did you swap the integral' boundaries ?
#help-10
alpine bison
robust raven
$$\begin{array}{l}
\int\limits_0^1 {\frac{{3xdx}}{{\sqrt {4 - 3x} }}} = = - \frac{1}{3}\int\limits_4^1 {\frac{{3\left( {\frac{{4 - t}}{3}} \right)dt}}{{\sqrt t }}} = \frac{1}{3}\int\limits_1^4 {\left( {\frac{4}{{\sqrt t }} - \sqrt t } \right)dt} = \
\
= \frac{1}{3}\left[ {8\sqrt t - \frac{2}{3}t\sqrt t } \right]_1^4 = \frac{1}{3}\left( {16 - \frac{{16}}{3} - 8 + \frac{2}{3}} \right) = \frac{{10}}{9}
\end{array}$$
warm shaleBOT
Joanna Angel
robust raven
plz analyse it
robust raven
when you change the sign of the integral, you reverse the limits of integration
it is very known formula
crude coral
i didnt know 
lone echo
technically the other way around. When you reverse the limits of integration you change the sign of the integral
crude coral
so if i change my integral bounds the sign changes?
lone echo
yes. If integral from a to b = K, then integral from b to a = -K
crude coral
so
$$\int^b_a f(x) = -\int^a_b f(x) a > b$$
warm shaleBOT
JustToPro
robust raven
so if i change my integral bounds the sign changes?
yes
lone echo
no, you dont need the condition a>b
alpine bison
for every a and b indeed
lone echo
that's true for every a and b
crude coral
if b is greater than a then we get negatives no?
lone echo
$\int^b_a f(x)dx = -\int^a_b f(x)dx$
warm shaleBOT
LordFelix
alpine bison
if b is greater than a then we get negatives no?
no depends on the function insid4e the integraò
ocean arrow
.
maiden bone
i am doing trigonometry for like 3 years now
i am also pretty good in solving "sums"
but all the time i feel like idk what is really going on there
i mean , i really think about very basic questions like why sin , cosec on 2nd quad is +ve , how trigno came in the world of graphs ? blah blah
i am really looking forward for the answers of these questions
is there any good resource where i can get answers of all these questions , who teaches trigno in deep?
obtuse pebbleBOT
i am doing trigonometry for like 3 years now
i am also pretty good in solving ...
@maiden bone Has your question been resolved?
maiden bone
<@&286206848099549185> please?
karmic ridge
What’s the question
lucid sparrow
?
maiden bone
i am doing trigonometry for like 3 years now
i am also pretty good in solving ...
this
i am really looking for a great resource
maiden bone
Refer to khAn academy
hmmm
have u learned trigno from there?
maiden bone
I use books
oh really interesting
karmic ridge
Or maybe try books from cengage
maiden bone
I use books
which book for trigno ?
also do u wanna recommend me some maths books which u loved?
karmic ridge
Bro cengage
maiden bone
umm hmm ?
i have done all this shit but i dont understand a shit
maiden bone
Bro cengage
ohk ohk but what is the name of the book ?
lucid sparrow
umm hmm ?
i have done all this shit but i dont understand a shit
well then I can't help you
karmic ridge
maiden bone
well then I can't help you
by the way , why did u share that pic with me ?
maiden bone
JEE ADVANCE 😵
holyyy shitt
lucid sparrow
by the way , why did u share that pic with me ?
I don't know, this is supposedly the answer to your question
maiden bone
I don't know, this is supposedly the answer to your question
oh lol
sorry for the confusion
i am really looking forward for some resources which teach trigno in depth
karmic ridge
Try learning from Indian teachers and authors
karmic ridge
oh lol
sorry for the confusion
i am really looking forward for some resources ...
Bro I just shared the pic of the book
maiden bone
Try learning from Indian teachers and authors
hmm
which one to be specific?
karmic ridge
Go order it
karmic ridge
hmm
which one to be specific?
Anyone u can find
maiden bone
Anyone u can find
can i ask u some basic question , to see if u understand trigno nicely?
maiden bone
Go ahead
from where did all the values on the trigno tables came from ?
like sin90 is 1
and yeah everything was going good with degrees , what is up with radians ?
hmm and yeah why the heck "pie" is everywhere?
and yeah what is sin , cos , tan ,cosec anyways
maiden bone
from where did all the values on the trigno tables came from ?
like sin90 is 1
...
<@&286206848099549185> hmm if u guys dont mind answering these stupid questions
burnt otter
from where did all the values on the trigno tables came from ?
like sin90 is 1
...
ok so radians is just another way of writing degrees in terms of pi
pi is essentially 180 degrees
maiden bone
pi is essentially 180 degrees
hmm interesting
aint pit 3.14 units , lol?
burnt otter
maybe someone else can explain that for you but you use pi when you look at the unit circle. sorry ToT
pi 3.14 is correct but we're not gonna use that form rn
maiden bone
pi is essentially 180 degrees
pie in trignometric terms is 180 degree (which basically came by converting 180 degree into radians (180* pie/180) )
but in another questions of circles its 22/7
maiden bone
maybe someone else can explain that for you but you use pi when you look at the ...
