and like how do u get these equations in the first place
#help-27
sand isle
graceful cosmos
Oh wait, so does c2 lol
sand isle
yuh apparently so
acoustic leaf
writing $\vec u$ as a linear combination of $\vec v$ and $\vec x$ is by definition finding scalars $c_1,c_2$ such that [ \vec u = c_1 \vec v + c_2 \vec x ]
woven radishBOT
cloud
acoustic leaf
so then when two vectors are equal each of their components are equal
acoustic leaf
so [ c_1(v_1, v_2) + c_2(x_1, x_2) = (c_1 v_1, c_1 v_2) + (c_2 x_1, c_2 x_2) = (c_1 v_1 + c_2 x_1, c_1 v_2 + c_2 x_2) ]
woven radishBOT
cloud
acoustic leaf
setting each component equal gives you two equations
devout snowBOT
idle carbon
What is the question being asked 💀
slow summit
wait wrong one
devout snowBOT
idle carbon
Make a nee channel and ping me
devout snowBOT
Available help channel!
devout snowBOT
devout snowBOT
@unborn comet Has your question been resolved?
smoky gyro
i took the first two derivatives but im not sure what to do next, i found that f'(0) = 0 so im assuming im supposed to look for a pattern but is there a less tedious way other than to differentiate f(x) a bunch of times
devout snowBOT
i took the first two derivatives but im not sure what to do next, i found that f...
lost laurel
Well, do you know the formula for the taylor seires?
smoky gyro
yes
smoky gyro
0
acoustic leaf
I feel like f^{14}(0) should be 1/6
you just need to equate the coefficients of x^(14) given by the formula and the problem
the formula has a factor of n! to account for
restive river
Just use
$$\frac{d^m}{dx^m}x^n = n!$$
for $m=n$
For $m<n$, at x=0, the derivative is $0$
woven radishBOT
à뜜
restive river
d/dx (x) = 1 = 1!
d²/dx² (x²) = 2 = 2!
d³/dx³ (x³) = 6 = 3!
smoky gyro
so im looking for m = 14?
restive river
Yep
smoky gyro
would that just get me 14! then?
restive river
For x¹⁴ yes
For f(x), -14!/6
Because the higher order terms (greater than 14) don't matter here since we're computing f^(14)(0)
smoky gyro
For f(x), -14!/6
why is there negative sign and where does 1/6 come from?
restive river
Coefficient of x¹⁴ term in the given taylor polynomial is (-1/6)
So the higher order terms (m>14), like x¹⁸ term do not influence the derivative of the taylor polynomial (and hence the function f(x)) at x=0
smoky gyro
so we are equating the 14th term in the taylor polynomial to the 14th derivative of f at x=0?
woven radishBOT
à뜜
acoustic leaf
you are equating the 14th term in the actual taylor polynomial to the 14th term in the general formula given here
smoky gyro
**à뜜**
oh i never knew you could do this like for example take the nth derivative e.g 14th derivative of for example x^5 or x^7 but it makes sense why they become 0
smoky gyro
you are equating the 14th term in the actual taylor polynomial to the 14th term ...
wait i dont know if i quite understand so what -x^14/6 = 14! ?
restive river
You're equating the coefficients:
(-1/6) (from given taylor polynomial) = f^(14)(0)/14! (from formula)
smoky gyro
You're equating the coefficients:
(-1/6) (from given taylor polynomial) = f^(14...
wait so then why isnt 14! on the denominator for f^(14)(0) ?
acoustic leaf
because you have to multiply both sides by 14! to solve for f^(14) (0)
restive river
wait so then why isnt 14! on the denominator for f^(14)(0) ?
water you doing 😭
smoky gyro
because you have to multiply both sides by 14! to solve for f^(14) (0)
ohh
i didnt know u could do that with derivatives
acoustic leaf
the value of the 14th derivative at 0 is a number like any other
devout snowBOT
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Available help channel!
vast eagle
can anyone help me with this proof , I tried reasoning by recurrence
solar minnow
do you know about probabilities ?
imagine 2 possibilities for each of the n elements of E, either you take it or you don't
so 2 × 2 × ... × 2 n times
which is the number of possible subsets of E
devout snowBOT
@vast eagle Has your question been resolved?
vast eagle
do you know about probabilities ?
I know about them but i think that's irrelevant
do you know wwhat does P(E) mean
solar minnow
the set of all subsets of E
and you fan try a proof by induction if this "combinatorics approach" didn't satisfy you
vast eagle
I have an idea, I wanna know ur thoughts
if we have a set E with n elements and we add to them an element , let's name it x° , then the set of all subsets of E will double because each subset will give itself and (itself + x°)
what do u think
solar minnow
that's the idea
eternal aspen
devout snowBOT
feral agate
Let $g(x)=f(x)+1-2^{f(x)}$, then what you want to prove is $g(x)>0$ for all $x>0$.
