$\vec{v}_1 = \vec{AB} = (x_2 - x_1)\mathbf{i} + (y_2 - y_1)\mathbf{j} + (z_2 - z_1)\mathbf{k}$
#help-27
dull parrot
woven radishBOT
MxRgD
dull parrot
and $\vec{v}_2 = \vec{AC} = (x_3 - x_1)\mathbf{i} + (y_3 - y_1)\mathbf{j} + (z_3 - z_1)\mathbf{k}$
woven radishBOT
MxRgD
white pawn
ah okay got it and for a vector to be on the plane it must be perpendicular to the normal vector?
dull parrot
ah okay got it and for a vector to be on the plane it must be perpendicular to t...
yes
basically $\mathbf{n} \cdot \vec{v} = 0$
woven radishBOT
MxRgD
dull parrot
where n is the normal
white pawn
Ah okay I see and for a point to be on a plane do you sub the points in the x+y+z=a dot n?
dull parrot
Ah okay I see and for a point to be on a plane do you sub the points in the x+y+...
yeah
white pawn
and it has to equal a dot n
dull parrot
yes
it's basically saying that every point (x, y, z) on a plane must form a vector that is perpendicular to the normal vector (a, b, c)
white pawn
because the r vector is the position vector on the plane, a vector is the direction vector and n is the normal
dull parrot
yep
white pawn
and for them to be on the same plane the direction vector and position vector must be normal to the plane
ohhhhh that makes so much sense
cause a plane is like an A4 piece of paper
dull parrot
yeah
white pawn
and you have a direction vector that goes along it
dull parrot
you can think of the normal vector as a “flagpole” sticking straight up out of the ground (the plane). Any line you draw on the ground will always be at a 90 degree angle to that flagpole, no matter what direction you draw it
white pawn
you can think of the normal vector as a “flagpole” sticking straight up out of t...
ahh I see
that makes a lot of sense
thank you so much
can I ask anothr question?
dull parrot
Sure
white pawn
why exactly does the cross product work in finding perpendicular vectors?
it kinda makes sense but like I dont think I have that full conceptual understanding of why it works
dull parrot
it kinda makes sense but like I dont think I have that full conceptual understan...
I guess one way you can think of it, is that the cross product is a way to find the “leftover” direction
If you have two vectors, they define a flat “floor” (a 2D plane)
In a 3D world, there is only one direction left that isn’t part of that floor: straight up or straight down and the dot product is basically used to find that
white pawn
ah okay but thetes no kinda proof on how turning into matrix form makes it perpendicular
dull parrot
There's an algebraic proof for it
white pawn
oh is thete
dull parrot
np
devout snowBOT
white pawn
.reopen
devout snowBOT
✅ Original question: #help-27 message
white pawn
.solve
dull parrot
was that accidental?
white pawn
How do I make it so it un occupied?
dull parrot
I think it's just .close again
coral zephyr
can i get help on this
devout snowBOT
can i get help on this
brave plaza
can i get help on this
what are you trying to do?
trail eagle
You've found a critical point in R so that's good.
Now you need to check the boundary.
What's the boundary of R?
coral zephyr
What's the boundary of R?
i mean idk tbh
coral zephyr
It's the boundary of a closed disk
trail eagle
Yes, so you want to investigate the extrema of f(x,y) on the boundary of R, namely when x^2 + y^2 = 4.
coral zephyr
Yes, so you want to investigate the extrema of f(x,y) on the boundary of R, name...
okay so next you want to test the boundries right
like the circle
trail eagle
yes
coral zephyr
yes
you wanna see if there exists f(x,y) thats a critical point such that it lies on the boundries of the circle
right
trail eagle
One thing you could try is to write points on the boundary of R in some other way.
So you need to write points (x,y) that satisfy x^2 + y^2 = 4 in some other way
coral zephyr
One thing you could try is to write points on the boundary of R in some other wa...
i was thinking of plugging x^2 + y^2 into f(x,y)
trail eagle
You can do that, but then you're left with a y in there.
You could solve for y in x^2 + y^2 = 4, but then you need to consider two parts to this boundary.
coral zephyr
You could solve for y in x^2 + y^2 = 4, but then you need to consider two parts ...
im tryna think of this similar to the last problem
like i had no issue with the last problem involving the triangle
and the sides
but idk whats different here \
trail eagle
You just want a way to parameterize the boundary essentially. The triangle one had to be split up because it had corners. This circle doesn't have corners, so you can parameterize it some way.
I would suggest considering x = 2cos(t) and y = 2sin(t).
coral zephyr
I would suggest considering x = 2cos(t) and y = 2sin(t).
hm i see
coral zephyr
I would suggest considering x = 2cos(t) and y = 2sin(t).
why did you make it t
trail eagle
You can call it whatever you want.
We just want a nice way of expressing the point on the boundary with one variable.
coral zephyr
We just want a nice way of expressing the point on the boundary with one variabl...
can i ask the intuition behind the way that the circle is represented that way?
for example i know that all points (x/r, y/r)
right
cos(t) = x/r , sin(t) = y/r
trail eagle
If you only had the unit circle, any (x,y) on the unit circle satisfies x=cos(t) and y = sin(t) for some t.
This is the same thing but scaled by 2.
t corresponds to the angle in radians around the circle, like on the unit circle.
coral zephyr
If you only had the unit circle, any (x,y) on the unit circle satisfies x=cos(t)...
yes but i mean the squaring and addition
like i can see why each point is cost sint
but what about cos^2(t) + sin^2(t)
trail eagle
You can think of them as such if you want.
Any coordinate is a vector.
