um
#help-49
quasi mauve
somber folio
I forgot, and therefore corrected it afterwards.
fallow scarab
!15m
quasi mauve
is that -cos(x+1) ?
midnight plankBOT
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quasi mauve
@somber folio is that -cos(x+1) ?
somber folio
No.
jolly pike
@somber folio multiply and divide the lhs by sinx-cosx+1
secx+tanx=1+sinx/cosx
try to make the left hand side that
quasi mauve
multiply 2 side by sinx - cosx + 1 ?
frozen finch
WHAT
frozen finch
whats A?
somber folio
Not very helpful.
dusty portal
Not very helpful.
You need to conjugate the denominator, essentially.
quasi mauve
multiply cosx on numerator
that means you need to mutiply cosx on 2 sides
jolly pike
yeah
on rhs u get 1+sinx
u want to manipulate the left hand side
y isnt the @somber folio not replying
midnight plankBOT
@somber folio Has your question been resolved?
quasi mauve
on rhs u get 1+sinx
how ?
wait
sec x = 1/cosx
tan x = sinx/cosx
oh ok
@jolly pike whats your answer ?
somber folio
Not very helpful.
somber folio
You need to conjugate the denominator, essentially.
Let me.
gaunt nimbus
blud provided evidence 🥀
somber folio
My teacher said it coudn't be done by multiplying the numerator by its conjugate.
I shall never trust him.
Thanks y'all, especially joshuva (and Nguyen Hoang Bach).
.close
midnight plankBOT
lyric charm
!nosols
midnight plankBOT
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.
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somber folio
As a helper, please do not give out answers that could be copied as a homework s...
.:this
silent magnet
Hello, I hope someone can help me with this assignment
midnight plankBOT
Hello, I hope someone can help me with this assignment
silent magnet
This is what I have gotten to.
The assignment has to be done without CAS or calculator etc. Yet I had to look up that phi(2)=0.9772, so Im not sure if I have shown correctly as the assignment wanted. But to my understand there's no way to actually calculate the probability 0.9545 without a calculator, or without looking it up in a table?
brittle grotto
it's because you're going to have some integral of $e^{-x^2}$ sort of form, and there's no nice closed-form representation of that over some arbitrary interval
grand pondBOT
00100000
silent magnet
The normal outcome is 0.9545, that's just an accepted fact that we're told and it says so in our books
brittle grotto
which makes it really hard to calculate by hand
but we have nice tools to transform our distributions to normal dsitributions with mean 0 variance 1, so we have tables of this
and also calculators can just use numerical tools to approximate the value well
silent magnet
yes, so that's why im asking. I THINK I have done it correctlt and that maybe the assignment is just not worded so well? Because it kinda says that I have to SHOW that the probability = 0.9545
brittle grotto
well, the probability does not equal 0.9545
that's almost certainly a false statement
it's approximately that probably
but I doubt it's exactly that
your work looks good to me 👍
silent magnet
In the interval from mean value - 2 x variance to the mean value +2 x variance, the area under the graph is indeed 0.9545. This is the normal outcomes.
brittle grotto
that's probably not true. it's probably not a rational number
silent magnet
This is true. it covers most of the area beneath graph for the bell curve for the normal distribution
more precisely it covers 95.45% of the area.
silent magnet
but I doubt it's exactly that
okay I understand what you mean 😉
brittle grotto
silent magnet
sure, you're right, it's not exactly that. And it's not so important either But yes you are correct.
brittle grotto
note that this integral is an equivalent expression for this
brittle grotto
sure, you're right, it's not exactly that. And it's not so important either But ...
oh, it's good that we're on the same page. my apologies, I thought you were under the impression that the exact value was that
silent magnet
Yeah the first equation you have we call in my language the "denseness function" and the second one the "distribution function". I dont the english terms
brittle grotto
"=" is a very strong symbol to use usually
you would generally only use it for an exact value
brittle grotto
Yeah the first equation you have we call in my language the "denseness function"...
probability density function and cumulative distribution function, respectively
sometimes shortened to "pdf" and "cdf"
silent magnet
ya, it's just a little besides the point I think, while technically not wrong. It's always good to remember, so as to not claim otherwise.
look here form our collection of formulas:
Stated as 95.45%; its not really the point whether it is EXACTLY precise. So yeah it is a close approximation.
Thanks for the help!
🙏
.close
midnight plankBOT
brittle grotto
there's just a delay to allow time for reopening, in the case of accidental closure
lyric charm
there's a grace period for reopens and also bc rate limiting
silent magnet
sorry
lyric charm
just leave it alone, it'll close by itself in 10 min or so
brittle grotto
no worries
silent magnet
ty
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safe oriole
$\int_{-a}^a\int_{-\frac ba\sqrt{a^2-x^2}}^{\frac ba\sqrt{a^2-x^2}}1\dd y\dd x=\int_{-a}^a\frac{2b}a\sqrt{a^2-x^2}\dd x$
midnight plankBOT
$\int_{-a}^a\int_{-\frac ba\sqrt{a^2-x^2}}^{\frac ba\sqrt{a^2-x^2}}1\dd y\dd x=\...
grand pondBOT
pirateking0723
safe oriole
now if i want to try to do something like $x=a\cos\theta$, will the bounds of integration be from $0$ to $\pi$ or from $-\pi$ to $\pi$
grand pondBOT
pirateking0723
safe oriole
if it is the latter then i know that i cant do this sub since in cos isnt bijective on the interval [-π,π] and so i cant use it as a sub i was informed earlier
if it is the former then i can certainly use it since cos is bijective on [0,π]
so what would the bounds be and why
safe oriole
0 to pi?
well i am not convinced that it is from 0 to pi
but when i tried it i got πab as the answer of the integral
which is the area of the ellipse
wait let me show you
gaunt nimbus
Is this not an ellipse tho?
safe oriole
**pirateking0723**
so from here this is $=\frac{2b}a\int_{\pi}^0-a\sin\theta\sqrt{a^2-a^2\cos^2\theta}\dd\theta=2ab\int_0^{\pi}\sin^2\theta\dd\theta=ab\int_0^{\pi}1-\cos(2\theta)\dd\theta=\pi ab$
safe oriole
Is this not an ellipse tho?