woven radishBOT
kheerii
feral agate
I think that should also work yeah
eternal aspen
But I can’t prove that it is >0
feral agate
If you can prove g(u) > 0 for all positive u and also prove f(x) is positive for all x > 0 then you're done
feral agate
But I can’t prove that it is >0
Prove that g(0)=0 and that g is an increasing function on (0, inf)
Actually wait
eternal aspen
Yeah that’s what I am trynna do , but i can’t get it to work
( the increasing function part)
feral agate
Ahh no we can't do that
feral agate
( the increasing function part)
Because it isn't increasing on (0, inf)
oak bane
But I can’t prove that it is >0
find the interval(of u) when it is >0 and show range of f(x) is that interval
eternal aspen
find the interval(of u) when it is >0 and show range of f(x) is that interval
I can’t find the interval of u
eternal aspen
Have to be a non graphic solution
oak bane
oh
oak bane
Have to be a non graphic solution
well then show that 2^x increases faster than x+1 for
and they intersect at x=1
eternal aspen
well then show that 2^x increases faster than x+1 for
and they intersect at x=1
That’s also shown from the graph no?
oak bane
That’s also shown from the graph no?
its easier from the graph but u can say that from the derivatives
eternal aspen
So do it u+1 > 2^u?
devout snowBOT
@eternal aspen Has your question been resolved?
north mountain
@eternal aspen try to use the derivative of g(t) = t - 2^t +1 ; with t = f(x)
just note that f(x) is in (0, 1)
humble tundra
try to find the derivative of f(x) and 2^f(x)-1
then set f'(x) to be greater than the other
then see if the solution set is x>0
devout snowBOT
@eternal aspen Has your question been resolved?
karmic bane
Question:
karmic bane
fin p and q when,
y = 3xp^p - qx^2
if dy/dx = 12x^q+1 - 4x
I found out that q = 2, but I have difficulty in finding p
pseudo basin
$3xp^p - qx^2$?
woven radishBOT
Ann
woven radishBOT
Ann
karmic bane
if dy/dx = 12x^q+1 - 4x
wait no
karmic bane
maybe send a picture
I can't, sorry!
pseudo basin
ok so the derivative of 3x^p is 3px^(p-1) yeah?
pseudo basin
I can't, sorry!
no camera?
karmic bane
no camera?
Yea
karmic bane
ok so the derivative of 3x^p is 3px^(p-1) yeah?
no it's 12x^q+1
karmic bane
if dy/dx = 12x^q+1 - 4x
this is dy/dx value☝️
karmic bane
you are meant to *match* these two.
yea and even replaced q with it's value to get: 12x^3
so it's prolly: 3x^p -> 12x^3
I have zero clue if this is right, even if it is idk where to proceed
@pseudo basin pls help
anyone?
<@&286206848099549185>
humble tundra
fin p and q when,
dy/dx = 3px^(p-1) - 2qx
dy/dx = 12x^(q+1) - 4x
you said you found that q = 2
so just sub in back to the equations
then compare both expressions and find the value of p which makes them equal
karmic bane
so just sub in back to the equations
that is where I am stuck
karmic bane
then compare both expressions and find the value of p which makes them equal
precisely where I am stuck
karmic bane
3x^p -> 12x^3
humble tundra
3x^p is the original equation
but the given is the derivative
so you need to differentiate it first
then compare
karmic bane
so you need to differentiate it first
i have to differentiate 3x^p?
karmic bane
ok thanks
karmic bane
sorry, was busy irl
i understand ann, bye!
devout snowBOT
@karmic bane Has your question been resolved?
crude frigate
Why is this true?
devout snowBOT
Why is this true?
devout snowBOT
devout snowBOT
Available help channel!
restive river
devout snowBOT
supple knot
Your graph is tan, not inverse tan
restive river
oh sorry
giv eme a sec
well still, the multiple choice doesn't include arctan at all
olive snow
Still not arctan
restive river
wait
olive snow
Whats the defintion of arctan
restive river
i thought both arctan and tan^-1 represent the inverse tan function?
olive snow
Yes
But what is the defintion of arctan
Where it takes values
And what does it output
Like set
Domain and range
restive river
it gives you like
so arctanx = y, and tany = x .. the domain and range is -inf, inf and -pi/2, pi/2
olive snow
Ok so is the range of the thing you draw is
-pi/2, pi/2?
restive river
;_; no it's -inf, inf but i recall what i drew being the inverse tan function, is it not?
olive snow
Ok
For tan having an inverse
Its restricted on -pi/2, pi/2
So you have the one on the middle and thats it
restive river
oh omg
olive snow
To have the inverse, it means that your inverse is symetric to the function wrt y=x
In terms of graph
restive river
yes, i see, but in this case what exactly did i draw ? i do recall it being inverse tan, or am i getting it mixed up with something erlse
devout snowBOT
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Available help channel!
real belfry
ive hit a brick wall in terms of substitution and im not sure where ive gone wrong
devout snowBOT
ive hit a brick wall in terms of substitution and im not sure where ive gone wro...
