or right are, then ...
it doesn't
oak chasm
OK, let's say I subtract two numbers and I don't tell you what the numbers are.
Like I subtract 10 - 5.
You're not going to know that I get 5 as a result, right?
hollow pelican
yea
oak chasm
OK, so if you don't know what numbers I'm subtracting, I might get any number.
If I subtract 1003403434 - 56, I'll get a really high number.
If I subtract 5 - 394834, I'll get a negative number.
I could get any number I wanted if I picked the right starting numbers.
Does that make sense?
hollow pelican
yes
oak chasm
OK, so you don't know what k is on the y side.
You don't know what k is on the t side.
With me so far?
hollow pelican
yea
oak chasm
So, when you subtract the left side k from both sides, you'll have an unknown constant on the right side.
y + k = ... + k
Let's use different variable letters since they can have different values.
hollow pelican
Ok
oak chasm
y + m = ... + n
Then we subtract m from both sides.
y = ... + (n - m)
And we have no idea what n - m is, right?
hollow pelican
yep
oak chasm
So, it's just another unknown constant.
Just like the y + k constant.
Just like the constant on the t side.
Does that make sense?
hollow pelican
yes
oak chasm
And we write constants that could be anything as k.
So, it's y = ⅓ ln(t)³ + k.
Does that make sense?
hollow pelican
yea
oak chasm
OK, so we're almost done.
Now we have to find out what our new k is.
It passes through (1, 2) according to the problem, right?
hollow pelican
yep
oak chasm
So, when t is 1, y should be 2.
So, we fill those in to our equation.
y = ⅓ ln(t)³ + k
2 = ⅓ ln(1)³ + k
Now we can solve for k.
hollow pelican
Ok
oak chasm
@hollow pelican Unfortunately, I have a meeting to go to in a few minutes. If you get stuck, ask, and if no one answers for 15 minutes, ping @Helpers once.
hollow pelican
<@249469789578330133> Unfortunately, I have a meeting to go to in a few minutes....
Okay, thank you very much!!
oak chasm
You're welcome.
lone heartBOT
@hollow pelican Has your question been resolved?
karmic rapids
how would I find the limit of this recursive sequence? $$g_t = 0, t = 2^n, n \in \mathbb{N}0$$$$g_c = 1 + g{f(c)}, c \in \mathbb{N}$$
ocean sealBOT
illuminator3
lone heartBOT
@karmic rapids Has your question been resolved?
lone heartBOT
@karmic rapids Has your question been resolved?
lone heartBOT
@karmic rapids Has your question been resolved?
lone heartBOT
@karmic rapids Has your question been resolved?
rose pumice
how would I find the limit of this recursive sequence? $$g_t = 0, t = 2^n, n \in...
what's f?
lone heartBOT
@karmic rapids Has your question been resolved?
lapis valley
lapis valley
Why is dot product not the second one
jagged imp
why would it be the first?
lone heartBOT
@lapis valley Has your question been resolved?
lone heartBOT
lapis valley
.reopen
lone heartBOT
✅
lapis valley
why would it be the first?
It is the first one
jagged imp
again, why do you think its the first? As far as i can tell it should be the second.
lapis valley
again, why do you think its the first? As far as i can tell it should be the sec...
Because it is the vector equation for r1
lone heartBOT
@lapis valley Has your question been resolved?
lone heartBOT
lone heartBOT
Available help channel!
native void
I need help.
I'm in third year of highschool, and I'm having a exam after new year break, but I don't understand a single thing because I was sick. I'm also a generally bad student considering maths, so could anyone please help and explain this to me?
native void
I remember having to like change sin to cos or something like that, but I'm completely stuck
oh isnt this
wait
Stuck at the beginning x,d
i literally know nothing
i think i have to like
decide in what quadrant everything is
and then decide if its +/-
wdym
If you mean Izračunaj, that just means calculate it lol
i mean i can't really understand or write a cheatsheet without knowing the basics
and isn't a cheatsheet kinda illegal
i wanna either do it normally or fail normally like i have been til now x,d
oh
I was not in school
I was sick
I got a few notes i rewrote from my classmates
but i can't really see the similarity between this
Yeah we have offline
this is the entire homework
Devil grandma teacher ;-;
ocean sealBOT
ocean sealBOT
native void
uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuh
4
4pi/6 is
less than 1 so that's in the second quadrant
And sin is
positive?
there
so that'd be +4pi/6
ocean sealBOT
lone heartBOT
@native void Has your question been resolved?
native void
One second, I have to help mom with something and I'll be right back
Alright im back
So i don't understand, is this the same as deciding if its +/- by quadrants or
wait that was a dumb question
ocean sealBOT
native void
i get that much
Ohhh
then
?