hmm you are the 3rd one denying , lol
burnt otter
like I was never really taught that
maiden bone
like I was never really taught that
and its my bad luck that i am taught like this
maiden bone
22/7 is 3.14...
yeah but whats up with 180 degress then
maiden bone
from where did all the values on the trigno tables came from ?
like sin90 is 1
...
maybe just these questions are dumb themselves
burnt otter
ok well you know what the unit circle is right
maiden bone
ok well you know what the unit circle is right
hmm which has radius 1 unit and angle subtended is 1 unit
it was smth like that
burnt otter
you know that points on the circle correspond with (cosx,sinx)
because the hypotenuse is 1, so cos corresponds with x value, sin corresponds with y value
maiden bone
like I was never really taught that
hmm u wanna recommend me some resources where u learnt?
burnt otter
my teacher ToT, i can give you some texbook introductions if that would help
maiden bone
my teacher ToT, i can give you some texbook introductions if that would help
yeah if they have good amount of theory stuff
maiden bone
from where did all the values on the trigno tables came from ?
like sin90 is 1
...
anyone else who want to try answering these?
<@&286206848099549185>
burnt otter
hopefully this is readable
basically unit circle with radius one has circumference 2π (2πr)
and half circumference would be pi
since half is also 180 degrees, pi arc length is therefore 180 degrees
maiden bone
its a bit blur
which book is it?
burnt otter
maiden bone
oh yeah this makes sense quite a bit
which book is it?
burnt otter
its called maths in focus year 11 chapter 5
maiden bone
its called maths in focus year 11 chapter 5
this one?
maiden bone
its called maths in focus year 11 chapter 5
or the yellow cover one?
maiden bone
.close
thick oracle
If i have f(x)=x²
obtuse pebbleBOT
If i have f(x)=x²
thick oracle
And im tryna find the points where the slope angle is 30
Degrees
Will i get 2 points?
2 symmetrical points?
Or will 1 have 30 degrees and the other one -30
crisp zephyr
One will be 30 degrees and the other one will be -30. You will get only one point where the slope angle is 30 degrees
inland matrix
Slope angle cannot be negative, it'll be between 0 and π
crisp zephyr
The derivative of x^2 is 2x which is an injective function, so every engle of a tangent is unique
inland matrix
Yeah they do but you want the angle of the slope line to be 30⁰, which is always measured wrt the positive x axis
In case of the triangle on the left, the slope line makes an angle 30⁰ but with negative x axis
And also the values of slopes of the two lines are different
obtuse pebbleBOT
@thick oracle Has your question been resolved?
thick oracle
Ok thanks
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
proper snow
how do u do 4b?
obtuse pebbleBOT
how do u do 4b?
proper snow
like i have P(A) so should i equate them or whaf idk im so confusee
pls ping me if you have an answer thank u
timid silo
pls ping me if you have an answer thank u
Whats the given answer ?
obtuse pebbleBOT
@proper snow Has your question been resolved?
scarlet gale
like i have P(A) so should i equate them or whaf idk im so confusee
Are those the correct answers, like P(B) = 0.7, or are they what you worked out?
proper snow
Whats the given answer ?
0.7, 0.5 and 0.06
proper snow
okay
scarlet gale
Then, factor the top.
proper snow
okay then
scarlet gale
What do you have for the factored parts?
proper snow
isnt it like 7/20, 14/25 and 2/5
scarlet gale
Oh.
proper snow
yeah
i tried equating p(a) but like it doesnt work out that way.
wait
hollup
i have p(a) no? so i can nus substitute them right?
nvn thaf diesnt work it'll cancel out 💀
scarlet gale
nvn thaf diesnt work it'll cancel out 💀
OK, so multiply P(A n B) P(A n C) P(B n C) = P(A)^2 P(B)^2 P(C)^2.
Then replace P(A) with its P(B) version (P(A) = 0.35/P(B)) and P(C) with its P(B) version (P(C) = 0.4/P(A) = 0.4 P(B)/0.35).
Then solve for P(B).
timid silo
OK, so multiply P(A n B) P(A n C) P(B n C) = P(A)^2 P(B)^2 P(C)^2.
This comes from them being independent
scarlet gale
Or I guess you could leave P(A) there and replace P(C) with 0.4/P(A).
timid silo
independent events P(A n B) = P(A)*P(B)
scarlet gale
P(A)^2 P(B)^2 P(C)^2 = P(A)^2 P(B)^2 (0.4/P(A))^2
P(A)^2 P(B)^2 P(C)^2 also is equal to 0.35 * 0.56 * 0.4.
proper snow
ahh oksy I'll try that
scarlet gale
Or, you could do it more directly.
P(A)^2 P(B)^2 P(C)^2 = (0.4/P(C))^2 P(B)^2 P(C)^2.
I think they're looking for something like that.