it is
grand pondBOT
pirateking0723
gaunt nimbus
So what’s wrong the integral?
safe oriole
i am not sure that i understand your question
safe oriole
**pirateking0723**
this integral is correct no ?
gaunt nimbus
Yes
safe oriole
**pirateking0723**
alright so then comes this
gaunt nimbus
Me being lazy I would replace the integrand with just a lol
gaunt nimbus
so it is the area enclosed by the ellipse of equation x^2/a^2+y^2/b^2=1
Yup
safe oriole
**pirateking0723**
alright so then whats wrong about this
gaunt nimbus
,w int from 0 to pi (1 - cos (2 theta))dtheta
grand pondBOT
safe oriole
this is crazy right ?
gaunt nimbus
Seems to be no mistake?
safe oriole
because i thought that the bounds should be from -pi to pi ?
why are the bounds from 0 to pi and not from -pi to pi
safe oriole
[-1,1]
gaunt nimbus
U would go from (-1,0) back to (0,0) and to (-1,0) again
safe oriole
the same as a [0,pi]
gaunt nimbus
That’s the whole circle!
safe oriole
but you traverse it 2 times
gaunt nimbus
So the area is doubled
safe oriole
That’s the whole circle!
well thats what we want
we want the area enclosed by the whole ellipse no ?
gaunt nimbus
You’ve already accounted for that
safe oriole
which is from -pi to 0 then to pi
gaunt nimbus
**pirateking0723**
safe oriole
from 0 to pi is only the top half unless i am tripping
safe oriole
ah i see
gaunt nimbus
Let’s say u do smth different
safe oriole
first i integrated wrt y from the top of the ellipse to its top
gaunt nimbus
first i integrated wrt y from the top of the ellipse to its top
Yes
safe oriole
after that i only need half a rotation and not a full one
i see
give me a sec i might have another doubt
i want to check something rq
gaunt nimbus
aight
safe oriole
if i instead try the sub x=asinθ then the bounds would be from -pi/2 to pi/2 right?
safe oriole
then so does -pi/2 to pi/2 right ?
gaunt nimbus
,w integral from 0 to pi of [(acosx sqrt{2^2 - 2^2sin^2x})] dx
grand pondBOT
midnight snow
safe oriole
0 to pi doesnt work
this should work too
gaunt nimbus
thats what i thought...
i mean
sin is always positive there
however
the added substitution
acos(theta)
is an asshole
and ruins my integral
safe oriole
$2ab\int_0^{\pi}cos^2\theta\dd\theta=ab\int_0^{\pi}1+\cos(2\theta)\dd\theta=\pi ab$
grand pondBOT
pirateking0723
safe oriole
**pirateking0723**
so start here, next do x=asinθ
set the boundaries to be 0 and pi instead of -pi/2 to pi/2
dx=acosθdθ
ah wait
gaunt nimbus
safe oriole
$\sqrt{a^2-a^2\sin^2\theta}=a\abs{\cos\theta}\neq a\cos\theta$ over the whole interval $[0,\pi]$
grand pondBOT
pirateking0723
safe oriole
bruh
safe oriole
one more thing
cos is periodic of period 2pi so the above integral over any interval of the same length with "distance" 2pi from the other interval is the same ?
gaunt nimbus
Yes
It can be 2pi to 3pi, 4pi to 5pi, etc
Eventually
During the evaluation step
Cos(3pi) = cos(pi) anyway
I’ve seen an integral problem involving periodicity of sinx
It asks for smth like
I(n) = $\int_{-\pi n}^{\pi n} |\sin x| dx$ for $n \in \mathbb{Z}$
grand pondBOT
k
midnight plankBOT
@safe oriole Has your question been resolved?
safe oriole
ohhh wait let me think about this for a second
safe oriole
$\int_{-\pi n}^{\pi n}\abs{\sin x}\dd x=\int_{-\pi}^{\pi}n\abs{\sin(nx)}\dd x=2n\int_0^{\pi}\abs{\sin(nx)}\dd x\=\begin{cases}2n\int_0^{\pi}\sin(nx)\dd x,\text{if}\ n=2k+1, k\in\mathbb{Z}\-2n\int_0^{\pi}\sin(nx)\dd x,\text{if}\ n=2k, k\in\mathbb{Z}\end{cases}\=\begin{cases}4,\text{if}\ n=2k+1, k\in\mathbb{Z}\0,\text{if}\ n=2k, k\in\mathbb{Z}\end{cases}$
let me recheck my work
grand pondBOT
pirateking0723
safe oriole
alright this is my final answer
midnight plankBOT
@safe oriole Has your question been resolved?
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@safe oriole Has your question been resolved?
tidal turret
midnight plankBOT
tidal turret
can i get some help with 4
let f(x) = 7cos(x) + 4, find Im(f) and thr maximums in the interval [π,3π]
🤔
night hawk
let f(x) = 7cos(x) + 4, find Im(f) and thr maximums in the interval [π,3π]
The function looks entirely real to me...
-# I'm assuming Im(a + bi) = b.
tidal turret
wdym?
blissful pier
Im(f)=image of f
tidal turret
The function looks entirely real to me...