@real belfry Has your question been resolved?
real belfry
<@&286206848099549185> if any of you are good with differential equations, your help would be much appreciated :'D
devout snowBOT
@real belfry Has your question been resolved?
devout snowBOT
@real belfry Has your question been resolved?
wicked aurora
@restive river bro just remember the range of arctan
wicked aurora
You did -tanx
like you mean cotx?
restive river
<@456226577798135808> bro just remember the range of arctan
??
restive river
oh iim knowii just don't know why im being pinged 2 hours later
restive river
ye dw
wicked aurora
also tan inverse in a one-one function
and always monotonically increasing
yk the 1st graph you drew had multiple values of x (x axis) giving the same values (y)
real belfry
.close
devout snowBOT
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Available help channel!
bleak arrow
Find the volume between z = sqrt(2-x^2-y^2) and z = sqrt(x^2+y^2). How do we find the limits for z, without 3D graphing? If we set the lower limit as sqrt(x^2+y^2), our region would be 1<x^2+y^2<2. If we set the lower limit as sqrt(2-x^2-y^2), our region would be x^2+y^2<1. Which option is correct and why?
devout snowBOT
Find the volume between z = sqrt(2-x^2-y^2) and z = sqrt(x^2+y^2). How do we fin...
restive river
Find the volume between z = sqrt(2-x^2-y^2) and z = sqrt(x^2+y^2). How do we fin...
x²+y²≤2 not x²+y²<1 in the end
bleak arrow
nope it's x^2+y^2<=1, i checked it on desmos
restive river
Sqrt(2-(x²+y²)) = z
Since z is Real, it implies x²+y²≤2
bleak arrow
sqrt(x^2+y^2) <= sqrt(2-x^2-y^2)
x^2+y^2 <= 2-x^2-y^2
x^2+y^2 <= 1
2-x^2-y^2>=0, x^2+y^2 <= 2
Final result: x^2 + y^2 <= 1
restive river
Oh mb the region yeah
bleak arrow
which option is correct? the hemisphere as the lower limit or the cone as the lower limit? how can we know without graphing
restive river
Had to go somewhere sorry @bleak arrow
So your approach is on the right way
Rather, find the intersection of the cone and hemisphere
which you kind of did, x²+y²=1
bleak arrow
I set the lower bound < the upper bound
restive river
So now you're sure that for x²+y²≤1, some volume of cone is contained in hemisphere, and for x²+y²>1, the remaining volume of hemisphere is contained within the cone
bleak arrow
there are two possible regions:
If the cone is the lower bound, we get 1<=x^2+y^2<=2
If the hemisphere is the lower bound, we get x^2+y^2<=1,
I assume the second case is correct, but why?
restive river
It's both since you need to find the volume contained in both regions
bleak arrow
one can't be the lower and upper limit at the same time
restive river
If you visualise the common volume is like an ice cream right?
Ice cream cone + filling
So the common volume is (ice cream cone) + filling informally
restive river
So now you're sure that for x²+y²≤1, some volume of cone is contained in hemisph...
Basically this
bleak arrow
Here is the region. You only need one triple integral to do it. The cone is below the hemisphere.
I was trying to do it without graphing in 3D
restive river
The actual cone is not below or above the hemisphere (it extends indefinitely above and below on z axis), but some part of it is contained within the hemisphere, which is shown there
As for the solving part, let me check
bleak arrow
yeah we're looking at the volume enclosed
restive river
So when cone is contained in hemisphere:
$$I_1 = \int_{0}^{2\pi}\int_{0}^{1}\int_{0}^{r}rdzdrd\theta$$
woven radishBOT
à뜜
restive river
Any questions on this?
bleak arrow
doesn't z go from r to sqrt(2-r^2)?
restive river
No this is when cone is inside the hemisphere
quartz island
so from sqrt(2-r^2) to r
other way
if u want to do this without a graph, simply just find the intersection which u guys did
now u have ur integral set up
bleak arrow
why not go from sqrt(2-r^2) to r
it would still give me a bounded region
(im doing this without graphing)
quartz island
why not go from sqrt(2-r^2) to r
thats what i said bro
bleak arrow
i meant the opposite
corrected it
it seems to me u have to graph to determine the bounds
because both of them give a bounded region
quartz island
because u integrating from z=r up to z = sqrt(2-r^2)
bleak arrow
why not use the limits in reverse
bleak arrow
i see
quartz island
the region z=r starts at the origin z=0 and z=sqrt(2-r^2) is centered at the oriigin but had radius root 2
just counter intuitive to go the other way
anyway because u want to do this without a graph
you find the intersection which is x^2+y^2<=1
so now u have ur integral
bleak arrow
No. The intersection is x^2+y^2 = 1 (a circle). The projection, when the lower bound is sqrt(x^2+y^2), is x^2+y^2 <= 1.
anyways
quartz island
mb i mean your region of integration is x^2+y^2<=1
and then the rest is just basic integration
restive river
@bleak arrow apologies to ping you what's the answer?
devout snowBOT
<@733113940367638539> apologies to ping you what's the answer?
woven radishBOT
bleak arrow
yes
devout snowBOT
devout snowBOT
Available help channel!
signal granite
how do i do this
frozen aurora
how do i do this
!xy
devout snowBOT
how do i do this
Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.
frozen aurora
you didn't show neither the problem nor a solution
pliant swan
devout snowBOT
frozen aurora
,rccw
pliant swan
Is work right
woven radishBOT
frozen aurora
Is work right
what's the question?
pliant swan
frozen aurora
does c refer to side AB?