That uses a part ii directly.
proper snow
yeh that could work too. i mean as long as i use the answer from a so yeah
omg thank you the answers matched up
tysmmm
.close
obtuse pebbleBOT
scarlet gale
You're welcome.
obtuse pebbleBOT
Available help channel!
fallow widget
A rectangle is inscribed in a semicircle y = √25 − x2 (with two of its vertices on the x-axis (4p)
and two on the semicircle). Determine the largest area the rectangle can assume.
obtuse pebbleBOT
A rectangle is inscribed in a semicircle y = √25 − x2 (with two of its vertices ...
fallow widget
$\sqrt{25-x^2}$
I have an idea on how to do this question but I will need verification if I’m thinking right
warm shaleBOT
Merineth
fallow widget
I understnad that i'm trying to fit a rectangle where a 2 corners are on the x-axel and 2 are on the green line
To get the biggest area possible, wouldn't the two corners need to be located halfway to the line of symmetry and the start of x roughly -5 and another corner located halfway towards 5 from the line of symmetry?
Since if i have my corners at -5 and 5 it will be extremely wide but height will be really low. Meanwhile if i had the x very small the y would become really large. So it has to be somewhere in the middle right?
So something like this?
I have another idea of how it might be done
So since the area is calculated using the 2x(y) https://gyazo.com/f326d334474f3cf2c71dee2e2ed600b5
I can plug in y into the equation A = 2x(sqrt(25-x^2))
past sand
Yes, you have the area as a function now, so how do you find the max of a function?
fallow widget
I did it quick in paint but i'm not sure if it's correct
or rather, i don't believe it's correct
Considering that 5 and 0 would make the rectangle extremely short in height
past sand
sqrt(25-x^2) is not sqrt(25) - sqrt(x^2) 
little ermine
i think you're finding the roots to A = 0 tho
past sand
i think you're finding the roots to A = 0 tho
No?
These would be -5, 0 and 5, for obvious reasons
little ermine
i mean what you did in the pic
lone python
consider this realations
fallow widget
$\sqrt{25-x^2} = \sqrt{25} - \sqrt{x^2}$
warm shaleBOT
Merineth
past sand
Well yeah that's the result they got but not for the correct reasons
fallow widget
**Merineth**
Aren't these equivalent?
little ermine
Well yeah that's the result they got but not for the correct reasons
mhm
past sand
Aren't these equivalent?
No
past sand
sqrt(25 - 16) = sqrt(9) = 3
sqrt(25) - sqrt(16) = 5-4 = 1
past sand
The square root is like an exponent, you can do that when you have a product, not when you have a sum
fallow widget
I'm trying to solve for x
lone python
you have 2 variables
fallow widget
So i know where the largest are will be
neon vector
But you are assuming A = 0?
fallow widget
Oh
neon vector
At least in the photo earlier
lone python
So since the area is calculated using the 2x(y) https://gyazo.com/f326d334474f3c...
That's your equation for any area under the circle. Now you have to find its maximum.
neon vector
https://gyazo.com/31d36ba61fa3b8492b902f19e84f3df7
Well here you get $2x \cdot (x-5) = A$
warm shaleBOT
USS-Enterprise
past sand
You're assuming the area is zero and then trying to find the maximum area?
neon vector
And then say $x_1 = 5$ , $x_2 = 0$
neon vector
Which is true when A = 0
warm shaleBOT
USS-Enterprise
fallow widget
So i keep A ?
past sand
past sand
or rather when it reaches its maximum
What's "it" in this sentence?
fallow widget
I find when the derivate is = 0. And then i take that x value and put it in the original function
past sand
Yes
fallow widget
Just trying to calculating the derivative, sorry
Got any tips for doing that? I'm currently trying chain rule
past sand
Product rule + chain rule I guess
fallow widget
$\frac{-2x}{2\sqrt{25-x^2}} * (\sqrt{25-x^2})$
warm shaleBOT
Merineth
past sand
That's wrong
fallow widget
How? Do i have to use the product rule here?
Why can't i just use chain rule all the way
or is chain rule only applied when we have + and / or -?
past sand
Specify the functions you're using
fallow widget
Well the chain rule is f(g(x)
ah
I see what you mean now
It's because we have f(x) * g(h(x)) ?
past sand
Yeah
warm shaleBOT
Merineth
fallow widget
I'm not sure if i can go further than this
warm shaleBOT
fallow widget
because f(x) = x^1/2
f'(x) = 1/(2x^1/2)
past sand
It's ok just put a 2 here
past sand
because f(x) = x^1/2
f'(x) = 1/(2x^1/2)
Yes but divided by 2 since this is right
past sand
So now to simplify this, you need the same denominator
fallow widget
$\frac{x^2}{\sqrt{25-x^2}} + 2$
warm shaleBOT
Merineth
fallow widget
That would be my final answer
past sand
No
fallow widget
past sand
Not sure what you tried to do here lol
fallow widget
i extend with 1 / sqrt(25 -x^2)
leaving the right term with +2
and left denominator one with 2*sqrt[25-x^2)
cancel out the 2:s in nominator and denominator
past sand
I don't understand
fallow widget
I'll show, one sec
i see what i did wrong
..
