help-23

[small sandal]

I’m having a hard time finding the correct domain and range for these types of questions, but what’s confusing in this question is the direction of the arrows, why are they going in opposite directions?

[errant rapids]

Are you trying to write $x$ as a function of $y$?

[flat frigate]

flynger

[errant rapids]

and finding the domain and range of $x=f(y)$?

[flat frigate]

flynger

[small sandal]

Uh, the Cartesian representation

Of the vector equation

[errant rapids]

Okay

[errant rapids]

if this is what you are doing

then the domain is correct

You can solve the domain here by checking what values of y you can get for your domain for t

[small sandal]

Yea, the confusing part is the direction of my line

Like… how can it go both ways

[errant rapids]

you should double check your range though

wait nvm

its correct i guess

[small sandal]

The domain and range and graph are correct

It’s from the answer

But idk how they get that graph

(The direction)

[errant rapids]

$y = \sin t$

[flat frigate]

flynger

[errant rapids]

so it should go from 0 to -1 to 0 to 1 to 0

as t goes from -pi to -pi/2 to 0 to pi/2 to pi

[small sandal]

But wouldn’t that be in one direction ?

[errant rapids]

also the graph doesn't show y crossing 0 more than once

[small sandal]

And I don’t understand the bounds for t and how they relates to the graph

[errant rapids]

oh wait

the graph is of the new function $x=f(y)$

[flat frigate]

flynger

[errant rapids]

which is different than the original parametric graph

because the original graph crosses over

the same x/y values

multiple times

which is not a cartesian function

[small sandal]

Uh what

It’s just the top part of a parabola

r(t) = cos(2t)i + sin(t)j ==> x = 1-2y^2

[errant rapids]

sorry bad wording

its more like

we can make this chart:

after graphing the parabola like you said

follow the path of the points

[small sandal]

Yea that makes sense

Goes left to right and back to left while going up

[errant rapids]

The "weirdness" is happening because you can have a path with any kind of movement when using parametric

unlike the left to right nature

of non parametric

[small sandal]

But from the lowest point to the middle, up to the top is a valid path or no

[errant rapids]

You can have that path but the ts would be different

so it corresponds to a different parametric function

[small sandal]

…

[errant rapids]

different t range

[small sandal]

It’s the same thing 😭

[errant rapids]

its not

[small sandal]

It looks the same

[errant rapids]

yes it does

thats because we are not graphing t

if you view t as time

then it makes sense why the two paths are different

the "cartesian graphs" are the same

so where the function can go is the same

but the actual pathing within that graph is different

[small sandal]

Hmm ok

[errant rapids]

its like if you have a loop and someone can go clockwise or counterclockwise but its still a loop

[small sandal]

Gotcha, guess I’ll have to practice it a bit

-# I’ve got to run btw

Thanks for the help

[errant rapids]

ok np

[small sandal]

.close

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[rich dune]

How fo i find matrix D

Basically

The lamda values are -1 -4 5

The eigenvalues

Part c just asks for the eigen values

P is

That

[light shoal]

well D should be a diagonal matrix with eigenvalues in the diagonal entries
(eigenvalues of A^-1)

[hollow dock]

D consists of your eigenvalues

[light shoal]

P should be a matrix with corresponding eigenvectors

in the columns

[rich dune]

Ye

It is

Wait so

How do i find D

[hollow dock]

Do you know your eigenvalues?

[rich dune]

Ye

-1 -4 5

[light shoal]

those are the eigenvalues of A

[rich dune]

How do i found the eigenvalues of a-1

[hollow dock]

Yeah, so D is basically the eigenvalues placed across the diagonal

[light shoal]

you need the eigenvalues of A^-1, those are the diagonal entries of D

[light shoal]

use this

[rich dune]

eλ^-1=A^-1e

Thats what i did for that one

[light shoal]

i mean, use the result

you know the values of lambda for A

[rich dune]

This is the result

Its what i wrote down

And the markscheme says so also

[light shoal]

ok, therefore the eigenvalues of A^-1 are?

[rich dune]

λ^-1;

?

[light shoal]

list them

[hollow dock]

And lambda is?

What values?

[rich dune]

-1 -1/4 1/5 ?