frozen aurora
Cause it’s opposite
why are you solving for AC then?
shut parrot
Use cosine rule..
shut parrot
Oh yeah, mb misread
restive river
np
shut parrot
81.2/sin 11.2 = c/sin131.6
C = 81.2 sin 131.6/sin11.2
frozen aurora
can't you solve for AB directly?
pliant swan
You can but I just did it this way
frozen aurora
$\frac{\sin(11.2)}{81.2} = \frac{\sin(131.6)}{AB}$
woven radishBOT
artemetra
frozen aurora
You can but I just did it this way
i mean
sure
,w sin(11.2 deg)/81.2 = sin(131.6 deg)/x
woven radishBOT
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Available help channel!
cursive shuttle
how did we end up with this expression(btw these r problems tht i didnt rlly take the time to understand and solve myself)thts why im askin now😭
devout snowBOT
how did we end up with this expression(btw these r problems tht i didnt rlly tak...
cursive shuttle
grand warren
the thing everyone knows
grim hound
do any of you need a affordable editor?
runic prawn
do any of you need a affordable editor?
no get out
cursive shuttle
wait why is it expressed like this tho
grand warren
wait why is it expressed like this tho
he just isolated the sine
runic prawn
wait why is it expressed like this tho
cos^2(t) + sin^2(t) = 1
cursive shuttle
why's there a 2 and a square root sign
runic prawn
which is the same as pythagoreas in this case
cursive shuttle
oh?
grand warren
why's there a 2 and a square root sign
isolate sin^2 and take square root
grim hound
no get out
ok have a nice day
potent tusk
err um ? uhhhhhhhhhhhh
devout snowBOT
err um ? uhhhhhhhhhhhh
void ember
yo bro
void ember
need some help? whats up?
potent tusk
i cant integrate ts no substitution in sight
queen hearth
Could you integrate sqrt(1 - x^4)?
void ember
yo bro, im still trying to decipher your diagram
potent tusk
Could you integrate sqrt(1 - x^4)?
no clue
devout snowBOT
@potent tusk Has your question been resolved?
acoustic leaf
that's an elliptic integral
no elementary antiderivative
,w integrate sqrt(1 + x^4) from 0 to 2
woven radishBOT
potent tusk
oh one of those integrals that pmo
lofty surge
on [0,1], there are some techniques here : https://math.stackexchange.com/questions/2311583/evaluating-int-01-sqrt1-x-4-d-x
potent tusk
on [0,1], there are some techniques here : https://math.stackexchange.com/questi...
well thanks
thanks cloud as well
.close
restive river
heyy
restive river
i got a 100% on the multiple choice section on my midterm
and
like 92% on my writing section
the mult choice section is worth 75%
the writing is 25%
how exactly . docicaltte usthis
deft herald
what
rare kernel
...
restive river
0.75 + (0.25*0.92)?
versed juniper
How exactly do I calculate this?
you do arithmetic
you can do 75 + 25 * 92 right? now just do it with decimal points
devout snowBOT
calm jolt
you do arithmetic
yo what anime is ur pfp from
restive river
yo what anime is ur pfp from
horimiya
calm jolt
oh ok
devout snowBOT
Available help channel!
green snow
I'm currently doing my homework and, I have completed the "easy ones", now I'm kind of confused about what they are asking of me.
devout snowBOT
I'm currently doing my homework and, I have completed the "easy ones", now I'm k...
lyric hornet
that tetrahedron will be bounded by the plane going through the points (3,0,0), (0,4,0), and (0,0,2) and is bounded by the first octant
you can use the said plane going through the three points to bound y
and the triangle formed by the projection of that plane onto the xz plane to bound x
green snow
it looks like it forms a 3d shape when formed, how is f(x,y,z) formed?
ig I don't need f(x,y,z) in this problem
lyric hornet
f(x,y,z) 🤔
green snow
so for v(x,z), I found a piece in my notes that my professor found that the line at which is in the xy plane offers the parameter of the dy integral
I tried this for this problem, but its saying its incorrect. I found that v(x,z) = -4/3x+4
lyric hornet
your professor is right but in this case you need the xz plane to offer the parameter for the dx integral
you need the plane to offer the parameter for the dy integral
green snow
you need the plane to offer the parameter for the dy integral
the xz plane?
lyric hornet
pardon in that msg I was referring to plane made by (3,0,0), (0,4,0, (0,0,2)
green snow
oh the top face plane
lyric hornet
top face? 🤔
green snow
like the plane that appears on top of the 3d shape
lyric hornet
ah yes, right
green snow
let me compute this real quick
ok
I found the second parameter for dx
0 <= x <= -3(z-2)/2
lyric hornet
and indeed that is :)
green snow
ah yes, right
is there a general formula for developing a equation of a plane for a triangle?
lyric hornet
there is... but you're better of knowing how to derive the plane for three points because it is much easier just outright using the formula
especially for calc III (which I assume this is??) bcz there are more important things to memorize
green snow
especially for calc III (which I assume this is??) bcz there are more important ...
yes
lyric hornet
do you remember from the beginning of calc III when you learned that a plane is defined by a point and a normal vector?