Hold on
Okay my final answer is
2/25 + 2
52/25
past sand
That's definitely wrong
past sand
Again you're doing all sorts of incorrect simplifications
$\frac{-2x^2}{\sqrt{25-x^2}} + 2\sqrt{25-x^2} \implies \frac{-2x^2}{\sqrt{25-x^2}} + \frac{2(25-x^2)}{\sqrt{25-x^2}}$
warm shaleBOT
Nel
past sand
$\implies \frac{-4x^2 + 50}{\sqrt{25-x^2}}$
warm shaleBOT
Nel
past sand
,w derivative 2x sqrt(25-x^2)
warm shaleBOT
past sand
Now set that to 0 and solve for x
past sand
$\sqrt{25-x^2} = \frac{\sqrt{25-x^2} \sqrt{25-x^2}}{\sqrt{25-x^2}} = \frac{25-x^2}{\sqrt{25-x^2}}$
warm shaleBOT
Nel
past sand
In general, sqrt(x) = x/sqrt(x)
fallow widget
But i didn't do
I did 1 / sqrt 25-x^2
$\frac{1}{\sqrt{25-x^2}}(\frac{-2x^2}{\sqrt{25-x^2}} + 2\sqrt{25-x^2})$
past sand
Well no that's just wrong
You can't just multiply A by B and expect the result to still equal A
warm shaleBOT
Merineth
past sand
Yeah but you're still doing A = A * B for no reason
You can only do that if B = 1
And 1/sqrt(25-x^2) does not satisfy that
fallow widget
$\frac{1}{\sqrt{25-x^2}} * \frac{-2x^2}{\sqrt{25-x^2}} = \frac{-2x^2}{25-x^2}$
warm shaleBOT
Merineth
fallow widget
But we agree on this?
past sand
Yeah
fallow widget
$\frac{1}{\sqrt{25-x^2}} * 2\sqrt{25-x^2}) = 2$
warm shaleBOT
Merineth
fallow widget
And we agree on this?
past sand
Yes if x=/=5
fallow widget
aaaaaaaaaa
Wait i think i understand
is it because when we have the same denominator
both terms will have the same divisor
I was trying to use the rules of * but it was + between them
let me try again
past sand
Right
fallow widget
-4x^2+50
why do we get a denominator
$\frac{-2x^2 * \sqrt{25-x^2} + 2 * \sqrt{25-x^2} * \sqrt{25-x^2}} {\sqrt{25-x^2}}$
warm shaleBOT
Merineth
fallow widget
This is what i have from the start right?
warm shaleBOT
fallow widget
This is what we have?
Ok
i finally got the right answer
holy smokes that was hard
fallow widget
Makes sense
So i put in the equation = 0
and got sqrt(12,5)
Meaning we have:
x1 = sqrt(12,5)
x2 = -sqrt(12,5)
My final answer is that A = 25
Do you think that sounds reasonable?
past sand
,w max 2x sqrt(25-x^2)
warm shaleBOT
past sand
Yeah
fallow widget
let's gooo!!!
Jesus this one was rough haha
Once again Nel, thank you so much for your help. I apprecaite it more than you know ♥️
My SO says the same
past sand
soft bear
can someone help me find the function for the integral? that's all i need
info: a circular cylinder, with a diameter of 12.0 cm and a height of 5.0 cm, is now a body that from the side looks like the picture below. Andreas observes that the "profile" of the lid, and also of the bottom quite precisely can be described by the graph of a quadratic function: 𝑦 = 𝑎𝑥² + 𝑏𝑥 + 𝑐. Calculate the increase in volume in percent, when the can is swollen so that it bulges out 1.0 cm on each side while the diameter and edge height are unchanged.
obtuse pebbleBOT
can someone help me find the function for the integral? that's all i need
inf...
@soft bear Has your question been resolved?
soft bear
<@&286206848099549185>
how do i find the function for the integral so i can calculate the increase in volume
i think the rotation volume is at the y axis since the can swolls from the top and bottom. but what about the other values?
ok but how do i get the interval? is it 0 and 1 or -1 and 1? and the function?
<@&286206848099549185>
:/
soft bear
can someone help me find the function for the integral? that's all i need
inf...
the information about the problem is here. how do i get the function for the integral as well as the interval for this
can the lower integral be negative? example: $\int _{-1}^1:$
warm shaleBOT
Crawling Ham
warm shaleBOT
Crawling Ham
soft bear
meaning we need to get x from 𝑦 = 𝑎𝑥² + 𝑏𝑥 + 𝑐. right?
sacred barn
My hint is to use symmetry to consider only the "upper" integral
Then, choose a coordinate system that will make things simple. I.e. cylindrical coords with z=0 at the intersection of the lid and the rest of the can
soft bear
huh
sacred barn
The lid and the bottom are the same, so the volume of their sum is 2*volume of lid/bottom
soft bear
i ned to get the volumes seperately? 2 integrals?
soft bear
but how do i get the volume if i don't know the function x for the formula
sacred barn
You do. They give you that the height of the shape is quadratic
soft bear
that's for y. the rotation volume is around the y axis, so i need x
i know the answer, but not the solution. will it help if i showed the answer?