-sin[4pi/6]?
oh yeah i got it mixed up with the previous thing
angle sum?
oh i see
Uhhhh
Well this is completely out of my reach
I don't want to be showed something while still being so clueless, so rather during this break I'll try learn what i can myself and come back and ask questions when i actually know what I'm doing
Seeing as how I'm just wasting yours and my own time lol
Thanks
.close
lone onyx
Hey guys, how do I find intersection point between 1/x and log2(x)?
woeful pulsar
hmm, looks like any closed form would likely be in Lambert-W
do you want to use Lambert W? or just find it numerically?
@lone onyx
lone onyx
actually, just proving that intersection exists is enough
woeful pulsar
ah, then get some intermediate value theorem
essentially we want to find an x such that 1/x-log2(x) is positive and an x so that it is negative
do you need to show intersection is unique?
lone onyx
yes
woeful pulsar
might want to differentiate and check the derivative to make sure
lime silo
<@&286206848099549185> im struggling with this problem cud anyone help me
fleet hearth
<@&286206848099549185> im struggling with this problem cud anyone help me
have you tried drawing it
severe sluice
<@&286206848099549185> im struggling with this problem cud anyone help me
off-topic: the question is weirdly simple...
clarification: some members have talked about the possibilities that terence tao is actually in this server.
lone heartBOT
@lime silo Has your question been resolved?
true viper
@lime silo I've sat here 10 minutes trying to rub my brain cells together to understand this and it's just not happeninng
severe sluice
<@!884818453737926687> I've sat here 10 minutes trying to rub my brain cells tog...
sum of ABD and CBD is........
true viper
100 degrees
severe sluice
and that is the measure of angle...
lone heartBOT
@lime silo Has your question been resolved?
lime silo
100 degrees
answer is 130 im not getting it
gray isle
show us your attempt in drawing a diagram
lime silo
this is my diagram
lone heartBOT
@lime silo Has your question been resolved?
lime silo
<@&286206848099549185> please help me
vale wigeon
where are your angle markings?
lime silo
havent written them
vale wigeon
then you do not have a complete diagram of the problem 
solid cargo
Actually im not really getting how acd will be 90...can someone help for any solution without trigo?
oak chasm
<@&286206848099549185> im struggling with this problem cud anyone help me
Simplify it by using the fact that moving it around and rotating it doesn't change it and the side lengths don't change it because all similar quadrilaterals will have the same angles, so set one of the side lengths to something easy.
Place B at (0, 0). Scale the figure to make BC = 1. Rotate it so that C is at (1, 0). AB = BC, so AB = 1. ∠ABC is 100°, so A is at (cos(100°), sin(100°)). Go from there.
jovial lava
.close
lone heartBOT
Available help channel!
vocal hawk
lone heartBOT
vocal hawk
my guess is that the answer should be no
cuz if there are two maximas, there should be a minima between them
and thus there can't be two maxima's with 1 minima, thus M > m +2 is impossible
is that right?
lone heartBOT
@vocal hawk Has your question been resolved?
echo socket
As an example you could give a cubic function
It can have just one (local) maxima and one (local) minima, which is what you need according to the problem
zenith pagoda
that doesnt seem right
it should have 3 maxima and 0 mimima, or else 4 maxima and 1 minimum
echo socket
Oh wait I thought m and M were values of the min/max points, my bad
zenith pagoda
it doesnt appear to be possible really
let's take a section of x² from [-1, 2]
it has a maximum at -1 and 2, and a minimum at 0
echo socket
Yeah it looks like it's impossible to me
zenith pagoda
now if you attempt to add a third maximum, where would it go?
noble sinew
It is possible - I explained 2 possible ways to construct such a function
zenith pagoda
ok and how many maxima does your function have
echo socket
I want to see a simple example of that
noble sinew
I can make it have 10000 local maximums and 0 local minimums if I want
shell widget
I read what u said scape, but isn't it true that for a extrema, the derivative has to exist and if it's not continuous, how is it differentiable? second of all, global extrema implies local extrema
zenith pagoda
I mean if you assume that an local is less than its neighborhood rather than less than or equal then yeah
noble sinew
I read what u said scape, but isn't it true that for a extrema, the derivative h...