We're just saying that the point (x,y) that lie on the circle x^2 + y^2 = 4 can be parameterized by (2cos(t), 2sin(t))
coral zephyr
You can think of them as such if you want.
why did you divide both sides by 2 tho
isnt this the direct trig representation
trail eagle
Your vector on this doesn't have magnitude 2.
coral zephyr
cos^2(t) + sin^2(t) is not equal to 4.
wait howww
wait were they supposed to be scaled
inside the sqrt
yeaahh
they were
trail eagle
Yes, so the points are at (2cos(t), 2sin(t)).
coral zephyr
Yes, so the points are at (2cos(t), 2sin(t)).
okay i agree
okay i see why the parameratization works now
trail eagle
So now you want to consider f(x,y) on this region, i.e. for f(rcos(t), rsin(t))
And find critical point on that.
coral zephyr
hope thats not wrong
do i even need to consider the inequality atp
because were just measuring the circle right
so we dont even need that right
trail eagle
You've defined those points so they lie on the boundary, so you don't need to check they're in R .
coral zephyr
its not that im checking
its that im representing it as an inequality still
which im saying isnt even right because we just wanna consider the circumference of the circle right
trail eagle
You don't need that. The points are on the boundary. The only thing you may want to do is ensure you're traversing the boundary once, so 0 <= t < 2pi.
coral zephyr
You don't need that. The points are on the boundary. The only thing you may want...
is this boundry you just wrote supposed to be for checking?
because if so cant you just plug the points into the circle
also that needs an angle
t
trail eagle
You use that to find critical points along the boundary yes.
f(rcos(t), rsin(t)) will give you a function of t.
You can find critical points of it.
trail eagle
No. Now you're just considering f(t, 4).
You need to compute $f(2\cos(t), 2\sin(t))$
woven radishBOT
Azyrashacorki
trail eagle
That's a function of one variable t.
So you take the derivative w.r.t. t and check where it vanishes.
devout snowBOT
@coral zephyr Has your question been resolved?
coral zephyr
That's a function of one variable t.
but like
its the shift between the original parametarization and the t parameterization thats kinda confusing
because technically even after parameterizing it with the trig were still considering that circle points from the output set of f(x,y)
so wouldnt you wanna deal with x and y
to find the instantaneous roc
trail eagle
Every point on the boundary corresponds to some t in the parameterization.
We want to find critical point of f(x,y) restricted to the boundary.
This corresponds to finding critical point of the function g(t) = f(2cos(t), 2sin(t)).
Okay let's do it some other way instead.
trail eagle
Okay, well it's like what I said then.
The value of f(x,y) on the boundary correspond to f(2cos(t), 2sin(t))
So we want to find critical point of f(2cos(t), 2sin(t)) w.r.t t.
coral zephyr
wait i think i get it
its kinda similar to how you take derivatives of vectors in spacew
where they are in terms of t
trail eagle
Yes it's a similar idea. You can think of $t \to (2cos(t), 2sin(t))$ as a curve parametrized by $t$, and you're evaluating $f$ on points of this curve.
woven radishBOT
Azyrashacorki
coral zephyr
Yes it's a similar idea. You can think of $t \to (2cos(t), 2sin(t))$ as a curve ...
i have a value for t
should i plug that in the circle
😵💫
trail eagle
What point does this correspond to?
coral zephyr
Sorry was away.
its okay
trail eagle
On the boundary
coral zephyr
What point does this correspond to?
uhhh
wdym
should i plug that into the function
pi/2
into f(2cost, 2sint)
my guess is yes i need to do that
trail eagle
t=pi/2 and x=2cos(t), y = 2sin(t).
You can plug it in there to get the x value
Also note that $cos(t) = 0$ gives $t=\frac{\pi}{2}$ and $t=\frac{3\pi}{2}.$
woven radishBOT
Azyrashacorki
coral zephyr
t=pi/2 and x=2cos(t), y = 2sin(t).
but what you can also do is plug t into the parameterized equation, get a point, plug that point into the unparameterized equation
right
but thats kind of an overkill
trail eagle
but what you can also do is plug t into the parameterized equation, get a point,...
That's the goal yes.
From t=pi/2 you get a value of x and y
From t=3pi/2 as well.
This gives you 2 points to look into on top of the initial critical point you've found
trail eagle
Yep that looks good.
coral zephyr
what why
trail eagle
**Azyrashacorki**
.
coral zephyr
.
can we do these? if you're available of course
also thanks for your help
i really appreciate it
trail eagle
Just to wrap up everything in the end you got points (0,1), (0,2) and (0,-2).
Your extrema must lie on those and you should find that you get (0,1) for you minimum and (0,-2) for your maximum.
See this https://www.desmos.com/3d/eofmxdlobq to see what we did.
coral zephyr
See this https://www.desmos.com/3d/eofmxdlobq to see what we did.
hm yeah
trail eagle
Anyways this is how you do it by parameterizing the boundary.
We could've actually proceeded by writing x^2 + y^2 = 4 so that f(x,y) = 4 -2y + 1 on this boundary. Then we would want critical points on there with respect to y between y=-2 and y=2. This would've given no points in the boundary since the derivative is constant -2, but we would have kept the points (0,2) and (0,-2) as they are endpoints of the interval [-2.2].
I personally find it more telling to parameterize directly.
Ok other problem.
coral zephyr
trail eagle
can we do these? if you're available of course
For 3. let's say x is the width, y is the height and z is the length of the package.
The wording is kind of weird, but can you write down what they mean by "the sum of the length and the perimeter of the cross section of the package cannot exceed 108 inches"?
coral zephyr
The wording is kind of weird, but can you write down what they mean by "the sum ...
when it comes to length
does it matter if you interpret the width as the length and vise versa
trail eagle
You just want to choose variables for each it doesn't matter which one.
coral zephyr
You just want to choose variables for each it doesn't matter which one.
yeah i getchu but does it matter tho?
if they were interpreted as the other
realistically speaking i mean
trail eagle
No
coral zephyr
The wording is kind of weird, but can you write down what they mean by "the sum ...
L + G =< 108?
trail eagle
Yes, but you'll want 3 variables, one for each side length
You'll need to express G with those.