well to get the plane passing through these three points
we take the two directions vectors and cross them (this is the normal vector to the plane)
green snow
yes
lyric hornet
so how would this look in your case?
green snow
I think it would be crossing the vectors in the form of the edges of the plane
lyric hornet
I think it would be crossing the vectors in the form of the edges of the plane
yep, and you have three points so you can indeed cross two vectors on the plane
green snow
yep, and you have three points so you can indeed cross two vectors on the plane
Are these accurate vectors that describe edges: <3,4,0> X <0,4,2>
green snow
I'm using the edges in XY and ZY
lyric hornet
oh right sorry made names in my head
yes be careful tho
(\vec{YX}=\vec{x}-\vec{y})
woven radishBOT
PajamaMamaLlama
lyric hornet
and the common point must come first
green snow
and the common point must come first
so the equation of the line would be -8(x-3)-6(y-0)-12(z-0) = 0
lyric hornet
solve for y and you know that must be bound form below 0 as well
and thus we have our y-bound :)
green snow
yes!
green snow
solve for y and you know that must be bound form below 0 as well
Thank you so much!
Also a big fan of your name!
lyric hornet
thank you, got it from a buddy of mine 😂
thin geode
devout snowBOT
rustic jetty
$\sin(\theta) = \sin(\pi - \theta)$
woven radishBOT
Dork9399
pseudo basin
replace pi with 180° if you are allergic to radians
devout snowBOT
pseudo basin
doesnt wait for an answer
thin geode
.reopen
devout snowBOT
✅
thin geode
>doesnt wait for an answer
mybad i thouhgt you wernt going to add anything more. you got anything more?
any tips?
sudden yarrow
see how the lines overlap?
pseudo basin
well i was just gonna say yes it holds for all theta
but i was just surprised to see you not be patient enough to wait more than 2 min
thin geode
see how the lines overlap?
ya there the same
sudden yarrow
the expressions are equivelent then
like how Ann say, for any value of theta (in this case x)
thin geode
but i was just surprised to see you not be patient enough to wait more than 2 mi...
well i usually dont bother to stick around all too long. usually when 2 minutes has passed, many people just dissapears and dont bother to help no more. but im glad you are diffrent
thin geode
like how Ann say, for any value of theta (in this case x)
ya makes sense
sudden yarrow
also
you can think of sine as the height of the point
let me find an image to examplify
thin geode
ratioed with hypotenuse?
id like to ask another question once you find the image. this one, for a)
thin geode
ya that makes it more clearer
sudden yarrow
for a)
thin geode
yes
sudden yarrow
what is the sum of the internal angles of a triangle?
thin geode
180 or pi
sudden yarrow
so if one of them is 60 and the other is lets say phi, how do you relate the two with theta?
thin geode
180 - 60 - phi?
sudden yarrow
yep, theta would be 180 - 60 - [the other angle]
they don't give us much more information tho...
sudden yarrow
see? and you knew it too!
sudden yarrow
thats why it's important to train and practice lots
thin geode
true true
mind taking me thorugh the other two?
for b and c
so im thinking of 1/2 ab sin c as general formula
sudden yarrow
ummmm can't you the law of sines and express the sides and angles in terms of theta and 60º and solve from there?
thin geode
angle l is theta + 60 but maybe we use sine rule to find the sides?
sudden yarrow
sorry I'm not very confident with this
thin geode
sorry I'm not very confident with this
no worries ill try it out
i think they must of forgotten to write 30 for jl
there isnt a way to derive that if yyou only have two angle and nothing else
devout snowBOT
@thin geode Has your question been resolved?
opaque talon
4c why cant i plug it straight into my calculator
devout snowBOT
4c why cant i plug it straight into my calculator
wicked turtle
they're expressing the probability in terms of an N(0,1) random variable
if your calculator can handle an arbitrary mean and variance then you can just do it directly
opaque talon
uh yeah Ill show what i typed in
cause i dont get the right answer
it should be left tail isnt it
d is everything below that shaded part
oh wait
the area would be 1 - then
yh nope 💀
ah im so lost
what shoulda i tped in here
wicked turtle
you want P(X > d), shouldn't you be examining the right tail?
opaque talon
oh i thought to get d we look at the left
wicked turtle
oh wait, you said 0.99
for the area
so that should be equivalent to the right tail having area 0.01
what answer is it giving you?
opaque talon
54.99
Fuck
I need to wake up
I put in variance instead of SD
😅
i got i tnow
.close
devout snowBOT
opaque talon
.reopen
devout snowBOT
✅
devout snowBOT
@opaque talon Has your question been resolved?
toxic grove
I think the assumption here is that customers will complain if and only if the volume of liquid is less than some defined value
@opaque talon
So you just need to find the volume of liquid such that 10% of the time, the machine will dispense less than that
opaque talon
oh ok so like P(X<x) =0.1?
opaque talon
Yeah I got X = -1.28
but the mark scheme says
I need to standardise it
like this
toxic grove
Well yes, it asks for a volume of liquid not a number of standard deviations
opaque talon
oh ok so like P(X<x) =0.1?