Differentiability is a point-wise property
So is continuity
And you wouldn’t call a global min/max a local min/max would you?
zenith pagoda
wut
noble sinew
Wut?
zenith pagoda
a global extemum is also a local extremum isnt it
noble sinew
Depends on definition ig - but my other example def works
lone heartBOT
vocal hawk
.reopen
lone heartBOT
✅
vocal hawk
interesting, the thing is the question didn't say anything about continuity or differentiability
although the subject is called maxima and minima
noble sinew
If nothing mentioned then everything goes
vocal hawk
If nothing mentioned then everything goes
I guess yeah
placid zinc
Unless you're in a class on analysis, assume differentiable
vocal hawk
yeah this is just calc 1
noble sinew
Yeah no disagree
zenith pagoda
well if its a Smooth function then theyre probably asking you to prove its not possible
noble sinew
If nothing mentioned then everything goes
.
placid zinc
Oh, fair point, you might need your function to not be differentiable
high badger
Could someone please explain to me why for 14 do we need to add a negative? Like could I have just not added it and leave it like it is
lone heartBOT
Could someone please explain to me why for 14 do we need to add a negative? Like...
@high badger Has your question been resolved?
lone heartBOT
@high badger Has your question been resolved?
high badger
.close
vital steeple
hi
ocean sealBOT
tarık b.
I know that the series $\sum\limits_{k=0}^{\infty} \fraq{a_k}/{4^{2k}}$ with $a_k = 4$ if k is even and $a_k = 12$ if k is odd can be converted into $a_k = 8+(-1)^{k+1} \cdot 4$
```Compilation error:```! Undefined control sequence.
l.55 ...e series $\sum\limits_{k=0}^{\infty} \fraq
{a_k}/{4^{2k}}$ with $a_k ...
The control sequence at the end of the top line
of your error message was never \def'ed. If you have
misspelled it (e.g., `\hobx'), type `I' and the correct
spelling (e.g., `I\hbox'). Otherwise just continue,
and I'll forget about whatever was undefined.
Preview: Tightpage -1310720 -1310720 1310720 1310720
[1{/usr/local/texlive/2020/texmf-var/fonts/map/pdftex/updmap/pdftex.map}]```
tarık b.
vital steeple
Correction: I know that $a_k$ in the series $\sum\limits_{k=0}^{\infty} \fraq{a_k}/{4^{2k}}$ with $a_k = 4$ if k is even and $a_k = 12$ if k is odd can be converted into $a_k = 8+(-1)^{k+1} \cdot 4$. What I would like to learn is if there is a general rule to convert the alternating $a_k$ into a normal formula?
ocean sealBOT
tarık b.
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
lone heartBOT
@vital steeple Has your question been resolved?
vital steeple
.close
lone heartBOT
gray isle
is there more information
what have you tried?
why did you set those equal to each other
wdym they're on opposite sides
if you marked the specified segments on the diagram, one is clearly longer than the other
especially since segment AV is part of segment XV
yes
is
yes, x=9
runic obsidian
'
runic obsidian
im not sure if this is either C, or D
runic obsidian
tbh im not sure
im pretty sure its not a or b
glass lichen
what does it mean for a domain to be all of the real numbers?
runic obsidian
pretty sure its all positive numbers on the x axis?
runic obsidian
fuck
glass lichen
and that "all real numbers", includes every number on the real number line..?
runic obsidian
ah
glass lichen
thus the domain being all of R means that every single number can be put into the function, and have an output.
hence which function has a problem at a/some x values?
runic obsidian
B?
wintry prawn
Hi, this is a probably a lack of understanding of some geometry but can someone tell me why the following is a valid description of an ellipse?
wintry prawn
The author says that the major and minor axis are of form R(thetaA) and R(thetaB). The area is then the following
Area of ellipse = pi/4 * R^2 * thetaA * thetaB
I am confused as to how the axises of that ellipse can be described as R(thetaA) and R(thetaB).