Since G isn't a side, it's the perimeter of a cross-section.
coral zephyr
Since G isn't a side, it's the perimeter of a cross-section.
i see what you mean, well in this instance G = 2h + 2W right
so all together L + (2h +2W) =< 108
trail eagle
Yes
trail eagle
so all together L + (2h +2W) =< 108
You want to maximize the volume V(L,H,W) = LHW given that constraint.
coral zephyr
You want to maximize the volume V(L,H,W) = LHW given that constraint.
yeah but i dont undrstand how youre "maximizing" anything
and how critical points will achieve that
trail eagle
Well V(L,H,W) = L*H*W is a function of 3 variables.
coral zephyr
Well V(L,H,W) = L\*H\*W is a function of 3 variables.
yeah
trail eagle
And you have a region given by L + (2H + 2W) <= 108 and the physical constraint being L >=0, H >=0, W >=0.
coral zephyr
And you have a region given by L + (2H + 2W) <= 108 and the physical constraint ...
yes
trail eagle
In fact, the largest volume will be attained by considering the largest perimeter available, so we may assume that L + 2H + 2W = 108 so that we're dealing with 2 variables.
We can still do the computation with the inequality though.
coral zephyr
In fact, the largest volume will be attained by considering the largest perimete...
how do partials here intuitively make sense
like previously they wanted you to see if you can find critical points in a bounded region
are we doing the same thing here
trail eagle
Yes it's the same thing.
Partials represent the same thing they represent for other variables.
You can write L=x, H=y and W=z for simplicity.
Let's go from the start with those variables. You have a function $V(x,y,z) = xyz$ which you want to find the extrema of in the region given by $x + 2y + 2z \le 108$, $x\ge 0$, $y\ge 0$ and $z \ge 0$.
woven radishBOT
Azyrashacorki
coral zephyr
Yes it's the same thing.
but what does that have to do with maximizing a volume
trail eagle
but what does that have to do with maximizing a volume
The volume is a function of the sidelengths x, y and z.
Maximizing the volume amounts to finding x,y,z such that V(x,y,z) is maximal.
And you find such x,y,z using the methods you've learned for finding extrema in a closed bounded region.
coral zephyr
Maximizing the volume amounts to finding x,y,z such that V(x,y,z) is maximal.
idk i thought we needed critical points right? we just have a box here
trail eagle
Yes. We can try to find critical point of V(x,y,z) = xyz
Now this is a function of 3 variables. It's the same thing as with 2. You take the gradient with respect to each variable and you set it equal to 0
coral zephyr
wait the graph of the volume wouldnt realy be a box
it would be
some other shape that represents the volume
does that have anything to do with it
trail eagle
Not much. You can't really represent a function of 3 variables well.
coral zephyr
oh yeah right
trail eagle
In any case, we won't have to deal with a function of 3-variables for long.
coral zephyr
Not much. You can't really represent a function of 3 variables well.
so whats the correlation of coordinates where theres a horizontal tangent lines and the maximum volume the package can be
i dont get how that tells you anything about it being maximized
trail eagle
They live on some 4D shape those tangent lines.
A critical point means the same regardless of the number of variables.
gradient = 0 means critical
coral zephyr
right
trail eagle
Can you compute the partial derivatives of V(x,y,z) = xyz?
coral zephyr
They live on some 4D shape those tangent lines.
whats the correlation between critical points and maximum volumes?
trail eagle
It's the same reason why critical points of a rabbit population gives you the maximum population at a given time.
Here critical points of V give the dimensions yielding the largest/smallest volumes.
trail eagle
What happens if you set them to 0?
coral zephyr
What happens if you set them to 0?
trail eagle
So do you see than in any case, the volume will be 0?
coral zephyr
So do you see than in any case, the volume will be 0?
the volume?
i just found critical points tho
trail eagle
I'm just saying that for any of the critical points you found, V(x,y,z) = 0.
coral zephyr
why
trail eagle
Well for your first partial to be 0, you need at least one of W or H to be 0.
Just from that you get that for any critical point V(x,y,z) = 0
In any case, this means that if we are to find a maximum, it won't be there. We have to look at the boundary of our region.
We had a region x>=0 y>=0, z>=0 and z + 2x + 2y >= 108.
The boundary would be something like x=0 or y=0 or z=0 or z+2x+2y = 108.
The first three give 0 volume as well.
So we're now looking at the boundary given by z + 2x + 2y = 108 with x>=0 and y>=0.
Since z >=0, you can check that this means that z = 108 -2x -2y >=0, so 108 >= 2x + 2y and this gives x+y <= 54.
This is what I meant by "we won't have to deal with a 3 variable function for long".
coral zephyr
Well for your first partial to be 0, you need at least one of W or H to be 0.
sorry but i got lost here
trail eagle
sorry but i got lost here
A critical point happens when all three partial derivatives are 0. If you just look at the first one, for it to be 0, you need W=0 or H=0.
In either case, V(W,H,L) = 0.
coral zephyr
A critical point happens when all three partial derivatives are 0. If you just l...
yeah but why does that make the volume itself always 0
trail eagle
Because on any one of those points one of the width or the height is 0
If you have a box with no width it's not a box
It's flat
With volume 0
Similarly with length or height.
coral zephyr
Because on any one of those points one of the width or the height is 0
oih so the maximum volume is 0
in all cases
okay i see
trail eagle
Not exactly. It means that in this region we've delimited initially, any critical point gives volume 0.
As V is nonnegative, this corresponds to point of absolute minimum.
Which makes sense : the smallest possible volume you can get is if you have no volume at all.
But the region we started with is closed and bounded, there has to be a maximum in there.
So we have to look at the boundary for a maximum.
coral zephyr
As V is nonnegative, this corresponds to point of absolute minimum.
ohh yeah what am i saying
youre infering that the critical point represents a minimum volume
because volumes of an existing package cant be maximally 0
so i guess what we can say here is that the minimum volume is 0.. but thats all what we did
right
so far
trail eagle
Yes. Although they specifically ask for the maximum volume.
And it has to be somewhere in the region.
It just can't be inside.
It has to be on the boundary.
trail eagle
It's ok. I will need to go in a minute.
trail eagle
We had a region x>=0 y>=0, z>=0 and z + 2x + 2y >= 108.
But the idea is this. You end up needing to maximize the function restricted to $z=108 - 2x -2y$, which means considering the 2-variable function given by $$v(x,y)= V(x,y, 108-2x-2y)$$ (you can just plug that in) and now you consider the region $x\ge 0$, $y\ge 0$ and $x+y \le 54$.
woven radishBOT
Azyrashacorki
trail eagle
So the problem transforms into a 2-variable problem now.