sacred barn
that's for y. the rotation volume is around the y axis, so i need x
If you want to call the height y, then you know y as a function of x
That's all you need to calculate the volume
soft bear
i don't think i can. see
sacred barn
Look up the shell method for solids of revolution
You absolutely can
And indeed, that is probably the easiest way to solve this problem since solving for x(y) as a function will be annoying
soft bear
should it look like this then? $\pi \int _{-1}^1\left(ax²+bx+c\right):::dy$
warm shaleBOT
Crawling Ham
sacred barn
no?
soft bear
then how
sacred barn
Read the link I sent
The shell method is usually taught along with the disk/washer method
soft bear
we haven't been taught that. just the rotation volumes i posted
sacred barn
Have you done all the readings? The shell method is almost always taught with the disk/washer method
They are very similar concepts
You should be able to learn it in 10 min with the link I sent
soft bear
ok. be back in 10 mins
soft bear
They are very similar concepts
i don't quite understand how i'm supposed to use that in this context
sacred barn
I've drawn the differential rectangle here
You integrate over x from 0 to the point where y intersects with the x-axis
soft bear
so?
sacred barn
Here's a very complete picture of the geometry and how the shell method would work here. I have to go, but hopefully you can use the link I sent and this picture to solve your problem
soft bear
<@&286206848099549185>
karmic palm
<@&286206848099549185>
I'm failing maths gng
soft bear
lmao. i barely passed last time
karmic palm
lmao. i barely passed last time
what grade u in
tardy epoch
https://math.libretexts.org/Courses/Monroe_Community_College/MTH_211_Calculus_II...
Did you read this link
soft bear
yeah
r(x) is the radius so it will be 6 cm. h(x) is the height = 5 cm.
$V=2\pi \int _a^b:r\left(6\right)h\left(5\right)dx$
warm shaleBOT
Crawling Ham
soft bear
Did you read this link
am i right so far?
tardy epoch
Why do you think the height is static
tardy epoch
can someone help me find the function for the integral? that's all i need
inf...
There's a bulge
soft bear
is it 7 then? or 6? idk
tardy epoch
You're calculating an increase in volume right?
soft bear
yeah
tardy epoch
You're told to model the bulge with a parabola
So you need to find the equation of the parabola
soft bear
"Andreas observes that the "profile" of the lid, and also of the bottom quite precisely can be described by the graph of a quadratic function: 𝑦 = 𝑎𝑥² + 𝑏𝑥 + 𝑐". so the equation is 𝑦 = 𝑎𝑥² + 𝑏𝑥 + 𝑐
hello?
<@&286206848099549185>
karmic palm
"Andreas observes that the "profile" of the lid, and also of the bottom quite pr...
okay
<@&286206848099549185> I need help with my 8th grade math homework
soft bear
solve the problem if you're so confident
deep summit
Дана функция f(x). Исследуйте её поведение при x → ±∞.
obtuse pebbleBOT
Дана функция f(x). Исследуйте её поведение при x → ±∞.
brazen gorge
Дана функция f(x). Исследуйте её поведение при x → ±∞.
mind providing a translation?
deep summit
The function f(x) is given. Investigate its behavior at x → ±∞.
deep summit
I don't really understand how to do it, since I'm a beginner in this
brazen gorge
I don't really understand how to do it, since I'm a beginner in this
will russian be better?
deep summit
no, thank you, I understand everything, thank you again, I will ask you for help, because you are very kind
fickle oracle
ty kind @brazen gorge 😄
deep summit
Prove the equivalence of the functions, then justify the corresponding approximate equality,
and use it to calculate the approximate number:
wouldn't it be too rude to ask you to solve this? The fact is that I'm a student and I'm a little late on deadline
deep summit
I'm solving another task
I haven't started this yet, but I'm in a hurry, so I'm unlikely to have time to start it (
deep summit
I haven't started this yet, but I'm in a hurry, so I'm unlikely to have time to ...
therefore, I ask for help from those for whom it is easy
obtuse pebbleBOT
@deep summit Has your question been resolved?
tulip hornet
Am i right?
fickle oracle
@mild sinew yo
obtuse pebbleBOT
<@884756951332974654> yo
fickle oracle
so if we calc all the 5 digits numbers we will get 2160 k?
fickle oracle
u agree with that?
mild sinew
yes
fickle oracle
k now lets calc all the odd numbers
we will have:
5 * 5 * 4 * 3 * 3
which is 900
now in order to find the even numbers we can just substract:
2160 - 900 =
which is 1260
fickle oracle
good?
mild sinew
yeah
mild sinew
well i could do by subtracting method
fickle oracle
explain
mild sinew
but cant we directly
fickle oracle
we can
mild sinew
find out the even number
fickle oracle
yh we can
mild sinew
can you tell method
fickle oracle
we will have something like this look
k if the last digit is 0:
then we will have :
6 * 5 * 4 * 3 * 1
you agree?
mild sinew
why at the the last place you put 1
i mean there are like 4 number that can satisfy for the even number criteria
so cant it be 4 ??
fickle oracle
if the last digit not zero then it will be : 2 or 4 or 6
so we will have:
5 * 5 * 4 * 3 * 3
its 1 because i told you
if the last digit is 0
so its only 1
I said it is zero
fickle oracle
now u can only do the +
5 * 5 * 4 * 3 * 3 = 900
6 * 5 * 4 * 3 = 360
so 900 + 360 = 1260
mild sinew
i see it now
fickle oracle
good? 🙂
soft bear
A circular cylinder, with the diameter 12.0 cm and the height 5.0 cm is now a body that looks like the picture from the side below. note that the "profile" of the lid, and also the bottom, can be described quite precisely with
the graph of a quadratic function of the type: 𝑦 = 𝑎𝑥² + 𝑏𝑥 + 𝑐.