Oh so solving for X is just the standard deviation
toxic grove
No
You've found the number of standard deviations less than the mean that will lead to a complaint
That's the equation you've solved
So to convert that into a volume of liquid you need to convert it
devout snowBOT
opaque talon
.reopen
devout snowBOT
✅
toxic grove
I think it means "fewer than 0.005 of men", aka "fewer than 0.5% of men"
It's not well phrased
Otherwise the answer just approaches infinity as the number of men approaches infinity
Which probably isn't the answer they're looking for because it's not helpful
sturdy mango
how?
sturdy mango
solving for a falling of a body in heliocentric reference frame without using Kepler's laws i found this relation wondering how to solve it
pseudo basin
what's y then i wonder
sturdy mango
real
pseudo basin
so you're trying to solve this rather nasty 2nd order DE then
sturdy mango
yeah i am wondering how to proceed
limits of integration are x= radius of earth to 0 and time=0 to any t
i know how to sovle first order and first degree differential equations only till now
pseudo basin
yeah i think this one is quite difficult to solve lmao
polar chasm
solving for a falling of a body in heliocentric reference frame without using Ke...
Neon helped me with this exact same thing in past
ill send you a link so that you can read through it
the first step was probably this
r = x, t = y
the solution is really ugly btw, i managed to only get a somewhat nice expression for the first derivative iirc
sturdy mango
the first step was probably this
this is trivial
polar chasm
this is trivial
well it gets you the expression for first derivative
polar chasm
after that, you can solve an extrmeely ugly integral and get to this
polar chasm
it introduces velocity
yeah, thats basically the first derivative
sturdy mango
im gonna send u what chatgpt sent i dont trust llms
so i wanna see if its correct
how to print latex here
polar chasm
I honestly dont know anything about DEs, i just sent you the link to share some of neon's wisdom lol
polar chasm
how to print latex here
$ latex $
sturdy mango
$
\begin{document}
% Step 1: Energy Conservation
For a particle falling radially from rest at ( r=R ), its total energy is
[
E = \frac{1}{2}\left(\frac{dr}{dt}\right)^2 - \frac{GM}{r}.
]
At ( r=R ), since (\frac{dr}{dt} = 0), we have
[
E = -\frac{GM}{R}.
]
Thus, the energy conservation equation becomes:
[
\frac{1}{2}\left(\frac{dr}{dt}\right)^2 - \frac{GM}{r} = -\frac{GM}{R}.
]
% Step 2: Solve for (\frac{dr}{dt})
Rearrange the equation:
[
\frac{1}{2}\left(\frac{dr}{dt}\right)^2 = \frac{GM}{r} - \frac{GM}{R},
]
[
\left(\frac{dr}{dt}\right)^2 = 2GM \left( \frac{1}{r} - \frac{1}{R} \right).
]
Since the particle falls inward (so (dr/dt) is negative), we write:
[
\frac{dr}{dt} = -\sqrt{2GM \left( \frac{1}{r} - \frac{1}{R} \right)}.
]
% Step 3: Separate Variables and Set Up the Integral
Separate variables to get:
[
dt = -\frac{dr}{\sqrt{2GM \left( \frac{1}{r} - \frac{1}{R} \right)}}.
]
To find the fall time ( t_{\text{fall}} ) from ( r=R ) to ( r=0 ), reverse the limits (which cancels the minus sign):
[
t_{\text{fall}} = \int_{r=R}^{r=0} -\frac{dr}{\sqrt{2GM \left( \frac{1}{r} - \frac{1}{R} \right)}}
= \int_{r=0}^{r=R} \frac{dr}{\sqrt{2GM \left( \frac{1}{r} - \frac{1}{R} \right)}}.
]
% Final Integral Expression
The final expression for the fall time is:
[
\boxed{t_{\text{fall}} = \int_{0}^{R} \frac{dr}{\sqrt{2GM \left(\frac{1}{r} - \frac{1}{R}\right)}}.}
]
\end{document}
$
woven radishBOT
The modest AI
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
polar chasm
https://discord.com/channels/268882317391429632/486841085210132490/1079810089457...
You can read it here
polar chasm
this looks nice
Well its only an expression for the first derivative
polar chasm
the first step was probably this
It's basically just this and then seperation or whatever is it called
v dv = -a / r^2 dr
and integrate both sides
that'll give you expression for velocity given r (or x in your case)
sturdy mango
got it
i had to use energy conservation
what the hell llm cooked
i never knew this
that this is a well known solution
devout snowBOT
@sturdy mango Has your question been resolved?
flat wolf
what is a general way to approach problem like this/
devout snowBOT
what is a general way to approach problem like this/
@flat wolf Has your question been resolved?
graceful cosmos
Do you know why the "common branch" Log(z) does not work?
devout snowBOT
@flat wolf Has your question been resolved?
opaque talon
devout snowBOT
supple knot
can someone explain to me 15 c
y is a whole number like 0, 1, 2, 3, ...
opaque talon
ahh I just tried using inverse normal to work out the area where its below 0.05
appar thats wrong
opaque talon
y is a whole number like 0, 1, 2, 3, ...