And it'll end up looking a lot like what you had with your triangle.
So now your problem is essentially just about finding extrema of the function $$v(x,y) = xy(108-2x-2y)$$ on the region $R = {(x,y) : x \ge 0, y\ge 0, x+y \le 54}$.
woven radishBOT
Azyrashacorki
trail eagle
I'll go now. Try and work it out like we did for those problems.
The pattern is always the same : find critical points inside the region and investigate the boundary.
Here there was extra steps because we started with a 3D region as a domain, whose boundary is a bunch of 2D regions.
Now you are investigating this 2D region, and then you'll investigate the boundary of the 2D region, which will be the perimeter of the triangle you get by drawing R above.
coral zephyr
It's ok. I will need to go in a minute.
No problem bro
coral zephyr
The pattern is always the same : find critical points inside the region and inve...
I’ll think about what you wrote, thanks a lot
devout snowBOT
@coral zephyr Has your question been resolved?
devout snowBOT
@coral zephyr Has your question been resolved?
coral zephyr
<@&286206848099549185> can i get help on this problem
steel cairn
whats the orig problem
coral zephyr
whats the orig problem
covert valve
let's see...
covert valve
we know girth is the width plus the height times 2...
steel cairn
we're given that the sum of the length and girth cannot exceed 108
coral zephyr
so lets start by writing an inequality
yeah wait so weve got here earlier
covert valve
we know that the maximum volume of anything must be a square, right @coral zephyr ?
coral zephyr
i have no idea how to continue from here
covert valve
i found the solution
coral zephyr
okay
covert valve
but i'll guide
coral zephyr
alright
coral zephyr
we know that the maximum volume of anything must be a square, right <@1402083806...
idk tbh
covert valve
idk tbh
notice something about some basic examples of squares where we try to get w*h with the same w+h
19=9
28=16
37=21
46=24
5*5=25
coral zephyr
wait i just wanna let you know that im trying to build off of the foundation of critical points and boundries, so it would be helpful if you could analogize whatever were doing here based off of that
covert valve
oh, I see...
lemme find a different way to get to the solution then...
are we using partial derivatives here?
or allowed to?
covert valve
alright.
so trying to solve for L gives us L=108-2(w+h)
so we should substitute
V=LWH
so V(w, h) = wh(108-2w-2h), right?
coral zephyr
wait what are we doing
covert valve
we notice that to get the max volume, instead of writing an inequality, we can just write that L+2(w+h) = 108, as that should be the way to maximize everything out
coral zephyr
we notice that to get the max volume, instead of writing an inequality, we can j...
yes sure
covert valve
so I just solved for L as 108-2(w+h)
coral zephyr
so I just solved for L as 108-2(w+h)
whats the reason tho
covert valve
and then plugged it in to the volume equation so we only have w and h
to get ready to take partial derivatives
coral zephyr
why solve in terms of L specifically
covert valve
idk I just chose it cuz it was the one on the left of my equations and that it wasn't really in the parenthesis of 2(w+h)
maybe it'll be easier, who knows
then, we should take partial derivatives of V/h and V/w
coral zephyr
idk I just chose it cuz it was the one on the left of my equations and that it w...
okay, i can see you have an equation that represents the grith in 3D, correct?
covert valve
yep
coral zephyr
okay i see and shouldnt we find critical points in the volume equation then plug the critical points into the inequality
because thats what weve been doing previously
with similar problems (not involving volumes tho)
covert valve
i see what you are getting at, but the problem says maximize V to a constraint, not over an interval
and this is an equation, not an inequality
so... lets continue
the partial derivative of V/h is h(108-4w-2h)
the partial derivative of V/w is w(108-4h-2w), if I didn't make a mistake, right?
coral zephyr
i see what you are getting at, but the problem says maximize V to a constraint, ...
oh okay but like hows that supposed to look like
coral zephyr
the partial derivative of V/h is h(108-4w-2h)
the partial derivative of V/w is w...
wait whyd you do it like that
im confused
covert valve
oh okay but like hows that supposed to look like
at a maximum, you can't increase v while staying on the contraint
so instead of delta V = 0
we use λ∇g
aka the langarange multipliers
coral zephyr
right so
coral zephyr
we use λ∇g
wait youre doing lagrange multipliers here
covert valve
where g = L + 2(w+h)
covert valve
wait youre doing lagrange multipliers here
yep
covert valve
should I not?
coral zephyr
because i was tryna understand lagrange multipliers and i was stuck on that too
covert valve
oh f**k
coral zephyr
should I not?
no i mean it would be great because then id understand lagrange multipliers but thats gonna require us to go over that 😂
covert valve
alright, I'll be the tutor for you
coral zephyr
thanks alot man
covert valve
lemme go over the whole thing
coral zephyr
say less
covert valve
if you'd like, I can give you my notes too?
coral zephyr
if you'd like, I can give you my notes too?
i wouldnt mind but im a sort of a person that understand when conversing about the topic gradually
covert valve
i see...
first things first.... Lagrange multipliers find interior constrained critical points only. You must still check boundary cases (like variables = 0). in this problem, that case is trivial and doesn't matter here.
coral zephyr
first things first.... Lagrange multipliers find interior constrained critical p...
okay
covert valve
but the core purpose is to find maximum or minimum values of a function subject to a constraint.
we want to optimize f(x, y, z...) subject to g(x, y, z...) = c: to find min/max values
coral zephyr
we want to optimize f(x, y, z...) subject to g(x, y, z...) = c: to find min/max ...
so the level curve is gonna give you a contraint
covert valve
yep
Instead of the usual condition:
∇f = 0
We replace it with
∇f = λ∇g
where we can interpret it simpler as this:
- ∇f = direction of greatest increase of f
- ∇g = direction perpendicular to the constraint
At a constrained max/min:
You cannot increase f without leaving the constraint,
so the gradients are parallel.
you got that so far?