Calculate the increase in volume in percent, when the jar swells so that it bulges out 1.0 cm on each side
while diameter and edge height are unchanged
obtuse pebbleBOT
A circular cylinder, with the diameter 12.0 cm and the height 5.0 cm is now a bo...
brazen gorge
they really put surströmming into the math problem 😵💫😵💫😵💫 the math problem stinks now
fickle oracle
lol
soft bear
apperently people were talking about this same problem in forums back in 2008. their thought process makes no sense tho
brazen gorge
anyways
brazen gorge
A circular cylinder, with the diameter 12.0 cm and the height 5.0 cm is now a bo...
do you know how to convert the 1 cm bulge out to a mathematical formula? hint: vertex form
soft bear
i dont
brazen gorge
i dont
well, the bulge means that the maximum of the parabola is at y=1
we will take these volumes one by one
so let's start with one of the bulges
fickle oracle
Use the general form of a quadratic function in vertex i would say
brazen gorge
precisely
brazen gorge
well, the bulge means that the maximum of the parabola is at y=1
so we have a(x-b)^2 + 1
for the vertex form
for the sake of simplicity, we can assume that b is 0
(the parabola is centered)
so we have ax^2+1
a must be negative
since the parabola is down
@soft bear i might be going quick so tell me if you are confused by something
soft bear
which parabola? the one facing down is the bottom one?
brazen gorge
here's what we have now
brazen gorge
which parabola? the one facing down is the bottom one?
the one facing down is the top one
soft bear
ok
brazen gorge
we are going to consider just one because the other is a copy of the first
soft bear
yeah
brazen gorge
here's what we have now
so we need to find a value of 'a' such that the roots are at 6 and -6
(the distance between them is 12)
makes sense?
soft bear
yeah
brazen gorge
so if we substitute x=6, we need to have a root in our equation, meaning it's going to be zero
so 36a+1=0
solve for a
brazen gorge
yes, with a minus in front
brazen gorge
saw it in the forums
ah lol
anyways
we have a parabola -1/36 x^2 + 1
we need to find volume of revolution around y axis
soft bear
yep
brazen gorge
do you know how to do that?
brazen gorge
yep
in general, what is the part you are struggling with?
i feel like you already know all of this / read it on the forum
soft bear
i read it but i don't get how they got this.
translation: You should be able to select the absolute bottom of the bulge as the origin. This is then exactly in the middle of the can in the x direction. Now you can write the quadratic function as y = ax² (this because the quadratic touches but does not intersect at the origin. You know that when x=6cm (farthest to the right of the can), the value of y (the difference in height) is 1cm.
1 = 36a => a = 1/36.
So now you have to find the volume of rotation of the function y = x² / 36 between x=-6 and x=6.
brazen gorge
i read it but i don't get how they got this.
translation: You should be able to...
they did essentially the same thing as me
in fact exactly the same thing
just for the bottom bulge
soft bear
why is the x² divided by 36?
soft bear
oh
brazen gorge
multiplicera med 1/36 är samma sak som dividera med 36
soft bear
yeah true lol
brazen gorge
so any other concerns, then?
soft bear
will the integral look like this? $\pi \int _{-6}^6:\left(\frac{x²}{36}\right)dy$
warm shaleBOT
Crawling Ham
warm shaleBOT
artemetra
soft bear
it does that with exponents for some reason
brazen gorge
will the integral look like this? $\pi \int _{-6}^6\:\left(\frac{x²}{36}\right)d...
uhh no
take a look at your formula for volume of revolution (around the y axis)
soft bear
$\pi \int _{-6}^6:\left(\frac{x^2}{36}\right)^2dy$
warm shaleBOT
Crawling Ham
soft bear
not **x** axis?
hello?
obtuse pebbleBOT
@soft bear Has your question been resolved?
obtuse pebbleBOT
@soft bear Has your question been resolved?
soft bear
<@&286206848099549185>
forest sinew
are you able to reframe this problem so i can come in and help
theres a lot of discussion
what is the problem as you are stuck on it now
soft bear
the volume of revolution around the y axis.
forest sinew
of what?
soft bear
we need to find volume of revolution around y axis
we got here
soft bear
A circular cylinder, with the diameter 12.0 cm and the height 5.0 cm is now a bo...
of this
the bulge on the top has the parabola -1/36 * x² + 1. and the bottom has x²/36
forest sinew
sorry im trying to catch up
soft bear
no worries
forest sinew
the bulge on the top has the parabola -1/36 * x² + 1. and the bottom has x²/36
so the top bulge is the same as the bottom right
yes
,w plot -1/36 x^2 + 1
warm shaleBOT
forest sinew
so, we can just use the top bulge
soft bear
isn't it? is the problem the function or the dy?