i have no clue what to do with that
supple knot
Find P(Y < y) until you get larger or equal to 0.05
if you have a calculator that finds inverse binomial that will get you within 1 of your answer
supple knot
Find P(Y < y) until you get larger or equal to 0.05
then do this
opaque talon
I thought we was using normal inverse
supple knot
yea inverse normal can give you the approximate solution
you transform Y to a z score then transform it back once you find the critical value
opaque talon
Yeah i tried that
I got z = -1.6
appar im wayyyy off tho
No clue what is wrong here
supple knot
I got z = -1.6
you didn't transform it back to y
your answer is in terms of z
the question asks for the answer in terms of y
this should look familiar
mu = mean, sigma = standard deviation
opaque talon
yeah no but i mean
opaque talon
I thought they used a instead of z here
and its not the same value i got
maybe im wrong about that tho
oh shoot
I got x = 166.24
or i mean
y = 166.24
supple knot
N is not defined anywhere
opaque talon
dyu reckon they j rounded it
supple knot
y is a whole number like 0, 1, 2, 3, ...
they did round because the question asks for a whole number
opaque talon
oooh right, how did u know it had to be a whole number
oh
cause of the contiunuity
hypothesis?
right?
supple knot
no idea what that is. it just follows because Y is binomial
opaque talon
hypothesis?
yeah i meant basically cause we converted from binomial to normal
ok thanks
.close
quaint rampart
Guys I’m cooked. For my test i need to know how to solve stuff on this lvl. Idk what to even do first
devout snowBOT
Guys I’m cooked. For my test i need to know how to solve stuff on this lvl. Idk ...
strange nimbus
[\frac{0,2^{x + 0,5}}{5} = \sqrt[3]{5} \cdot 0,04^{x - 2}]
woven radishBOT
Chai T. Rex
strange nimbus
OK, so the first thing to do is to get rid of what's added to and subtracted from x.
quaint rampart
So half and half-2?
strange nimbus
Yeah, you have x + 0.5 and x - 2.
quaint rampart
Ok
strange nimbus
Do you know how to do that?
quaint rampart
No
strange nimbus
OK, so (a^{b + c} = a^b \cdot a^c).
woven radishBOT
Chai T. Rex
strange nimbus
So, if you have (0,!2^{x + 0,!5}), how would you separate it using that rule?
woven radishBOT
Chai T. Rex
quaint rampart
0,2^x * 0,2^0,5
woven radishBOT
Chai T. Rex
quaint rampart
0,04^x * 0,04^-2
strange nimbus
[\frac{0,!2^x \cdot 0,!2^{0,5}}{5} = \sqrt[3]{5} \cdot 0,!04^x \cdot 0,!04^{-2}]
quaint rampart
The 0,2 stuff
woven radishBOT
Chai T. Rex
strange nimbus
OK, let's work to get all the x stuff on the left side and all the rest on the right side.
quaint rampart
Ok so basic equation stuff
strange nimbus
Right.
strange nimbus
You can write either cbrt(5) (for cube root of 5) or you can write 5^(1/3).
strange nimbus
Yes, but you can also move the 5 to the right side.
quaint rampart
U can?
strange nimbus
Yes, you multiply both sides by 5.
quaint rampart
Oh
strange nimbus
That gets it out of the denominator on the left side.
quaint rampart
How do u multiply the cube root of 5
strange nimbus
Oh, you would just leave that on the right side.
quaint rampart
Would u just add a *5 in the right?
strange nimbus
Right.
woven radishBOT
strange nimbus
[\frac{0,!2^x \cdot 0,!2^{0,5}}{5} = \sqrt[3]{5} \cdot 0,!04^x \cdot 0,!04^{-2}]
[\frac{0,!2^x}{0,!04^x} = \frac{5 \sqrt[3]{5} \cdot 0,!04^{-2}}{0,!2^{0,5}}]
woven radishBOT
Chai T. Rex
strange nimbus
Right, dividing by something is the same as multiplying it into the denominator.
quaint rampart
Yeah
strange nimbus
OK, we have another rule: (\qty(\frac{a}{b})^x = \frac{a^x}{b^x}).
woven radishBOT
Chai T. Rex
strange nimbus
How can we use that to get only one x on the left side?
quaint rampart
How can we use that to get only one x on the left side?
How would that help tho
Oh
U can divide then put x on the result
,w (0,2/0,04)^x
woven radishBOT
quaint rampart
Bru
strange nimbus
[\frac{0,!2^x}{0,!04^x} = \frac{5 \sqrt[3]{5} \cdot 0,!04^{-2}}{0,!2^{0,5}}]
[\qty(\frac{0,!2}{0,!04})^x = \frac{5 \sqrt[3]{5} \cdot 0,!04^{-2}}{0,!2^{0,5}}]
quaint rampart
What’s qty
woven radishBOT
Chai T. Rex
quaint rampart
There you go
strange nimbus
\qty(stuff) makes the parentheses big enough to surround what's inside.
quaint rampart
Oh
strange nimbus
One thing about Wolfram Alpha (the ,w stuff) is that it wants . instead of , as the decimal separator.
,w (0.2/0.04)^x
quaint rampart
Oh
woven radishBOT
quaint rampart
I see
strange nimbus
Oh, just left.
quaint rampart
0,2/0,04 = 20/4 = 5.