I found the book, here is the exact wording:
<@&286206848099549185>
lone heartBOT
@wintry prawn Has your question been resolved?
wintry prawn
<@&286206848099549185>
hearty summit
@wintry prawn an ellipse is a slice of a cone at an angle. i can't tell exactly what is happening in the picture but you can see where the cone comes from
wintry prawn
yes I can see how the ellipse is formed but not how the radius R of the sphere can be related to its axises
OH wait the slant height R and the altitude of the cone would form a right triangle
hearty summit
i really need to know like what exactly a radiation sphere and the attenna beam is
wintry prawn
is that correct?
hearty summit
OH wait the slant height R and the altitude of the cone would form a right trian...
maybe? idk
wintry prawn
So the radiation sphere is just the energy distributed by the antenna and the beam is just where the energy in the spherical space is being focused by the antenna
but that shouldnt affect any of the parameters in the ellipses' equation
hearty summit
i think in this case the radius is just being used as how far away it is from the source
the more distance it travels
the bigger the ellipse
wintry prawn
hearty summit
so the foci would depend on R in someway
wintry prawn
Yeah so if this cone was part of the bigger cone (aka the beam)
and pretending that this cone's base was the slice of the bigger cone at a slant aka it forms the ellipse you said
and the slant height is S
how would the axises of the ellipse look?
I assume the major and minor axises could be calculated using two seperate right triangles and they would look like he said:
minor & major would be Rsin(theta)
But it still doesnt explain the 1/4 factor he got.
hearty summit
what is m in this case?
wintry prawn
oh thats the units meters^2
hearty summit
im not sure where the 1/4th is coming from either
lone heartBOT
@wintry prawn Has your question been resolved?
wintry prawn
<@&286206848099549185>
lone heartBOT
@wintry prawn Has your question been resolved?
lone heartBOT
@wintry prawn Has your question been resolved?
lone heartBOT
@wintry prawn Has your question been resolved?
lone heartBOT
turbid tulip
I have tried finding the curve of intersection to see if that would help, but it's not making it any clearer for me .
echo socket
Did you find the intersection tho?
turbid tulip
Did you find the intersection tho?
(-2, 2) I believe
echo socket
(-2, 2) as (x, y)?
Uh, haven't done triple integrals for a while, I'll think for a minute
turbid tulip
no, sorry, -2 <= x <= 2 and y = +/- sqrt(4-x^2)
echo socket
I think I got somewhere
So z definitly goes from 0 to 8 here
All we need to do if divide the omega 3d region into slices
Just like how the integral works
And up until z=4, those slices have form of x^2 + 2y^2 <= 4
turbid tulip
i absolutely suck at finding limits of integration
echo socket
Don't worry, it took me quite a while to get any intuition for them too
devout summit
Better to consider domain to integrate over as the disc x²+y²≤4
turbid tulip
I literally sketch them out as you should and I still struggle to come up with them
echo socket
Better to consider domain to integrate over as the disc x²+y²≤4
Shouldn't it be x²+2y²≤4?
devout summit
You need to find the volume of Omega correct?
echo socket
Omega is the region of the triple integral I think
turbid tulip
You need to find the volume of Omega correct?
The part I sent in is part of a bigger question, omega represents the domain in R^3
of yes, the region of a triple integral
but the integration is trivial
I just cannot form these limits
devout summit
What function f(x,y,z) do you have?
turbid tulip
sqrt(x^2+y^2)
devout summit
It would be much easier if you work with cylindrical coordinates
turbid tulip
I believe I am supposed to, but isn't that to do with calculating the integral itself rather than finding the limits?
devout summit
You would still have to write out the limits of integration but it would be in cylindrical coordinate form (but it would be much more Intuitive)
turbid tulip
right yeah that makes sense
turbid tulip
You would still have to write out the limits of integration but it would be in c...