[hollow dock]

Yes

[rich dune]

So D would be

On a diagonal

-1
. -1/4
. . 1/5

[hollow dock]

Yes

[rich dune]

With 0s in the surrounding

Ahhh

[hollow dock]

Remember that the positioning of your eigenvalues is based on the position of your eigenvectors

[rich dune]

Yea that makes sense lol

Thank you guys

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[ionic blaze]

I''ve evaluated this as integral of dx/(3+1/(sinxcosx)^2) and is there like an easier way to do this bc I don't wanna do weierstrauss

[ionic blaze]

hmmm wait

sinxcosx=sin2x/2

4/(sin2x)^2

sin2x^2 / (3sin2x^2+1)

now sin2x^2 = (1-cos4x)/2

(1-cos4x)/ (3-3cos4x+2) => 1-cos4x / 5-3cos4x
weierstrauss 🥀

could I just multiply with the conjugate: probably actually

oh wait that doesn't work

was it (1-z^2)/(1+z^2),

[ionic blaze]

that

doesn't look like it wants to be done

[simple galleon]

you could also let u=tan x

[ionic blaze]

2u^4+3u^2+2
ain't nicely factorable hm

[simple galleon]

you get this

[simple galleon]

yeah true

maybe this calculator wasn't a good idea

[ionic blaze]

oh right the ^4 thing

other dude some long time ago taught me that

quadrics make no sense smh smh

quartic

quartic polymials make no sense smh smh*

[simple galleon]

link to the site if you want it

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@ionic blaze Has your question been resolved?

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[loud halo]

When I enter the equation y + |y| = 0 in desmos, I expect a line coinciding with the x-axis AND a line coinciding with the negative y-axis. This is because for all values of y < 0:

-y + |-y| = -y + y = 0

The same is not true for positive values of y:

y + |y| = y + y = 2y = 0 => y = 0

But the graph I get includes only a line coinciding with the x-axis. What am I doing wrong?

[safe radish]

@loud halo Has your question been resolved?

[wild copper]

Why are you expecting a line if the expression literally does not contain x?

Your expression is basically "solve y, and then make a horizontal line at that solution"

[loud halo]

Oh yeah, didn't think about that

But then shouldn't the entire lower half of the graph be the solution?

[wild copper]

hmm, yeah. Probably because Desmos sees the equality sign and decides this has to be a curve.

[golden nebula]

I think its due to the fact that when y is less than 0, the equation simplifies to 0=0. I dont think desmos can plot that

[loud halo]

I see now. Thanks Xwtek!

[safe radish]

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[untold yarrow]

@uneven agate HElp

[untold yarrow]

Help me

[uneven agate]

what?

[untold yarrow]

Oh this thing is pretty nice

it shows my name upo there

and everything

[uneven agate]

.close it

smh

[untold yarrow]

what if someohne has

a name with strange characters

[uneven agate]

bruh the test time is over this is actually a question channel you're occupying

.close

[untold yarrow]

There is an additional one available now, so I am not any rush to experiment

Let me try somethign

[uneven agate]

I am not associated with this don't ban me

[untold yarrow]

If I am banned because of curiosity, then so be it

.open

.pin

.close

[safe radish]

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[warped roost]

@untold yarrow do not occupy channels for no reason.

theres an inherent cooldown because of discord rate limits

[untold yarrow]

I didn't realize

Noted

[warped roost]

so you put the channel out of commission for at least ~15 mins

[untold yarrow]

My apologies, won't happen again

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[crude merlin]

how would this be done?

[merry swift]

The solution set would be the intersection of the solution sets of 9-2/3x<5x-11 and 5x-11<17-x/4

[crude merlin]

ah alright

oh I understood now

thanks!

[merry swift]

nw mate

[crude merlin]

.close

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[bronze rivet]

for a cotangent function if the asymptotes occur at -2,2,6,10,14,18 how can I make a general “equation” for jt?

[bronze rivet]

n(pi) would be the equation for a non transformed cot function so what would it be in my case

<@&286206848099549185>

so sorry for ping

[frosty kestrel]

so in general, you want a normal cotangent function but with asymptotes at 4n+2 for any integer n

or the new period would be 4

since the difference in that arithmetic sequence is 4

[bronze rivet]

okay thanks

.close

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[lean otter]

Hey guys I really don't know how can I draw two different sizes of square with the same vertices

[steel stag]

here are 2 ways

[lean otter]

U can't do that with these parameters

Cause it is gonna make a rectangle

[steel stag]

looks fine to me

[lean otter]

Yeah

Alright wait a minute lemme see

[lean otter]

How to get the length from just using the coordinates

[steel stag]

which color square

[lean otter]

Red

We don't know the length of the sides though

[steel stag]

squares have the property that the diagonal is sqrt(2) times its side

[lean otter]

I still don't know the sides

[steel stag]

do you know the distance from (1,1) to (3,5)?