Mhm
strange nimbus
Now let's simplify the right.
quaint rampart
I forgot how to put the 5 inside the root
strange nimbus
Oh, you can say cbrt(5) or 5^(1/3).
strange nimbus
Or, for the bot, \(\sqrt[3]{5}\).
woven radishBOT
Chai T. Rex
quaint rampart
U can also swap the 0,04 to make exponent positive
strange nimbus
Right.
quaint rampart
Oh, you can say cbrt(5) or 5^(1/3).
What does this help with though
Can u multiply that
strange nimbus
[\qty(\frac{0,!2}{0,!04})^x = \frac{5 \sqrt[3]{5} \cdot 0,!04^{-2}}{0,!2^{0,5}}]
[5^x = \frac{5 \sqrt[3]{5}}{0,!2^{0,5} \cdot 0,!04^2}]
woven radishBOT
Chai T. Rex
strange nimbus
OK, one thing about that is that 0,2 and 0,04 are related.
strange nimbus
No, you need the same base to add the exponents.
quaint rampart
Uh I’d need to try multiplying a few times for that
strange nimbus
Well, you can turn it into a fraction to make it easier.
2/10
Squaring gives 4/100.
quaint rampart
How does that make it easier
strange nimbus
And what's 4/100 in decimal?
quaint rampart
0,04
strange nimbus
Right.
quaint rampart
U mean if u multiply 0,2 by itself?
strange nimbus
Right, that's what squaring is.
quaint rampart
Oh ok
strange nimbus
But see how that turns 0,2 into 0,04?
strange nimbus
Not yet.
quaint rampart
Oh
How do u get it then
U need to root by 4?
To get rid of the 2 and then have a square root left
strange nimbus
We can change (0,!04) to (0,!2^2).
woven radishBOT
Chai T. Rex
strange nimbus
So, if we do that, we get:
[5^x = \frac{5 \sqrt[3]{5}}{0,!2^{0,5} \cdot 0,!04^2}]
[5^x = \frac{5 \sqrt[3]{5}}{0,!2^{0,5} \cdot (0,!2^2)^2}]
quaint rampart
Would it just be 0,2^4
woven radishBOT
Chai T. Rex
strange nimbus
Yes, it would be.
quaint rampart
Ok
strange nimbus
So, what does the bottom simplify to?
strange nimbus
[5^x = \frac{5 \sqrt[3]{5}}{0,!2^{0,5} \cdot 0,!04^2}]
[5^x = \frac{5 \sqrt[3]{5}}{0,!2^{0,5} \cdot (0,!2^2)^2}]
woven radishBOT
Chai T. Rex
strange nimbus
We have this after we change (0,!04) to (0,!2^2).
woven radishBOT
Chai T. Rex
quaint rampart
Yeah
strange nimbus
So, you said we could simplify it to this:
[5^x = \frac{5 \sqrt[3]{5}}{0,!2^{0,5} \cdot (0,!2^2)^2}]
[5^x = \frac{5 \sqrt[3]{5}}{0,!2^{0,5} \cdot 0,!2^4}]
woven radishBOT
Chai T. Rex
quaint rampart
Yeah
strange nimbus
OK, and how can we further simplify the denominator?
strange nimbus
Right.
quaint rampart
U add then not multiply
strange nimbus
[5^x = \frac{5 \sqrt[3]{5}}{0,!2^{0,5} \cdot 0,!2^4}]
[5^x = \frac{5 \sqrt[3]{5}}{0,!2^{4,!5}}]
woven radishBOT
Chai T. Rex
quaint rampart
I don’t remember how to do the top part though
strange nimbus
Well, (\sqrt[n]{a} = a^{\frac1n}).
woven radishBOT
Chai T. Rex
strange nimbus
But we're not done with the bottom yet.
quaint rampart
Oh we need to multiply still
woven radishBOT
Chai T. Rex
quaint rampart
I mean multiply the exponent
strange nimbus
Oh, with what?
strange nimbus
You mean in the numerator?
quaint rampart
No
strange nimbus
Oh, let's not do that.
There's a better thing to do.
Let's change 0,2 to a fraction.
quaint rampart
2/10
strange nimbus
Right, and simplify that.
strange nimbus
Right, so:
[5^x = \frac{5 \sqrt[3]{5}}{0,!2^{4,!5}}]
[5^x = \frac{5 \sqrt[3]{5}}{\qty(\frac15)^{4,!5}}]
woven radishBOT
Chai T. Rex
quaint rampart
Yeah
strange nimbus
How can we move that to the numerator?
strange nimbus
Yes, but we'll see that that's a good thing here.
quaint rampart
Wait but would it even go to the numerator
strange nimbus
[5^x = \frac{5 \sqrt[3]{5}}{\qty(\frac15)^{4,!5}}]
[5^x = \frac{5 \sqrt[3]{5}}{5^{-4,!5}}]
[5^x = 5 \sqrt[3]{5} \cdot 5^{4,!5}]
quaint rampart
All good
woven radishBOT
Chai T. Rex
strange nimbus
There we go.
strange nimbus
Right.