and here is my current problem
forest sinew
hmm i think what youre describing there is about the x axis
soft bear
how do i get the x for the y-axis then?
forest sinew
well we can make it the other way around, is the way id probably do it
make it x=function of y
soft bear
x² = y/36?
warm shaleBOT
jan Niku
soft bear
$+=6\sqrt{-y+1}$
warm shaleBOT
Crawling Ham
soft bear
is this incorrect
forest sinew
,w plot 6 sqrt(1-y) = x
warm shaleBOT
forest sinew
looks good to me
okay, we moved to y-world
well, i guess we can back up a little
when were doing this solids of revolution thing, do you believe the method?
like if i say were moving along the y axis and getting radii
do you get what i mean
soft bear
no
soft bear
yes
forest sinew
so what we wanna think of when were getting volume is that were adding up a bunch of little circles
that are all stacked on top of each other
wish i could find a nice image
soft bear
yeah?
forest sinew
kinda like this
so what you wanna do, is move along the axis youre rotating about
and you grab the radius
calculate the area of a circle with that radius
and move on to the next circle
does that make sense?
soft bear
yeah
forest sinew
so here were moving along the y axis
because y is our independent variable
and we have a function now
forest sinew
**Crawling Ham**
here
soft bear
yep
forest sinew
this gives us our radius, given some y
so we can get the area of all these little triangles
$\int \pi \qty(R (y) ) ^2 \dd y$ say
warm shaleBOT
jan Niku
forest sinew
but what are the bounds?
soft bear
-6 and 6
soft bear
idk
forest sinew
look at our picture
we wanna move along the y axis
as bounded by the x axis and the top of the can
soft bear
yea
forest sinew
so, where should we start?
soft bear
0
soft bear
1
warm shaleBOT
jan Niku
forest sinew
Nothing to it but to do it now
whats your radius function?
plug it in, square it, and integrate
brb
back
warm shaleBOT
Crawling Ham
soft bear
you put y instead of x tho
forest sinew
$R(y) = 6 \sqrt{1-y}$
warm shaleBOT
jan Niku
forest sinew
you put y instead of x tho
so?
soft bear
y = -1/36*x²+1 and x = 6 square(1-y)
forest sinew
so
soft bear
so i used the y
forest sinew
we want to integrate out the independent variable
soft bear
but i guess i missunderstood
forest sinew
see here
were thinking of the radius from the y axis
thats why it makes sense our function is a function of y
independent y
we pick a y, and R(y) gives us the radius there
so, we can just loop over all the y values, getting the radius, using pi r^2 to get the area
add them all up
but R is a function of y
soft bear
i get that now
forest sinew
we chose both like
we chose bounds of y
and were really using dy as the height of these circles
so were restricted in this choice of the radius function
it has to be a function of y
so what you want is $\int _0 ^1 \pi R(y) ^2 \dd y$
warm shaleBOT
jan Niku
forest sinew
where $R(y) = 6 \sqrt{1-y}$
warm shaleBOT
jan Niku
forest sinew
can you try this integral?
soft bear
$\int _0^1\pi \left(6\left(6\sqrt{1-y}\right)\right)^2dy$
warm shaleBOT
Crawling Ham
forest sinew
extra 6
warm shaleBOT
Crawling Ham
soft bear
like that?
forest sinew
okay, finish it out
soft bear
it says $18\pi $
forest sinew
it says 
soft bear
wdym
forest sinew
,w Integrate[ Sqrt(1-y) ^2, {y,0,1} ]
warm shaleBOT
soft bear
you forgot the pi?
soft bear
increase in volume when the can is swollen so that it bulges out 1.0 cm on each side while the diameter and edge height are unchanged.
soft bear
doesn't mention it
forest sinew
Pretend you are the mathematician in the scenario 
I measure the can and give you the measurements
you have to find everything else
forest sinew
but whats the volume?
is everything okay
we can find the volume of the can using some fancy method if you like @soft bear but its totally acceptable to just model it using a basic shape instead
soft bear
180 pi
soft bear
is it what we've calculated with the rotation volume?
forest sinew
what physical volume did we calculate when we did that calculation
what physically did 18 pi correspond to?
soft bear
oops it's 180π. it's the volume of the cylinder without the bulges
forest sinew
right yea
and we did the other calculation about the bulge and got 18 pi
what does 18 pi mean? what physically did we calculate when we did that
soft bear
i actually don't know
soft bear
with the bulges? or just the top bulge
forest sinew
heres what our cross section looked like
im just asking for you to connect the pieces together but maybe it wasnt clear what we were doing
or youve been working on this for too long 
lemme think uhh
soft bear
10 hours long
forest sinew
heres how i interpreted what you did
lets say were talking about the top
you were looking for a function to model the shape of the bulge at the top of the can
you found a function that gave you the height of the bulge above the top of the can
we used geometric reasoning to create an integral that gives the volume of this top bulge
then, we found the volume of the original can
does that help? @soft bear
so we have the volume of the bulge, from the stuff you and me did initially
you got 18 pi
and after that, we found the volume of a normal can, 180 pi
so whats the volume of the original can, plus a top bulge, and a bottom bulge?