-# I'm assuming `Im(a + bi) = b`.
image i mean
not imaginary part
night hawk
I recommend first finding the maximums and the minimums of f(x).
tidal turret
,w derivative of 7cos(x) + 4
grand pondBOT
tidal turret
,w -7sin(x) = 0
grand pondBOT
tidal turret
I recommend first finding the maximums and the minimums of `f(x)`.
how?
night hawk
how?
$$x = \pi n$$
Use this to find possible critical values on $x \in [\pi, 3\pi]$, then test for their $f(x)$ values.
grand pondBOT
@night hawk
tidal turret
$$x = \pi n$$
Use this to find possible critical values on $x \in [\pi, 3\pi]$, ...
n is within [1,3]
night hawk
n is within [1,3]
So find $f(\pi n)$ for all $n \in [1, 3] \cap \bZ$.
grand pondBOT
@night hawk
tropic oyster
$$yeah$$
grand pondBOT
girlyboss90
tidal turret
So find $f(\pi n)$ for all $n \in [1, 3] \cap \bZ$.
mate cos(x) image is [-1,1]
-1 <= cos(x) <= 1
-7 <= 7cos(x) <= 7
-7 +4 <= 7cos(x) + 4 <= 7 + 4
@night hawk
night hawk
-7 +4 <= 7cos(x) + 4 <= 7 + 4
I assumed you were doing it the Calculus way because you were differentiating the function and setting that derivative equal to 0.
tidal turret
-1 <= cos(x) <= 1
-7 <= 7cos(x) <= 7
-7 +4 <= 7cos(x) + 4 <= 7 + 4
-3 <= 7cos(x) + 4 <= 11
f(x) = 7cos(x) + 4
Imagen(f) = [-3,11]
tidal turret
I assumed you were doing it the Calculus way because you were differentiating th...
well, inquality is mlre elegant
and faster
and cosine has a known image, so this inequality i think is valid unless I made some mistake
night hawk
,w range of y = 7cos(x) + 4, pi <=x <= 3pi
grand pondBOT
tidal turret
,w range 7cos(x) + 4
grand pondBOT
tidal turret
,w range of y = 7cos(x) + 4, pi <=x <= 3pi
how do I find the maximums?
night hawk
how do I find the maximums?
night hawk
can you explain with words
f(2pi) = ?
f(3pi) = ?```well
tidal turret
```f(pi) = ?
f(2pi) = ?
f(3pi) = ?```
first, one second
f'(x) = -7sin(x)
,w sin(x)=0
grand pondBOT
tidal turret
```f(pi) = ?
f(2pi) = ?
f(3pi) = ?```
how do you know sin(π)=0 , sin(2π) = 0, sin(3π)=0
night hawk
how do you know sin(π)=0 , sin(2π) = 0, sin(3π)=0
You would have to use the Unit Circle.
tidal turret
You would have to use the Unit Circle.
no
we know sin(0)=0
is there any identities we can use
night hawk
is there any identities we can use
I guess you could use analyze how each cycle of the sin wave works. But it's overkill compared to memorization or the Unit Circle.
night hawk
tidal turret
sin(0)=0
sin(π)=0
@night hawk
k ∈ Z
π <= 0 + 2kπ <= 3π
π <= 2kπ <= 3π
1 <= 2k <= 3
1/2 <= k <= 3/2
k=1
@night hawk
,w maximums of 7cos(x)+4 in [pi,3pi]
grand pondBOT
Requested by 938c2cc0dcc05f2b68c4287040cfcf71
tidal turret
,w Maximize 7cos(x) + 4 on the interval [pi, 3pi]
grand pondBOT
tidal turret
,w Find all local maxima of 7cos(x) + 4 between pi and 3pi
grand pondBOT
tidal turret
,w critical points f(x)=7cos(x)+4 where x in [pi,3pi]
grand pondBOT
tidal turret
,w Minimize 7cos(x) + 4 on the interval [pi, 3pi]
grand pondBOT
tidal turret
,w Find the global maximum and minimum of 7cos(x) + 4 on [pi, 3pi]
grand pondBOT
tidal turret
.close
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viral dagger
let PA and PB be tangents to a circle with center O, where A and B are on the circle, a line from P intersects the circle at C and D, and intersects AB at Q, prove that PQ^2=PC×PD-QC×QD
midnight plankBOT
let PA and PB be tangents to a circle with center O, where A and B are on the ci...
viral dagger
im guessing your supposed to play with poap somehow?
midnight plankBOT
@viral dagger Has your question been resolved?
lethal path
im guessing your supposed to play with poap somehow?
it's definitely a combination of QC * QD = AQ * QB
PC * PD = PA^2 = PB^2 and so on
viral dagger
yeah but i feel like you need to change PQ^2 to something for it to fall in place
viral dagger
.close gtg
molten bay
midnight plankBOT
molten bay
here by derivative i got 2x+b=0 so b=-2x and x=-b/2
and by condition we can assume that it is symmetric about line x=2
lyric charm
so you can find b.
lyric charm
no need at all
you only need to compare values of f at various points against each other
molten bay
Ohh
f(1) will be
1-4+c =-3+C
f(2) will be 4-8+C=-4+C
Not understanding
Depends on the sign of c because c can be zero positive and negative
woeful turret
That won't matter
molten bay
you only need to compare values of f at various points against each other
How can I compare?
woeful turret
2>1
2+c>1+c
woeful turret
So u understand by adding or subtracting a constant the inequality doesn't change?
woeful turret
2>1
if C is negative
Suppose C=-10
2-10>1-10
-8>-9
Yes..-8>-9
molten bay
Hmm
lyric charm
2>1
if C is negative
Suppose C=-10
2-10>1-10
-8>-9
yes, comparisons survive subtraction of same number from both sides
midnight plankBOT
@molten bay Has your question been resolved?
thorn mirage
aide grand oral maths
midnight plankBOT
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twilit field
Just to be sure. The way to do this is
$\lambda(x,y)=(-3y,x)$ and I attempet to solve that , right
midnight plankBOT
Just to be sure. The way to do this is
$\lambda(x,y)=(-3y,x)$ and I attempet to...
grand pondBOT
wai
grand pondBOT
wai
twilit field
which gives \lambda^2=-3, which doesn't make much sense...
novel sedge
.. why?
twilit field
yes, I meant from R^n
novel sedge
well thats simply not true
for example a rotation by any non-multiple of pi will not have real eigenvalues
twilit field
so the eigenvalue is (3i)?
novel sedge
for example a rotation by any non-multiple of pi will not have real eigenvalues
you can see this easily geometrically, since no vector will be parallel to its image after rotation
quiet hinge
how did you get \lambda = 3i from \lambda^2 = -3??
twilit field
how did you get \lambda = 3i from \lambda^2 = -3??