Still unable to form the limits. Not sure what to do specifically
<@&286206848099549185>
devout summit
I suppose it should look like this:\$\int_0^{2\pi}\int_0^2\int_{r^2+r^2\sin^2(\theta)}^{8-r^2\cos^2(\theta)}r\sqrt{r^2},dz,dr,d\theta$
turbid tulip
oh jeez
devout summit
Wait no its wrong
ocean sealBOT
Euclid31415
devout summit
Integrand is just r^2
turbid tulip
how would I go about finding r and theta to be able to evaluate the definite integral?
devout summit
The sqrt(r^2) is just the sqrt(x^2+y^2) part and an extra r is due to jacobian determinant
devout summit
how would I go about finding r and theta to be able to evaluate the definite int...
You know how to evaluate triple integrals right? Starting with the inner part integrating wrt z then r and then theta
turbid tulip
yeah I'm fine with that part
devout summit
how would I go about finding r and theta to be able to evaluate the definite int...
r and theta are the variables we are integrating with respect to
turbid tulip
r and theta are the variables we are integrating with respect to
I was wrong, I am unsure. Which integral corresponds to which function
I keep getting 0
devout summit
In cylindrical coordinates x is replaced by rcos(theta) and y by rsin(theta). The functions r^2+r^2sin^2(theta) and 8-r^2cos^2(theta) are just from the original functions z=x^2+2y^2 and z=8-x^2
turbid tulip
In cylindrical coordinates x is replaced by rcos(theta) and y by rsin(theta). Th...
don't know how to use latex so here
is this right?
oh hold on
nope, still
devout summit
First of all, they are dz dr and d(theta). Secondly, z varies from r^2+r^2sin^2(theta) to 8-r^2cos^2(theta)
turbid tulip
First of all, they are dz dr and d(theta). Secondly, z varies from r^2+r^2sin^2(...
yes my apologies on those limits
turbid tulip
First of all, they are dz dr and d(theta). Secondly, z varies from r^2+r^2sin^2(...
integrate(integrate(integrate(r*sqrt(x**2+y**2),(z,r**2+r**2*sin**2(t),8-r**2*cos**2(t))),(r,0,2)),(t,0,2pi))
I've resorted to sympy
because I cannot seem to do it by hand, and it still doesn't work
devout summit
**Euclid31415**
Observe that sqrt(x^2+y^2) turned to sqrt(r^2) according to the cylindrical coordinates. Which means the integrand is basically r*r=r^2
turbid tulip
I see
that still doesn't explain sympy's inability to calculate it, or am I being dense?
devout summit
` integrate(integrate(integrate(r*sqrt(x**2+y**2),(z,r**2+r**2*sin**2(t),8-r**2*...
How would it be able to calculate if you write the integrand in terms of x,y after using cylindrical coordinates
turbid tulip
integrate(integrate(integrate(r**2,(z,r**2+r**2*sin**2(t),8-r**2*cos**2(t))),(r,0,2)),(t,0,2pi))
It's in terms of dz dr and dt now
devout summit
I don't know how integration works in sympy, but the bracketing seems off. And maybe it does not know how to deal with variable upper and lower limits
turbid tulip
I don't know how integration works in sympy, but the bracketing seems off. And m...
I think I have it now
devout summit
Nope. Integrand is r^2 not r^3
devout summit
Looks promising
turbid tulip
Still have little to no idea how those limits were constructed fully. There is another question in this textbook that's turning my stomach
turbid tulip
@devout summit do you have a general way that you, personally construct limits? Do you sketch?
devout summit
Still have little to no idea how those limits were constructed fully. There is a...
Well, yes sketching is a good way.
The 0 to 2 limits for r and 0 to 2pi limits for theta are just due to integrating over a disc of radius 2 centered at origin. (Projection of the region we are integrating over on the xy plane is a disc)
devout summit
The curve z=8-x^2 bounds the region from above and z=2y^2+x^2 from below.
So, z ranges from 2y^2+x^2 to 8-x^2. sin, cos and r is due to conversion to cylindrical coordinates.
turbid tulip
So, z ranges from 2y^2+x^2 to 8-x^2. sin, cos and r is due to conversion to cyli...
That makes sense. Now it's time for spherical coordinates!