covert valve
ngl i wouldnt say so
oh noes
i'll help.
maybe tommrrow actually I sadly have to go to bed :(
but I'll helper ping for someone who understands it better
<@&286206848099549185> for @coral zephyr
coral zephyr
maybe tommrrow actually I sadly have to go to bed :(
its okay i appreciate your time
have a great night
cursive sable
<@&286206848099549185> for <@1402083806034268161>
??
covert valve
have a great night
thank you.
summer summit
btw how does someone obtain a helpers role
scroll all the way up in the list of channels. theres a thing named "Channels & Roles" select accordingly
coral zephyr
scroll all the way up in the list of channels. theres a thing named "Channels & ...
i dont see an option for that
theres a role where it says "i wanna help others" but it doesnt give me anything when selected
restive coral
It will give you the helper role as seen on the right side
coral zephyr
It will give you the helper role as seen on the right side
ohh i see it now thanks
restive coral
Have you tried finding the derivative for the function girth times length. I am not that good in calculus but I just want to see hoe to solve the problem too
coral zephyr
Have you tried finding the derivative for the function girth times length. I am ...
no i havent done that
i dont even know what i should do here
restive coral
assuming that the girth is the perimeter of the cross section of a regular polygon
I wont concern with wich configuration of sides based on the perimeter results in maximal area
then we could take a derivative of length* Girth and equate it to zero
where girth = 108 -length
that should be sufficient\
devout snowBOT
@coral zephyr Has your question been resolved?
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unique cove
hloooo
restive coral
bro 7 is prime
unique cove
bro 7 is prime
so
restive coral
I thought of using FLT
but you now mention that it is 2^x + 3^x and not 3^x - 2^x
Ok
use FLT to get that 2^6 and 3^6 are 1 (mod 7)
then (2^6)^m is 1 mod 7
hence we have two cases here
case 1: 2^x = 4 mod 7 and 3^x = 3 mod 7
last parrot
Hello, I just have a quick question, if f: R-> Z, and f continuous, how is f must be constant? ;-;
devout snowBOT
Hello, I just have a quick question, if f: R-> Z, and f continuous, how is f mus...
last parrot
Wait i think i figured out how to solve this but i have other question tho
gloomy aurora
IVT???
last parrot
This one basically using iVT and the fact that there exists a non-integer between fx1 and fx2
Ye ye
Wait i send my wuestion
last parrot
.pin
devout snowBOT
gloomy aurora
what have you done so far
trail eagle
Well the first thing you need to do is show that A+B is bounded above.
The rest of the question gives a good hint as to which upper bound you could consider.
gloomy aurora
most of such bound-questions require either a construction or a contradiction.
last parrot
Wait how would we start to show A+B is bounded above
Thank you for your help
<@&268886789983436800>
trail eagle
Well you know that A and B are bounded above.
trail eagle
Not unless you show it's bounded above
last parrot
Hmm
trail eagle
The fact that A and B are bounded above should give you a way of finding an upper bound for A+B.
trail eagle
Yeah, those are in particular upper bounds of A and B respectively
How do you think you can bound any element of A+B using those?
last parrot
We let x is the element of A+B?
trail eagle
Yes. What form do elements of A+B have?
last parrot
What do you means form here?
trail eagle
Well an element x in A+B arises in a particular way from the way A+B is defined.
last parrot
If x is element of A and y is element of B then x+y<= sup A + sup B
This is what i think of
trail eagle
If x is element of A and y is element of B then x+y<= sup A + sup B
That's the idea, yes. An element x of A+B has the property that x=a+b for some a in A and b in B. You can show that x = a+b <= sup(A) + sup(B).
last parrot
Alright i have shown one direction
trail eagle
This tells you that A+B is bounded above, so now it makes sense to think of the number sup(A+B).
trail eagle
Yep
trail eagle
Well now you still need to show that sup(A+B) = sup(A) + sup(B).
last parrot
Hmm true as stated in the question
trail eagle
In particular, show that sup(A+B) <= sup(A)+sup(B) (this should be fairly simple if you recall what sup(A+B) means). And then show the other direciton
last parrot
Do i show supA +supB>= supA+B and vice versa
Yes
Alright I have shown 1 direction
I am on bed so the handwriting looks pretty chaos
trail eagle
Hahah it's readable
So yeah now you can show that $\sup(A+B) \ge \sup(A) + \sup(B).$
woven radishBOT
Azyrashacorki
last parrot
My friends told me thats doctor handwriting but anyways, how would it be possible to approach other way?
I am thinking of sth like epsilon
trail eagle
There's a few different ways you can go at it. I think the easiest would be to use the epsilon characterisation of supremums of A and B separately to get nice inequalities to work with.
last parrot
But the problem is... I never done using epsilon show for this sup bound
trail eagle
Well let's say you fix $\varepsilon > 0$. What do you get from $\sup(A)$ and $\sup(B)$, respectively?
woven radishBOT
Azyrashacorki
last parrot
**Azyrashacorki**
Then I can get a point lower than sups that is a+epsilon?
last parrot
I can share my proof if you want
Wait dont i want to think
I need to understand the concept for exams
spring oasis
In case you get stuck, here is the thread
last parrot
But if we want a nice epsilon for the sum supA and supB
I think of epsilon/2?
Or sth like that
trail eagle
Yeah you can do that to get a cleaner expression in the end.
trail eagle
Then I can get a point lower than sups that is a+epsilon?
Just write this down as inequalities for A and B with epsilon / 2.
last parrot
Alright i came up to sth like this
I should say for a fix epsilon >0
Let me scrap that out
I think i messed up in the last line ;-;
trail eagle
Yes. The only other thing is that your inequalities added don't say what you wrote down.
trail eagle
Yes that's it
last parrot
Thank you so much
Wait can I have a last question? This one quite long a bit, i dont know if you have time
trail eagle
Ask away!
last parrot
Okay let me screenshot
This is my advanced worksheet for workshop its hard to understand 😂
last parrot
.pin
devout snowBOT
last parrot
For the first one i think about making a helper function
trail eagle
For the first one i think about making a helper function
Sorry made coffee. Yes considering some other function may help
trail eagle
Yep that's it
last parrot
I will think of b a bit, might need epsilon delta here
I solved like this
Alright lets do (3), so far so good
last parrot
Yep that's it
Do i use contradiction in C here? Or i just prove straightaway?
trail eagle
Do i use contradiction in C here? Or i just prove straightaway?