I took the root of both sides?
novel sedge
apparently you didnt
because it should be \lambda = \pm\sqrt{3}i 😄
you will get two solutions
twilit field
right, but none of the eigenvalues are real?
novel sedge
sure
novel sedge
so the eigenvalue is (3i)?
this is still wrong, though
twilit field
this is still wrong, though
eigenvalues are ±3i
novel sedge
..., you just agreed they are ±\sqrt{3}i
twilit field
this is still wrong, though
why
novel sedge
because \lambda = ±3i implies \lambda^2 = -9
twilit field
oops
novel sedge
that is not the original equation... 😄
twilit field
$\lambda = ±√3 i$
grand pondBOT
wai
novel sedge
yes
twilit field
Okay. Thanks
novel sedge
there are no real eigenvalues for this
twilit field
I guess the more correct way to do it would be to show T -\lambda I is injective for all \lambda
no?
hmm
,w solve -3y=ax; x=ay for a
grand pondBOT
twilit field
.close
midnight plankBOT
novel sedge
but keep in mind- the transformation T itself will not have any eigenvalues, since the scaling factor must be from R (by definition of eigenvalue). so the final answer is- there are no eignevalues
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quiet hinge
but keep in mind- the transformation T itself will not have any eigenvalues, sin...
read this @twilit field . i also found the same question here https://math.stackexchange.com/questions/1987042/find-eigenvalues-of-tx-y-3y-x
midnight plankBOT
read this <@796939110815629322> . i also found the same question here <https://m...
quiet hinge
.close
midnight plankBOT
twilit field
read this <@796939110815629322> . i also found the same question here <https://m...
Thanks!
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steel rivet
If the direction of velocity is along acceleration then velocity will increase in the that direction. why?
midnight plankBOT
If the direction of velocity is along acceleration then velocity will increase i...
steel rivet
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dusty portal
$\int_1^{\infty}\qty(\frac{\ln(x)}{x})^{2024}dx$
midnight plankBOT
$\int_1^{\infty}\qty(\frac{\ln(x)}{x})^{2024}dx$
grand pondBOT
;(
dusty portal
\begin{align*}=\int_1^{\infty}\frac{(\ln(x))^{2024}}{x\cdot x^{2023}}dx\ \overset{u=\ln(x)}{=}\int_0^{\infty}u^{2024}e^{-2023u}du\ \overset{v=2023u}{=}\frac1{2023}\int_0^{\infty}\qty(\frac{u}{2023})^{2024}e^{-u}du\ =\frac1{2023^{2025}}\Gamma(2025)\ =\boxed{\frac{2024!}{2023^{2025}}}. \end{align*}Is this correct?
twilit field
,w integrate (ln(x)/x)^{2024} from 1 to infty
dusty portal
,w integrate (ln(x)/x)^{2024} from 1 to infty
1 to infinity
but still, doesn't work
grand pondBOT
Requested by math_rocks
twilit field
It can't calculate it lol
I think It's right
dusty portal
Ok
twilit field
why gamma function though
twilit field
unnessary (I don;t now it, hence I hate it)
grand pondBOT
;(
fallow scarab
use \newline
dusty portal
use `\newline`
I should have used \begin{align*} but whatever
fallow scarab
yea that too
dusty portal
It looks right but shitty?
grand pondBOT
;(
dusty portal
i verified it numerically for smaller powers, it works
Ok!
midnight plankBOT
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fading slate
Is this correct? Or does g have to be surjective as well?
midnight plankBOT
Is this correct? Or does g have to be surjective as well?
prime hornet
Is this correct? Or does g have to be surjective as well?
wait, I'm confused by the question's wording
what's it asking you?
is h = g o f?
are g and f surjective? 
fading slate
we have surjective composition h o g o f and we are supposed to say if f, g or h have to be surjective
prime hornet
do we know anything about h, g, or f?
prime hornet
we have surjective composition h o g o f and we are supposed to say if f, g or ...
aha, I see
prime hornet
Is this correct? Or does g have to be surjective as well?
g need not be surjective.
consider A = {a}, B = {b_1, b_2}, C = {c_1, c_2}, D = {d}
we can have f(a) = b_1, g(b_1) = g(b_2) = c_1, and h(c_1) = h(c_2) = d.
then the composition h o g o f is surjective, because h(g(f(a))) = d, but g isn't surjective
you're right that h needs to be surj though
fading slate
Then just h has to be surjective right?
prime hornet
yeah
fading slate
okayyy
what if the composition h o g o f was injective?
which one of the functions would have to be injective
wouldnt it be h only again?
prime hornet
wouldnt it be h only again?
unfortunately, this isn't the case 
h could be injective on the image of g o f, but not injective on all of C
likewise, g could be injective on the image of f, but not on all of B
f however, f needs to be injective 
if it weren't, then we could find two elements a_1 and a_2 in A with a_1 neq a_2 but f(a_1) = f(a_2)
and then we'd have that h(g(f(a_1))) = h(g(f(a_2)))
injectivity of the composition would force us to have a_1 = a_2, which is a contradiction
injectivity of h on the image of g o f and injectivity of g on the image of f are needed for the composition to be injective, but they don't guarantee that h or g themselves are injective
fading slate
okay thank you
prime hornet
midnight plankBOT
@fading slate Has your question been resolved?
fading slate
yes
prime hornet
yes
you gotta react to the bot with the checkmark emote 
in general though, if your question has been resolved, you can close the channel by typing .close or .solved 
that way it quickly goes back into the available category and others can use it c:
safe oriole
given a cone of equation x^2/R^2+y^2/R^2=<z^2/h^2, where R,h>0. isnt the projection of this cone onto the xy plane just the point (0,0,0)?
midnight plankBOT
given a cone of equation x^2/R^2+y^2/R^2=<z^2/h^2, where R,h>0. isnt the project...
safe oriole
because its projection onto the xy plane is exactly the set of points (x,y,0) that satisfy the above equation right ?