.close
dusky osprey
lone heartBOT
shell widget
try drawing a diagram first
dusky osprey
i drew a diagram but i dont think its right
shell widget
show it
shell widget
can u find the radius or diameter given the area of the circle?
you should be able to
dusky osprey
i tried rearranging pie r squared to find r on its own
but idk if thats what ur supposed to do
shell widget
yeah you have 49 = pi r^2
dusky osprey
yea
dusky osprey
ocean sealBOT
shell widget
ok so r = 3.95
shell widget
sec lemme draw a diagram too
dusky osprey
i only know r tho
dusky osprey
why is it x/2??
shell widget
(x/2)^2 + (x/2)^2 is just like a + a, so you have 2a.
Or 2[(x/2)]^2
so 15.6 = 2[(x/2)]^2
dusky osprey
4x/4 *
shell widget
so you have (15.6/2) = (x/2)^2
7.8 = (x/2)^2
then take square root on both sides
you have sqrt(7.8) = (x/2)
Then multiply both sides by 2
so you have x = 2sqrt(7.8)
alpine sable
shell widget
,w solve (3.95)^2 = (x/2)^2 + (x/2)^2
ocean sealBOT
dusky osprey
You both have done if different ways
shell widget
hmm
shell widget
,w sqrt(31.2)
ocean sealBOT
dusky osprey
I think i’ve got it now
shell widget
yeah
dusky osprey
How would you know how that diagram would look like tho
I just wouldn’t picture that
dusky osprey
ya i did that but then like how u linked the diagram to pythagoras
I just would literally never think of that
shell widget
well now u know
alpine sable
Calculate the value of k if the graph of f(x) = kx2 passes through the point T(-8; -24).
severe sluice
Calculate the value of k if the graph of f(x) = kx2 passes through the point T(-...
kx^2?
i mean
$kx^2$
ocean sealBOT
:rice_scene: Gang ¬Sam
alpine sable
here orginal photo
echo socket
Calculate the value of k if the graph of f(x) = kx2 passes through the point T(-...
If you meant kx^2, then you need to plug in x=-8 and y=-24 into y=kx^2
So we have -24 = k*64
crude quartz
If you meant kx^2, then you need to plug in x=-8 and y=-24 into y=kx^2
Well this is still true even if he meant $2kx$
echo socket
k = - 3/8
ocean sealBOT
Siupa
severe sluice
He meant that the method would still be the same
oh-
ocean sealBOT
Touch Her Beans
upper ravine
lone heartBOT
upper ravine
what integral rules are used here?
which ones do i need to learn for this
https://www.youtube.com/watch?v=o75AqTInKDU im watching this rn bc i dont remember enuf about integrals
but would this be enuf
velvet magnet
all you need is the integral for dx/x and how to substitute limits of integration
i dont see anything else related to integration being used here
velvet magnet
no, substitution is a different thing
it is mostly used in indefinite integration
substituting limits is the basis of definite integration
proper ibex
why does mathematica not evaluate function properly
lone heartBOT
why does mathematica not evaluate function properly
proper ibex
ok so it looks like I had to put a dot after one of the numbers, however I do not get why that is
lone heartBOT
@proper ibex Has your question been resolved?
lone heartBOT
@proper ibex Has your question been resolved?
bleak ridge
It's prob has to do with the decimal data type or something
Because it is a programming language
opaque rapids
can someone help me prove the definition of hyperbola |d1-d2|=2a for any point in the hyperbola the distances for the forci are constant
vale wigeon
this is a definition
opaque rapids
A hyperbola is the set of all points in a plane such that the difference of the distances from two fixed points (foci) is constant.
opaque rapids
|d1-d2|=2a
and i need to prove it
vale wigeon
what do you mean
vale wigeon
you just stated the definition once in symbols and once more in words
opaque rapids
see in the url
vale wigeon
you just stated the definition once in symbols and once more in words
vale wigeon
what?
vale wigeon
i don't get what you're talking about at all.
it sounds as if you want to prove that a hyperbola is a hyperbola.
opaque rapids
this thing
opaque rapids
i don't get what you're talking about at all.
kind of
severe sluice
if you write this in symbols, you get |d1-d2|=2a
d1 is distance from 1st focus
d2 is distance from 2nd focus
vale wigeon
i would argue that the symbolic version could be made more precise
severe sluice
2a is the constant
vale wigeon
but taking sam's clarification into account, it becomes perfectly clear
severe sluice
although it can be a as well...