I think you can prove it straight away using the hint with continuity. Assume it's positive, get a contradiction. Assume it's negative, get a contradiction.
Am I misreading or is d) asking the same thing as the second part of c)?
last parrot
I am doing c
Wait i think i got something
I got something like thís, quite unsure about the domain (delta-alpha, alpha] part
Is this correct?
Looks so messy 
trail eagle
The main idea is there
magic briar
try using proof by contradiction
last parrot
Yes but i dont know where to start with
I think i will continue this tomorrow, thanks guys
.close
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azure galleon
need help solving this integral
vital sedge
complex analysis?
halcyon apex
Hey
halcyon apex
I can't understand how an area function is equal to the anti-derivative
Someone please link it for me
Computing is really easy though
I understand the Riemann sum/integral well though
raven bramble
I also wanna know 
mild sorrel
Is that the Fundamental Theorem of Calculus?
mild sorrel
That requires a proof. It's not necessarily trivial to prove it.
halcyon apex
i only want to understand
why the antiderivative of a function is the area
the prof has been saying that all the time
mild sorrel
Understand it intuitively or by proof?
halcyon apex
A'(x) = f(x) and A(x) = F(x)
i understandn the rate at which a function's area increases
is just the height
that is somewhat intuitive
but not that the area of a function is just the antiderivative
should i just keep watching lectures...
the answer lies in what i just said:
given the rate of change in area is just the height of the original function
then to find the area of f(x)
we need to find the antiderivative
lunar harbor
halcyon apex
because the derivative of the antiderivative is just the rate of change in are of the original function
i.e. f(x)
@mild sorrel
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white pawn
I don't understand question (b)
devout snowBOT
I don't understand question (b)
white pawn
It used vector projection but I don't know how and why they used it like this?
trail eagle
It used vector projection but I don't know how and why they used it like this?
Take a vector from the origin to a point on the plane.
Then project that vector onto the normal vector to the plane. The length of this projection is the distance from the plane to the origin.
It's useful to think of the 2D case as it's the same idea.
white pawn
ohh okay so you find the projection of teh normal vector that is closest to the origin
since the shortest distance from origin to plane must have a vector that is perpendicular to the plane
trail eagle
Yep
white pawn
how do you find the evctor with shortest distance to the plane knowing that information?
would it have to do with 2 vectors that are perpendicular having a dot product of 0?
oh so you project A onto the unit evctor of n
and magnitude it
trail eagle
Yes, in particular the projection of the vector $OA$ onto the normal vector is $$\frac{OA \cdot \vec{n}}{\vec{n}\cdot \vec{n}}\vec{n}$$.
woven radishBOT
Azyrashacorki
white pawn
and you can use any point on the plane as one of the components will be the vector with minimum distance away from origin
trail eagle
You can pick any A on the plane, yes
Using the one they provide you seems like a good fit.
trail eagle
But say in this image you could pick any A on the line and get the same projection onto n.
trail eagle
**Azyrashacorki**
Well once you get this you'll take its magnitude
And magnitude is always positive.
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lunar harbor
This
I assume the last sentence of (c) is not supposed to be there?
Since $f$ is continuous at $\alpha$, what does the definition of continuity give you?
woven radishBOT
Civil Service Pigeon
last parrot
I solved it, for a small delta >0 there exists an area around alpha such that f non positive
From yesterday
arctic field
α-δ?
arctic field
now copy it for (d)?
arctic field
it's basically the same as what you've got for (c)
last parrot
Do i assume f<0 here or ff>0?
arctic field
you're asked to show f(α) cannot be negative
last parrot
Then assume f<0 here for contradiction
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agile crag
devout snowBOT
last parrot
Why are you closing and reopening?
agile crag
I closed 1 because I didnt want to ping anyone yet but im really stuck apologies
woven vale
agile crag
I have to specifically use u-sub method here all I know is that I identify the inside function (1+x^2) but idk what after
narrow plank
Yes, u = 1 + x^2. Then what do you need to find given u in terms of x?
agile crag
1/2 but why do we need that I’m sorry I’m just trying to understand it
I meant by 2 apologies
last parrot
no problem
you want to make the 2x appears on the numeratror
to make into du
du = 2xdx, you agree?
bronze void
$u = 1+x^2, du=2xdx \implies dx=\frac{du}{2x}$ what can we do from there?
woven radishBOT
KB
agile crag
yes I agree that du = 2xdx
last parrot
so you want to make du appears in your equation
so you can calculate 1/sqrt (u)
right?
agile crag
so I want to get rid of the x in the numerator?
to only have it in terms of u and du?
last parrot
correct
bronze void
substitute what we know and see what happens
last parrot
take 1/2 out of the integral since its a constant
now, we have 2x/sqrt(u) dx
which is 1/sqrt(u) 2xdx = 1/sqrt(u) du
now can you find integral?
bronze void
did you already integrate it?
last parrot
if you integrate then remove the integral sign and du
agile crag
was I supposed to put my original I back in first?
agile crag
if you integrate then remove the integral sign and du
oh okok
last parrot
last 2 lines are unnecessary
bronze void
but yea it looks correct, just cancel the 2's and get u^1/2 + C
last parrot
scrap those 2 and write without the signs
bronze void
yes
agile crag
ok thanks yall
bronze void
and then dont forget to sub in u = 1+x^2 in final solution
agile crag
so final answer would be (1+x^2)^1/2. or sqrt(1+x^2)
bronze void
depending on how the original question was given either with idex for radical form give answer in the same form as the original question
in this case it should be in radical form
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last parrot
Hi, how do i prove c? I shown that b is false
devout snowBOT
Hi, how do i prove c? I shown that b is false
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gritty saddle
Ive shown that X:GLnR→TGLnR=GLxMatnR with X(A)=A•ξ where ξ is a constant matrix is a smooth vector field, with flow Φ(t,A)=Aexp(tξ). Then i had that X on O(n,R) which is subset of GlnR, is also a smooth vector field. How can i find the flow of X on O(n, R) (i had trouble cause i dont know something about OnR atlas and so i asked for any other way).
devout snowBOT
Ive shown that X:GLnR→TGLnR=GLxMatnR with X(A)=A•ξ where ξ is a constant matrix ...
last parrot
can you show the original question?
gritty saddle
Well yes its on greek, let me translate it
gritty saddle
no problem bro
Let me do it now
We take Mat(n, R)=Rn² and u can see that the flow is Φ=Aexp(tξ)
All those in 5
In 6 is my question
last parrot
Can you have a quick translate?
gritty saddle
Can you have a quick translate?