obsidian drum
because its projection onto the xy plane is exactly the set of points (x,y,0) th...
no, the projection should be the set of all (x,y) which (x,y,z) satisfies the equation for some z
safe oriole
so it is set of points forming the disk given by x^2+y^2=<z^2R^2/h^2 where z is a constant ?
last slate
In this case it would be entire plane if z is unbounded
safe oriole
so it is set of points forming the disk given by x^2+y^2=<z^2R^2/h^2 where z is ...
not exactly a constant
last slate
No z is the third axis in most cases
safe oriole
more like a parameter?
last slate
Yes
safe oriole
so here if 0=<z=<h then the projection onto the xy plane is 0=<x^2+y^2=<R^2
but then when i found the intersections of the cone with the planes z=0 and z=h i got the same result
is this just a coincidence?
so the intersection with z=0 is the origin of the space and the intersection with z=h is the disk of radius R centered at (0,0,h)
ie the disk given by x^2+y^2=<R^2, z=h
along with the (x,y)=(0,0) from the intersection with z=0
safe oriole
so here if 0=<z=<h then the projection onto the xy plane is 0=<x^2+y^2=<R^2
i got this
i dont think it is just a coincidence, is it ?
because i just took the intersection of the cone with the boundaries of z that enclose the intended region right?
safe oriole
given a cone of equation x^2/R^2+y^2/R^2=<z^2/h^2, where R,h>0. isnt the project...
wait i thought that i mentioned integration but i didnt . So i am trying to find the volume of this using a triple integral
last slate
Projection onto plane won't work in this case because area will be distorted
safe oriole
i am not sure that i follow
safe oriole
so here if 0=<z=<h then the projection onto the xy plane is 0=<x^2+y^2=<R^2
so is this projection correct to begin with
safe oriole
Projection onto plane won't work in this case because area will be distorted
if yes then can you explain to me what exactly you mean by this
safe oriole
i am trying to find the volume of the cone that is given by x^2/R^2+y^2/R^2=<z^2/h^2 , 0=<z=<h using a triple integral
last slate
Then do you know it's formula
safe oriole
the formula of what ?
last slate
∭_E dV, where E represents the solid region.
safe oriole
i want to find $\iiint_E\dd V$
grand pondBOT
pirateking0723
safe oriole
i rewrote as E to have the same notation
yea i know about this
so $\frac{x^2+y^2}{R^2}\leq\frac{z^2}{h^2}\implies\frac Rh\sqrt{x^2+y^2}\leq z$ (the other possible inequality is omitted since $z\geq 0$) and $z\leq h$ so $\frac Rh\sqrt{x^2+y^2}\leq z\leq h$
grand pondBOT
pirateking0723
safe oriole
so $\iiint_E\dd V=\iint_D\int_{\frac Rh\sqrt{x^2+y^2}}^h\dd z\dd A=\iint_Dh-\frac Rh\sqrt{x^2+y^2}\dd A$
grand pondBOT
pirateking0723
safe oriole
now i want to find region D which is the projection of the cone onto the xy plane
which is done above
but then comes my question
which is if i find the intersections of the cone with z=0 and z=h i am getting the same bounds
so is it a coincidence or am i interpretting something wrong after doing this or what
last slate
Yes that's what bounds mean
So the intersection points can be taken as bounds in this case
safe oriole
alright so why project the cone on the xy plane
last slate
No reason why would you that
safe oriole
So the intersection points can be taken as bounds in this case
exactly thats the normal thing to do usually right
safe oriole
No reason why would you that
i see
last slate
exactly thats the normal thing to do usually right
I find them that way
safe oriole
alright so why project the cone on the xy plane
now i am talking about this btw
last slate
No need to project
safe oriole
yes there is no need
safe oriole
but it works since you are projecting with the bounds of z ?
but it is correct because of this right?
or is the reason something else
obsidian drum
I guess this is what you want:
The volume of the cone can be interpret as $\iiiint_E \dd V=\int_0^h \iint_D \dd A\dd z$ here D is the projection to the plane z = a
grand pondBOT
Micni9
safe oriole
ohh i see
safe oriole
**pirateking0723**
so here there is no projection
safe oriole
**pirateking0723**
D here is just the intersection of the cone with the planes z=0 and z=h
safe oriole
**Micni9**
while here D is the projection of the cone onto the xy plane
is that it ?
obsidian drum
D here is just the intersection of the cone with the planes z=0 and z=h
no, it change as z changes
imagine you are summing the area of the projection from z=0 to z=h
safe oriole
no, it change as z changes
isnt this change taken into consideration with the integrand ?
thats why the integrand depends on x and y
no ?
safe oriole
imagine you are summing the area of the projection from z=0 to z=h
yes
but the bounds are z=0 and z=h
ah well what i said is wrong
what is correct to say is that the intersections with z=0 and z=h are the boundaries of D (?)
is that it ?
is that what you were trying to tell me ?
last slate
D is the projection
safe oriole
which D
last slate
**Micni9**
z=0 and z=h are bounds of a
safe oriole
**pirateking0723**
this?
safe oriole
**Micni9**
or this ?
last slate
Yes
last slate
**Micni9**
This
safe oriole
**pirateking0723**
and this D here is the region bounded by the intersections of the cone with the planes z=0 and z=h?
last slate
Nope
safe oriole
**pirateking0723**
then what is D here
safe oriole
and this D here is the region bounded by the intersections of the cone with the ...
so the slice of the cone bounded by these intersections ?
last slate
No cone's height is bounded by those intersections
safe oriole
because if no then what slice ? how is it determined ?
safe oriole
ok so D is the part of the cone bounded by the intersections with z=0 and z=h
because it starts at z=0
the intersection with z=0
the plane goes up
and keeps intersecting the cone
it stops at z=h
last slate
Yeah
But those are not bounds of D (ie slice itself)
Those are bounds of the blue plane
But the slice is region between the circle
And it's bound is the boundary of the circle
safe oriole
And it's bound is the boundary of the circle
how do i find the equation of this circle ?
isnt it changing ?