soft bear
180π + 18π +18π = 216π
soft bear
the increase of volume in percentage
soft bear
216π/180π i recon to get the percentage?
forest sinew
i dont want to tell you yes or no
what answer does it give you?
does this do anything to confirm the work we did? or do we need to work something else out
soft bear
if we divide the original by the sum, we get 0.833...
forest sinew
if you calculate 216pi/180pi?
this ratio gives us a number
what does the number mean
soft bear
that's the total divided by the original. i don't think it will give the percentage
forest sinew
well lets think
say you divide 180pi / 180pi
lets say the volume didnt change at all
no bulge
180pi/180pi
this gives us 100%
what does it mean
soft bear
it means the volume neither increased nor decreased
warm shaleBOT
jan Niku
forest sinew
this with the thought that if new=old
we want 0% out
we can check this
$\frac{old}{old} -1 = 1-1 =0$
warm shaleBOT
jan Niku
forest sinew
okay, this is our working formula for percent change
apply our formula to new = 216pi and old=180pi
soft bear
so it's correct. 0.2 = 20%
forest sinew
soft bear
lmao
forest sinew
congrats
its not healthy to work on problems this long man
you accumulate too many pieces to keep track of whats going on
youll trick yourself into thinking youre dumber than you are
but you made it so same difference I guess
not that you asked for my feedback
soft bear
appreciate it regardless. i felt myself getting crazy after this long
forest sinew
yea, itll happen
soft bear
thanks very much. ❣️
forest sinew
soft bear
.close
quasi ferry
obtuse pebbleBOT
kind hawk
if h is any element of G_x, then ghg^-1 y = ghg^-1 g x = ghx = gx = y
so ghg^-1 is in G_y
jaunty reef
Hello, I am having trouble with this question:
Express m^2 in the form a(m-1)^2 + b(m-2)^2 + c(m-3)^2
I don’t really have a clue about what to really do with the coefficients
obtuse pebbleBOT
Hello, I am having trouble with this question:
Express m^2 in the form a(m-1)^2 ...
@jaunty reef Has your question been resolved?
jaunty reef
im completely lost lmfao
forest sinew
okay
well, lets starts here
you want m^2 right
just m^2, and nothing else
so lets look at $a(m-1)^2$
warm shaleBOT
jan Niku
forest sinew
when you expand this, what is it equal to?
jaunty reef
uhhh
a(m^2-2m+1)
forest sinew
so $am^2-2am + a$
warm shaleBOT
jan Niku
jaunty reef
yep
forest sinew
okay
can you see the problem?
specifically, say we only had this a term
and we want $m^2 = am^2 - 2am + a$
warm shaleBOT
jan Niku
forest sinew
whats the issue?
jaunty reef
There’s no m^2?
forest sinew
okay here lets step outside of this problem
lets look at a simpler example
lets say i want $x = a(x-1) + b$
warm shaleBOT
jan Niku
forest sinew
and you have to find a and b
heres one to do that
you could distribute the right hand side, and group it up in a different way, to get
$x = ax + (b-a)$
warm shaleBOT
jan Niku
warm shaleBOT
jan Niku
forest sinew
and we would take this equation to mean that $a=1$ and $b-a=0$
warm shaleBOT
jan Niku
forest sinew
does that make sense?
jaunty reef
epic
forest sinew
specifically
distribute, and group by powers of x
well, in your case, powers of m
can i ask uhh
we can use a calculator to do some algebra here
are you comfortable with that?
jaunty reef
Sure
forest sinew
i mean if you had to distribute and group without a calculator
you could accomplish that?
distribute $a(m-1)^2 + b(m-2)^2 + c(m-3)^2$ and group in powers of $m$ i mean
warm shaleBOT
jan Niku
jaunty reef
i don’t know how to do that sorry
forest sinew
you did one of them
forest sinew
uhhh
a(m^2-2m+1)
here
youd do something similar to each one
distribute all the way out
jaunty reef
I see
forest sinew
maybe its a good exercise
jaunty reef
Sorry the term tripped me up lol
forest sinew
ah
jaunty reef
Holdup my grandma’s calling me
obtuse pebbleBOT
@jaunty reef Has your question been resolved?
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
timid silo
Assuming T is temperature and t is time, how would i calculate the intervals of time when T=is less than 0
obtuse pebbleBOT
Assuming T is temperature and t is time, how would i calculate the intervals of ...
obtuse pebbleBOT
@timid silo Has your question been resolved?
timid silo
trying to compute the probability that a geometric series H ends at a value greate than or equal to n
got a negative probability...
obtuse pebbleBOT
trying to compute the probability that a geometric series H ends at a value grea...
still walrus
trying to compute the probability that a geometric series H ends at a value grea...
break it down first
timid silo
idk what that means
still walrus
yes you can lol
timid silo
(1-p)^(n-1)
oml
do you know where the proof went wrong tho
its pretty damn close
eh not important
.close
raw torrent
$\frac{sinxtan^2x+sinx}{y}=\frac{secx}{cscx}$
obtuse pebbleBOT
$\frac{sinxtan^2x+sinx}{y}=\frac{secx}{cscx}$