Thats 5, ive already have is ok: We denote by
M
a
t
(
𝑛
;
𝑅
)
Mat(n;R) the set of all
𝑛
×
𝑛
n×n matrices with real entries. Consider the subset
G
L
(
𝑛
;
𝑅
)
{
𝐴
∈
M
a
t
(
𝑛
;
𝑅
)
:
det
𝐴
≠
0
}
⊂
M
a
t
(
𝑛
;
𝑅
)
,
GL(n;R)={A∈Mat(n;R):detA
=0}⊂Mat(n;R),
of real invertible matrices.
(a) Show that
G
L
(
𝑛
;
𝑅
)
GL(n;R) is an open subset of
M
a
t
(
𝑛
;
𝑅
)
Mat(n;R).
(b) Show that the maps
G
L
(
𝑛
;
𝑅
)
×
G
L
(
𝑛
;
𝑅
)
→
G
L
(
𝑛
;
𝑅
)
,
(
𝐴
,
𝐵
)
↦
𝐴
𝐵
,
GL(n;R)×GL(n;R)→GL(n;R),(A,B)↦AB,
and
G
L
(
𝑛
;
𝑅
)
→
G
L
(
𝑛
;
𝑅
)
,
𝐴
↦
𝐴
−
1
,
GL(n;R)→GL(n;R),A↦A
−1
,
are smooth. Thus
G
L
(
𝑛
;
𝑅
)
GL(n;R), with matrix multiplication and inversion, becomes a Lie group.
(c) Show that for each
𝜉
∈
𝑇
𝐼
G
L
(
𝑛
;
𝑅
)
M
a
t
(
𝑛
;
𝑅
)
ξ∈T
I
GL(n;R)=Mat(n;R), the map
𝑋
𝜉
:
G
L
(
𝑛
;
𝑅
)
→
𝑇
G
L
(
𝑛
;
𝑅
)
≃
G
L
(
𝑛
;
𝑅
)
×
M
a
t
(
𝑛
;
𝑅
)
,
𝐴
↦
𝐴
𝜉
,
X
ξ
:GL(n;R)→TGL(n;R)≃GL(n;R)×Mat(n;R),A↦Aξ,
(where the product is the usual matrix multiplication) is a vector field with flow
𝜑
(
𝑡
,
𝐴
)
𝐴
𝑒
𝑡
𝜉
,
𝑡
∈
𝑅
.
φ(t,A)=Ae
tξ
,t∈R.
(d) Show that the Lie bracket satisfies
[
𝑋
𝜉
,
𝑋
𝜂
]
𝑋
[
𝜉
,
𝜂
]
,
[X
ξ
,X
η
]=X
[ξ,η]
,
where
[
𝜉
,
𝜂
]
𝜉
𝜂
−
𝜂
𝜉
[ξ,η]=ξη−ηξ
is the commutator of
𝜉
,
𝜂
∈
𝑇
𝐼
G
L
(
𝑛
;
𝑅
)
M
a
t
(
𝑛
;
𝑅
)
ξ,η∈T
I
GL(n;R)=Mat(n;R).
Oh
Let it go. Thanks
.close
devout snowBOT
faint gorge
ti egine 💀
last parrot
Let it go. Thanks
Bruh
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Available help channel!
wintry sand
Quick question — is this considered order (n) or order (2n)?
$$
\cos(x) = \sum_{k=0}^{n} \frac{(-1)^k}{(2k)!} x^{2k} + o(x^{2n})
$$
I’m confused because the highest power here is (2n), not (n).
woven radishBOT
Klein Bottle
wintry sand
and if I had to write the taylor expansion in young's form of order n, would I be writing it like this:
$$
\cos(x) = \sum_{k=0}^{\left\lfloor N/2 \right\rfloor} \frac{(-1)^k}{(2k)!} x^{2k} + o(x^N)
$$
woven radishBOT
Klein Bottle
wintry sand
Thanks in advance. Feel free to ping me
faint gorge
**Klein Bottle**
I'd say 2n
wintry sand
are you fasho?
stone stump
when you tell someone to give you the taylor poly of cosine of order 2 they would give you 1-x^2/2 not 1-x^2/2+x^4/24
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willow helm
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finite fable
devout snowBOT
finite fable
!status
devout snowBOT
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
What do you know about logs
finite fable
What do you know about logs
hmm. well, the basic rules
i learned 11 properties
i tried make if log(a)(a) , so it becomes 1
but can't figure out a away
willow helm
Right
willow helm
i tried make if log(a)(a) , so it becomes 1
How about if we try with 3 a's
willow helm
i tried make if log(a)(a) , so it becomes 1
What did you mean by this then?
finite fable
What did you mean by this then?