obsidian drum
it changes as z change, so it can be expressed in terms of z
safe oriole
it changes as z change, so it can be expressed in terms of z
i did that at first but then deleted it
because now there is no z right?
there shouldnt be z anymore
we integrated wrt z at first
obsidian drum
there is
safe oriole
now there is only x and y , there shouldnt be z in the boundaries of the integral
last slate
Integrate dA first
obsidian drum
dz is the outermost integral which you integrate last
safe oriole
are we talking about this
ah i thought we were talking about this
safe oriole
Integrate dA first
in this case let me do it
each disk here is given by $x^2+y^2\leq\frac{z^2R^2}{h^2}$ so for each z, this is D and thus its boundary is given by $x^2+y^2=\frac{z^2R^2}{h^2}$
grand pondBOT
pirateking0723
safe oriole
now we can use polar coordinates with 0=<r=<zR/h and 0=<θ=<2π
so that our integral can be rewritten as $\iiint_E\dd V=\int_0^h\int_0^{2\pi}\int_0^{\frac{zR}h}r\dd r\dd\theta\dd z=2\pi\int_0^h\frac{z^2R^2}{2h^2}\dd z=\frac{\pi R^2h}{3}$
grand pondBOT
pirateking0723
last slate
Yessss!!!!
obsidian drum
Congratz
safe oriole
**pirateking0723**
can we talk about this
obsidian drum
this is also doable
safe oriole
give me a sec to think about it
last slate
Yes
obsidian drum
it should be $\iiint_E\dd V=\iint_D\int_{\frac hR\sqrt{x^2+y^2}}^h\dd z\dd A=\iint_Dh-\frac hR\sqrt{x^2+y^2}\dd A$
grand pondBOT
Micni9
obsidian drum
you make a mistake on the lower bound of z
safe oriole
you make a mistake on the lower bound of z
yes R/h instead of h/R
so $\iiint_E\dd V=\iint_D\int_{\frac Rh\sqrt{x^2+y^2}}^h\dd z\dd A=\iint_Dh-\frac hR\sqrt{x^2+y^2}\dd A$
grand pondBOT
pirateking0723
safe oriole
what i am about to say is probably somehow inaccurate. So we know that $x^2+y^2\leq\frac{z^2R^2}{h^2}\leq\frac{h^2R^2}{h^2}=R^2$ so $D$ is the disk given by $x^2+y^2\leq R^2$
grand pondBOT
pirateking0723
safe oriole
i think that this is somehow incorrect/inaccurate
last slate
You're assuming z=h
safe oriole
because i disregarded the lower bound of z for example
last slate
z=a, where a is the height of disk
obsidian drum
same trick
safe oriole
shouldnt D here be the projection onto the xy plane?
safe oriole
**pirateking0723**
which is this disk here
obsidian drum
now you assume z changes as D changes, D changes from a dot to a big circle
this is similar to previous case that we say z changes from 0 to h
safe oriole
**pirateking0723**
D is indeeed this disk here right ?
because assuming that D is this then the integral becomes $\ \iint_Dh-\frac hR\sqrt{x^2+y^2}\dd A=\int_0^{2\pi}\int_0^Rhr-\frac{hr^2}R\dd r\dd\theta=2\pi(\frac{hR^2}2-\frac{hR^3}{3R})=\frac{\pi R^2h}3$
grand pondBOT
pirateking0723
safe oriole
**pirateking0723**
but this reasoning is wrong, right?
safe oriole
this is similar to previous case that we say z changes from 0 to h
can you explain a bit more
obsidian drum
in previous case, z is a moving coordinate point, we didn't treat it as a line segment
similarly, we treat D as a moving, expanding disk instead of simply the base of the cone
the base is indeed the "upper bound" of D
safe oriole
the base is indeed the "upper bound" of D
and i suppose that the lower bound is (0,0) at z=0?
obsidian drum
yes
safe oriole
i think that i see the reason why D is the projection onto the xy plane
so we are moving up in disks
now if we draw all of these disks on the same plane
the boundaries of each of these disks will form concentric circles
with the largest circle being of radius h and the lowest of radius 0
and so this disk is formed
obsidian drum
yup
safe oriole
the same thing is going to happen for any surface if integrated first wrt z
the region D will be its projection to the xy plane
similarly integrating first wrt x will lead to D being the projection onto yz plane and integrating wrt y first will lead to D being the projection onto the xz plane
obsidian drum
as long as you can write out the innermost integral in terms of the plane
safe oriole
so what if you cant
then what should be done
first of all what is an example of this case
midnight plankBOT
@safe oriole Has your question been resolved?
safe oriole
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midnight plankBOT
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Available help channel!
twilit field
My idea is to prove it's injective. So the only reasoning I have is T(v)=0 only if v=0 by defn all elements z_1 and onwards must be 0. The only vector with this property is the 0 vector.
midnight plankBOT
My idea is to prove it's injective. So the only reasoning I have is T(v)=0 only ...
hard umbra
injective things can have eigenvalues
twilit field
injective things can have eigenvalues
I;m using this
hard umbra
but this is for T - lambdaI not T
twilit field
so I consider T- \lambda I and show it's always injective then, for any lambda in R
okay, so this won't work for an infinite dim vector space
$\lambda(z_1,z_2,\dots)= (0,z_1,z_2,\dots)$
grand pondBOT
wai
grand pondBOT
wai
midnight plankBOT
midnight plankBOT
Available help channel!
gaunt nimbus
midnight plankBOT
gaunt nimbus
got stuck
for (b)
Suppose ${a,b \in (n)}$ where ${(n) = {a: n|a}}$. Then, for some ${k, j \in \mathbb{Z}}$, ${a = nk}$ and ${a = nj}$. Therefore, ${ax + by = (nk)x + (nj)y = n(kx + jy)}$, meaning ${n | ax + by}$. Consequently, ${ax + by \in (n)}$. ${\square}$
grand pondBOT
k
gaunt nimbus
this is for (a)
Assume ${(n) \subseteq (m)}$. Then, for all ${x}$ s.t. ${x = nk}$ for some ${k \in \mathbb{Z}}$, we also have ${x = mj}$ for some ${j \in \mathbb{Z}}$. Therefore, ${nk = mj}$.