opaque rapids
if you write this in symbols, you get |d1-d2|=2a
thats right
but what is the prove
severe sluice
it's THE DEFINITION
vale wigeon
the proof of WHAT???
vale wigeon
are you asking why it's 2a and not just a?
severe sluice
argh
vale wigeon
no, sorry.
severe sluice
it's the definition-
opaque rapids
i dont realy know what it means
severe sluice
(wait why am i posting this while ann is here? is this offensive to her?)
vale wigeon
no, it's ok.
opaque rapids
my proffesor just gave
a project and told me
prove the defintion of the thyperbola |d1-d2|=2a
which is const
vale wigeon
they really said exactly that?
severe sluice
again
*please read and understand what i am saying. it is literally the definition, there is no proof, it is the DEFINITION
severe sluice
vale wigeon
do you have the instructions EXACTLY as your prof stated them to you?
opaque rapids
why d1-d2 looks like that is hyperbola
vale wigeon
and i mean EXACTLY even if it's in another language
severe sluice
why d1-d2 looks like that is hyperbola
uh, because, definition.
opaque rapids
he told it verbally but that was exactly and i am realy broke my head about it
vale wigeon
so
opaque rapids
when i gave this solution
vale wigeon
there is a possibility you are misremembering what you were instructed to do
opaque rapids
he told me its good but i need
vale wigeon
talk to your professor again
opaque rapids
this is a good soultion in which i am going too 2a
but he wants it for every point
i mean
d1= when i take point for the hyperbola
severe sluice
you claim a statement is true at the start of this thing
at the end, you say that that statement is true
this doesn't prove anything
argh
opaque rapids
i need to
severe sluice
who is your prof
opaque rapids
i need to true statment
severe sluice
and why did he write this question
opaque rapids
to reach true statment
severe sluice
and why did he accept your thing
opaque rapids
d1=sqr(y^2+(x+c)^2), d2=sqr(y^2+(x-c)^2)
d1-d2=const- is there a way to prove that it will be const ?
and the const will be 2a
thats what i mean
echo socket
x^2 - y^2 = a^2
y^2 = x^2 - a^2, sub this into the equation
vale wigeon
vegeta you're misunderstanding things massively here...
severe sluice
d1-d2=const- is there a way to prove that it will be const ?
you claim that the statement is true
then use confusing distance formulas (okay, not that confusing, but it's not necessary)
then come up with an equation which literally represent "sum of the distances is constant"
opaque rapids
you claim that the statement is true
then use confusing distance formulas (okay...
thats true, but i dont come up with an equation which literally tells me that it is constant
severe sluice
bruh
opaque rapids
d1=sqr(y^2+(x+c)^2), d2=sqr(y^2+(x-c)^2)
if i do d1-d2 how can i say it will be const
severe sluice
2a is constant
opaque rapids
i understant that already
but how can i say it will be const
when it looks like that d1=sqr(y^2+(x+c)^2), d2=sqr(y^2+(x-c)^2)
severe sluice
d1-d2 is constant ACCORDING TO THE CLAIM YOU MADE AT THE FIRST SENTENCE OF YOUR PSEUDO-PROOF
THAT YOU USE TO DERIVE THE REST OF THE PROOF
proper ibex
why are people screaming
severe sluice
why are people screaming
"prove the definition"
proper ibex
I mean you could always link them to one of those books where they take 200 pages to prove 1+1=2
if that's what they want
severe sluice
I mean you could always link them to one of those books where they take 200 page...
but "prove the definition"
argh
opaque rapids
man i am more than 3 hours trying to to figure out this thing
severe sluice
man i am more than 3 hours trying to to figure out this thing
okay, i suggest that you stop and call your prof, saying that you can't prove a definition
proper ibex
or say that the definition of hyperbola covers the required proof
it is up to prof to decide what works
lone heartBOT
@opaque rapids Has your question been resolved?
plucky storm
do you mean that "with the definition of a hyperbola derive an equation for it"?
@opaque rapids
severe sluice
<@&286206848099549185> can anybody teach this @opaque rapids guy the principle of explosion