5
willow helm
Ah
Right
Yes
We kind of use this property
Together with (3)
I don't really see any other way than doing this trick
As changing bases and whatever complicates things drastically
finite fable
well. how can i make 5root(2)-7 into x^y
willow helm
But you do agree that $\log_a{a^2} = 2$
finite fable
yeah
woven radishBOT
willow helm
Okay
So we just try to find
If the argument can be expressed
As a power of the base
Because that simplifies everything neatly
If we can get a^2, or a^3, or a^4 in the argument right
That's just 2, 3 or 4 respectively
finite fable
yeah
willow helm
Okay
finite fable
but. i tried that. but couldn't get the argument in terms of base raised to some power
finite fable
3-2root(2)
willow helm
Okay
So squared doesn't work
3-2sqrt(2) and 5sqrt(2)-7 clearly have nothing in common
So we go higher
How about cubed
willow helm
Yeah
finite fable
willow helm
😂
willow helm
Yeah
finite fable
you amazing
willow helm
This is the trick I do
With these kind of problems
Obviously as I said changing base, and using other properties
Just results in a mess
willow helm
you amazing
🙂
finite fable
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tawdry meteor
How do I…input this answer
devout snowBOT
How do I…input this answer
trail eagle
You're not done.
faint gorge
How do I…input this answer
I wouldn't yet, it seems you only computed the roots?
tawdry meteor
Oh-
Well I did all that was on the example area-
And checked the answers
What else do I need to do?
trail eagle
Make a sign table to check when the function is positive or negative in between and outside the roots
tawdry meteor
Is that…not what I did?
trail eagle
You found the roots, which by definition are those x's so that f(x) = 0 and you then checked whether f(x) > 0 at those points.
tawdry meteor
In that lower area
tawdry meteor
Make a sign table to check when the function is positive or negative in between ...
That’s not in the examples-
To my knowledge
trail eagle
Can you show what the examples do?
tawdry meteor
Yeah sure
trail eagle
Because if they're solving inequalities then they're not just checking the roots.
tawdry meteor
trail eagle
Any number you want between the roots
tawdry meteor
The roots being….
trail eagle
You have roots -5 and 6, so you need to check for a number <-5, a number between -5 and 6 and a number > 6
tawdry meteor
Just….anything…
trail eagle
Well you'll need to plug them in, so try and keep them small hahha
tawdry meteor
Well it has to be less than 6 so…
trail eagle
Yeah well you can still pick numbers which are not like -1000 and 1000
tawdry meteor
Why would I tho
You said it needs to be between -5 and 6
-1000 isn’t between -5 and 6
trail eagle
You have roots -5 and 6, so you need to check for a number <-5, a number between...
.
You have 3 numbers to check
trail eagle
Pick any 3 numbers you want that respect those conditions.
The point is you'll need to plug them inside the inequality to see if it holds
tawdry meteor
Less than…-5?
trail eagle
Can you give me a number which is less than -5
tawdry meteor
-6
trail eagle
Ok that's your first number.
tawdry meteor
So it needs to be outside of -5 and 6????
tawdry meteor
You said between -5 and 6?
trail eagle
Can you give me a number between -5 and 6
tawdry meteor
Hold on one sec
Different class
So…..2 numbers on either outside and 1 in the middle?
trail eagle
one for every region on this line
tawdry meteor
So…what I said
trail eagle
Yes, what you said.
tawdry meteor
Ok
devout snowBOT
@tawdry meteor Has your question been resolved?
tawdry meteor
So like…. -6, 1, and 7?
mystic scarab
Sure, they are fine
tawdry meteor
Ok
And just plug that into the og equation
Right….and so what answers am I looking for? Smth between -5 and 6?
tawdry meteor
Cus none of these work….
tawdry meteor
<@&286206848099549185> how do I figure out the direction the answer goes
thick lotus
<@&286206848099549185> how do I figure out the direction the answer goes
hi
you there
tawdry meteor
Hi
thick lotus
where are you stuck
thick lotus
ok what have you tried
tawdry meteor
Uhm…nothing yet because I don’t know the rules of moving things across inequalities
I lowkey forgor
thick lotus
ok , there are few simple
-
you can add/subtract like you do for an equation on both sides and the inequality sign will remian same
-
when you multiply/divide by something negative the inequality sign flips
tawdry meteor
Ok cool
thick lotus
for eg you have x+1 >y , multiply whole inequality by -1
-x-1 < - y
tawdry meteor
So once I have it equal to 0….
Should I….factor out a 3?
So it’s 3(-x^2 -2x + 3)?
thick lotus
So once I have it equal to 0….
you jave >= here but yes factor the 3
tawdry meteor
I know that lol
thick lotus
xD
tawdry meteor
So how does the 3 come into play when I factor it as a trinomial
thick lotus
So how does the 3 come into play when I factor it as a trinomial
you can divide both sides by 3 to remove it
thick lotus
But it doesn’t change?
no it doesnot here
tawdry meteor
Ok
thick lotus
it doesnot affect the sign of expression
thick lotus
Alright I’ve got it to x=1 and x= 2
for what ?
tawdry meteor
The roots
ocean haven
imback
ocean haven
no
uneven coral
well, I'd expand the det and do it the long way, IDK if there's another method
uneven coral
you can wait for someone else, genuinely
in case there's a shorter method I'm unaware of
winter torrent
it looks like a lot of stuff might cancel out
uneven coral
it looks like a lot of stuff might cancel out
yeah I hope it does, otherwise it'd be really long
ocean haven
C3-> C3 +C2
,w
,w Det[{{Cos[x] + 1, Cos[x]^2, 1 + Sin[x]^2}, {-Sin[x], -Sin[2x], Sin[2x]}, {Sin[x], 2^2 * Sin[2x], -2^2 * Cos[2x]}}
<@&286206848099549185>
ocean haven
no
woven radishBOT
faint gorge
The author surely had something in mind
uneven coral
The author surely had something in mind
must have had an argument with their partner or some shit to conjure up this
faint gorge
Bold of you to assume such people have partners
ocean haven
fr
i skip
btw
if ts does not match with option istg ill tear them
,w 1/30 (-60 x - 5 cos(3 x) - 15 cos(4 x) - 3 cos(5 x) - 120 sin(x) + 35 sin(3 x) + 15 sin(4 x) + 3 sin(5 x)), x = 0
woven radishBOT
ocean haven
,w 1/30 (-60 x - 5 cos(3 x) - 15 cos(4 x) - 3 cos(5 x) - 120 sin(x) + 35 sin(3 x) + 15 sin(4 x) + 3 sin(5 x)), x = pi/2
woven radishBOT