grand pondBOT
k
gaunt nimbus
Seems like i have to show that ${k / j \in \mathbb{Z}}$. However, division is not an operation here.
grand pondBOT
k
gaunt nimbus
help b
fair sleet
K
frozen finch
nope im out
gaunt nimbus
chrome rain
hint - n belongs to (n)
frozen finch
<:blobcry:971797574916907028>
im sorry, i just didnt do set theory yet
chrome rain
hint - n belongs to (n)
this implies n belongs to?
gaunt nimbus
$0 = mj - nk}$?
grand pondBOT
k
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
chrome rain
ye, so m divides n
gaunt nimbus
how so?
chrome rain
you just said that n is a multiple of m, right?
gaunt nimbus
yup
chrome rain
what does m divides n mean?
gaunt nimbus
(20) = ..., 20, 40, 60, ...
(4) = ..., 4,8,12,16, ...
gaunt nimbus
what does m divides n mean?
n = mk for some k
chrome rain
n = mk for some k
yes so m divides n is same as n is a multiple of m...
chrome rain
yeah
gaunt nimbus
and by being in m, it is multiple of m
gaunt nimbus
so m divides n
chrome rain
yesss
gaunt nimbus
for (c)
if m|n, then mk = n.
wait
if m divides n
then n is a multiple of m
since also n is a multiple of n
we have that if n is a multiple of n, then n is a multiple of m
thus the definition of subset?
is this right?
chrome rain
no here you have to prove this :
let x be ANY number in (n), then x will also be in (m)
you have only proved that the number 'n' will be in (m)
gaunt nimbus
ok
so
if x is any number in (n)
then x is a multiple of n
also by defn of subset
x is a multiple of m
so x = mj for some j and x = nk for some k...
now
n is also in m
as proven
so
m = nb for some b
hence we have x = nbk = n(bk)
chrome rain
yes that's right now
basically
n=mk and x=an so
x=an=(ak)m
so x is a multiple of m => x is in (m)
gaunt nimbus
oh ok
let me formalise it
Suppose ${(n) \subseteq (m)}$. Then, for all ${x}$ in ${(n)}$, we also have ${x \in (m)}$. In particular, for ${x=n}$, we have that ${n = mk}$ for some ${k \in \mathbb{Z}}$. By defn, ${m | n}$. ${\square}$
this is for b
grand pondBOT
k
gaunt nimbus
For (c)
Suppose ${m|n}$ and let ${x \in (n)}$. Then, ${mk = n}$ for some integer ${k}$ and ${x = an}$ for some integer ${a}$. Therefore, ${x = an = a(mk) = (ak)m = mj}$ for some integer ${j}$. By defn ${m|x}$, so ${x \in (m)}$. Consequently, for any ${x \in (n)}$, also ${x \in (m)}$. Thus, ${(n) \subseteq (m)}$. ${\square}$
@chrome rain
chrome rain
yep 👍
gaunt nimbus
thank u so much
gaunt nimbus
oopsie
grand pondBOT
k
chrome rain
jaunty canopy
how do you do the last part of this q?
midnight plankBOT
how do you do the last part of this q?
jaunty canopy
i've found
(r^2-3)f(r)
(x+yi)^2f(r)
(x-yi)^2f(r)
(x+yi)zf(r)
(x-yi)zf(r)
but i still need a last basis vector
jaunty canopy
i've found
(r^2-3)f(r)
(x+yi)^2f(r)
(x-yi)^2f(r)
(x+yi)zf(r)
(x-yi)zf(r)
also how do i check that they're eigenstates of L^2 quickly? ig i could probs compute L_x and L_y for each of these 3 eigenstates but that feels like it would take quite a long time
fallow scarab
uhhh did you already try https://discord.gg/physics
jaunty canopy
uhhh did you already try https://discord.gg/physics
i'm not in that server but i can try
i mean my degree is just maths
hence why i posted in maths
fallow scarab
then you should be explaining all the definitions in terms of math behind the problem. "angular momentum operator" is not defined in many math courses.
jaunty canopy
then you should be explaining all the definitions in terms of math behind the pr...
yh fair
L_x, L_y, L_z are the components of this operator
L^2 = L_x^2 + L_y^2 + L_z^2
midnight plankBOT
@jaunty canopy Has your question been resolved?
midnight plankBOT
midnight plankBOT
Available help channel!
midnight plankBOT
twilit field
like can;t T have atmost dim range T eigenvectors
sharp coral
[0] has range {0} and n eigenvectors
twilit field
[0] has range {0} and n eigenvectors
nmm
flat spire
wai
twilit field
hi
midnight plankBOT
subtle blaze
I would think about a matrix but maybe you don’t want to do that
midnight plankBOT
Available help channel!
night gyro
bruh
midnight plankBOT
@night gyro Has your question been resolved?
night gyro
.close
fair frigate
can someone help me
midnight plankBOT
can someone help me
plucky lark
Uh so like what u want?
midnight plankBOT
@fair frigate Has your question been resolved?
fair frigate
Uh so like what u want?
help
fair frigate
Is this a test?
nope
fair frigate
Is it a quiz?
no homeowrk
plucky lark
Do u know how to do Gaussian elimination?
fair frigate
Do u know how to do Gaussian elimination?
not reslly
my teacher didnt teach
anything\
fair frigate
i did
midnight plankBOT
@fair frigate Has your question been resolved?
tawdry ermine
i did
Instead of posting the whole list of questions into one help channel, you may wanna post one by one instead
narrow edge
why is the 3rd graph the graph of this inequality
midnight plankBOT
why is the 3rd graph the graph of this inequality
livid python
midnight plankBOT
midnight plankBOT
Available help channel!
strong oak
Hello everyone. I am having difficulty specifying the nature and geometric characteristic elements of the attached set E.
midnight plankBOT
Hello everyone. I am having difficulty specifying the nature and geometric chara...
mental shuttle
,rotate ccw
grand pondBOT
mental shuttle
you are doing E and F, right?
strong oak
you are doing E and F, right?
Yes, but it's E that's the problem for me.
mental shuttle
there's probably a few ways you could go about this
do you know how to classify a conic section?
you could either square both sides and expand it out in terms of x and y, where z = x + iy, that will give you a conic section equation in general form
or you can simplify it a bit and interpret it geometrically. for example, |z + 1 + i| is the distance to (-1 - i)
strong oak
do you know how to classify a conic section?
Yes
mental shuttle
ok, then if you square both sides, you should get a quadratic in x and y on each side, and that'll give you a conic section
since |z|^2 = x^2 + y^2
might not be what they are looking for nor the easiest solution though
strong oak
or you can simplify it a bit and interpret it geometrically. for example, |z + 1...
Ok I didn't understand it like that. In fact, by interpreting this geometrically, we will end up with the ratio of the geometric definition of an ellipse with eccentricity e = 1/2 (because 1/2 is positive and less than 1
mental shuttle
i don't think it is an ellipse tbh
an ellipse would be like |z - a| + |z - b| = c right? sum of distances to foci is constant
the RHS has z and z^bar though
strong oak
might not be what they are looking for nor the easiest solution though
I spoke of ellipse for two reasons. The first is the fact that at the end of the manipulations we have a ratio of the form MB/MH=1/2 and since H is the orthogonal projection of M on a straight line (the axis of t which is in fact an orthogonal symmetry in the plane) and that B does not belong to this line, we can conclude from the geometric definition of a conic that (E) is an ellipse of Directrix the axis of orthogonal symmetry, with focus B. The second thing (which follows from the first) is that the eccentricity which is the result of this ratio is positive and less than 1
strong oak
an ellipse would be like |z - a| + |z - b| = c right? sum of distances to foci i...
That's true, but I would like to obtain the geometric definition, that is to say with the points in the plane.
mental shuttle
you're correct H is the orthogonal projection of M on a straight line yes, good job
but from what i remember, the set of points equidistant from a point (focus) and a line (directrix) is a parabola, i could be wrong though
strong oak
you're correct H is the orthogonal projection of M on a straight line yes, good ...
Oh thank you very much
strong oak
but from what i remember, the set of points equidistant from a point (focus) and...
This is also true
strong oak
ah there's a factor of 1/2 though
Exactly what made me say it's an ellipse
Thank you very much for your valuable help.
mental shuttle
no problem, good luck
strong oak
midnight plankBOT
@strong oak Has your question been resolved?
signal rock
idk how to do part i
midnight plankBOT
idk how to do part i
signal rock
ik the 10 + t is the amount of water
and 2Q comes from the original 20kg salt divided by 10L, so it is the concentration?
or should i start to make an equation for Q first
or should i seperate the variables
midnight plankBOT
@signal rock Has your question been resolved?
frozen finch
idk how to do part i
mhm
d/dt, or d/dx, or dy/dx, all the d/d forms are derivatives
midnight plankBOT
@signal rock Has your question been resolved?
signal rock
.close
vocal raft
@signal rock
midnight plankBOT
<@1126009713419620353>
vocal raft
.close
midnight plankBOT
midnight plankBOT
Available help channel!
cursive swan
PWICFOP on this polynomial:
x^3 - 48x + 128
midnight plankBOT
PWICFOP on this polynomial:
x^3 - 48x + 128
slender walrus
what's that acronym
tawdry ermine
what's that acronym
Idk bout PWI but i know whhat CFOP iis tho
cursive swan
It's quite famous:
PWICFOP = Peform What Is Called "Factorization of Polynomials (FOP)"
slender walrus
never seen that acronym before in my life
why not just use one word: factorise
anyway what gave yo tried?
cursive swan
never seen that acronym before in my life
why not just use one word: factorise
It's written that way in my book.
midnight plankBOT
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
start with rational root theorem
cursive swan
+- p/q is a root.
Then p divides 128
q divides 1
q = 1
128 = 2 * 2 * 2 * 2 * 2 * 2 * 2
Possible (rational) roots:
1, 2, 4, 8, 16, 32, 64, 128.
And the negative version of these.
Now what.
slender walrus
test too see if they're roots,
usually start small
after you find one do polynomial division
lyric charm
It's quite famous:
PWICFOP = Peform What Is Called "Factorization of Polynomials...
it is decidedly not famous, and your jocular insistence to the contrary is decidedly unfunny.
cursive swan
x = 4 works.
Now I can divide the cubic by x - 4.
And get a quadratic equation to be solved.
And use the quadratic formula.
And get the other fruits.
Thanks.
I think I should close this channel now.
.close
midnight plankBOT
cursive swan
Thanks for your help.
midnight plankBOT
Available help channel!
molten bay
midnight plankBOT
molten bay
@lyric charm i got 11 answer
Can we solve it by geometrically too?
Just curious to know
.close
molten bay
midnight plankBOT
https://math.stackexchange.com/questions/4322288/whats-the-value-of-the-area-of-...
molten bay
Can anyone explain how did they find the points over grid?
In the second comment
Suppose if I am given three length √20,√26,√34
woeful turret
can someone explain how the torque about the midpoint is that
midnight plankBOT
can someone explain how the torque about the midpoint is that
woeful turret
shouldnt it be 2 into that
verbal pumice
Should be twice right 
woeful turret
yeah thats what i thought as well
verbal pumice
Oh actually
Let's say you have a single rod of uniform charge density lambda
What would be the torque about an end point?
You might have to do a bit of integration
verbal pumice
can someone explain how the torque about the midpoint is that
because each point on the rod has some force on it
its not point masses at the ends
verbal pumice
yeah its continuously distributed here
manic bison
I have no clue
midnight plankBOT
I have no clue
manic bison
and find the pattern
There's none
woeful turret
what series did you get
woeful turret
whats the next term
manic bison
a2 is 1
woeful turret
26 should be
manic bison
Probably