[lean otter]

No

You have the question there

I found this

https://www.wikihow.com/Use-Distance-Formula-to-Find-the-Length-of-a-Line

[steel stag]

right that's how do get the distance

since this is a right triangle, just use pythag

[lean otter]

Yeah

[steel stag]

sqrt(20)

[lean otter]

I still don't know how to draw the triangle

The side length will be 2sqrt(5)/sqrt(2)

How can I get the parameter though when I just have a diagonal

[lean otter]

<@&286206848099549185>

[safe radish]

@lean otter Has your question been resolved?

[lean otter]

Guys anyone can help with that

?

[weak shuttle]

you found the length of the smaller square

[lean otter]

I need it's parameters too

[weak shuttle]

to get the perimeter, multiply it by 4

[lean otter]

What..
How ?

[weak shuttle]

perimeter is the sum of all sides' lengths

[lean otter]

How can that even get us parameters

[weak shuttle]

since a square has 4 equal sides, you just multiply its length by 4

[lean otter]

I said parameters not perimeter

[weak shuttle]

what parameter are you talking about?

[lean otter]

I need the coordinates

[lean otter]

Please look at the question

[weak shuttle]

(1,1) and (3,5)

for both

squares

[lean otter]

Having those vertices

Smallest and largest possible squares

[pure pendant]

hm

[lean otter]

The question is not easy.... its how difficult it is that makes it look easy

[pure pendant]

any clearer images?

[weak shuttle]

the distance between (1,1) and (3,5) is 2sqrt(5). For the bigger square, this distance is already the side length. And for the smaller square, it is the diagonal

and you found what's the side length for the smaller one: 2sqrt(5)/sqrt(2)

now just take both perimeters and get the difference between them

this will be the answer

[lean otter]

How can you get the coordinates from just a diagonal

[lean otter]

Here it is

[weak shuttle]

the coordinates were given

[pure pendant]

i assume the largest is 1, 5 and smallest is 3, 5 but without any explaination that will be hard

for the side of the square

but what kind of large was it talking about? is it perimeter or area

[lean otter]

Ok how?

[pure pendant]

oh

[lean otter]

Largest square

I think size

Area and perimeter

[pure pendant]

im pretty sure we can use the pythagorean theorem to calculate the size of the largest square

[lean otter]

Not the size we need the parameters

[pure pendant]

i mean

not the size

i mean the length of the side

sorry

[lean otter]

How to get the parameters from just knowing the length

[weak shuttle]

I still don't understand what parameters you are talking about

[pure pendant]

parameters are perimeters kek

[pure pendant]

we are calculating square parameteres right?

[weak shuttle]

they said it's not

[lean otter]

I need to know the coordinates guys

The coordinates of other vertices

[weak shuttle]

the coordinates were given

[weak shuttle]

you don't need them

[pure pendant]

what

[lean otter]

Oh I read it wrongly... sorry

Oh that makes things much easier

[pure pendant]

@lean otter

so its $\sqrt{(2^2+4^2)}$

wtf how do i do this again

[weak shuttle]

\sqrt

[pure pendant]

oh yeaah thank you

[flat frigate]

alephcomputer

[pure pendant]

which is sqrt(20)

[lean otter]

How do you get the side of a square from diagonal again

[weak shuttle]

you divide it by sqrt(2)

[pure pendant]

no no don't relate to any square diagonal here

what i am doing is using a right triangle

to calculate the longest side

which happens to be the side of the largest square

and from that calculate $4(\sqrt{20})$

[flat frigate]

alephcomputer

[lean otter]

But he said I should do it

[weak shuttle]

i think he asked already for the smaller square

yeah

[pure pendant]

to get the perimeter of the largest square

[lean otter]

How about the smallest one

[pure pendant]

ok

so

[weak shuttle]

for the smaller one, that sqrt(20) will be the diagonal

so you divide it by sqrt(2) to get the side length

[pure pendant]

now we are going to calculate B

and we already know diagonal side of the B square

so to calculate the side of the B square we have

$\sqrt{(\sqrt{20})/2}$

[flat frigate]

alephcomputer

[pure pendant]

is that?

@weak shuttle @lean otter

[weak shuttle]

no

should be sqrt(20)/sqrt(2)

[lean otter]

I don't think that that's right

[pure pendant]

what im doing here is that to calculate the side length of B

[lean otter]

Why

[weak shuttle]

because sqrt(20) is the diagonal

[pure pendant]

divide by sqrt(2)?

tbh i have a way to do this really simple

and visiually

[weak shuttle]

and if a square has side x, the diagonal is equal to x . sqrt(2)

[pure pendant]

@lean otter do you want to know?

[weak shuttle]

so x . sqrt(2) = sqrt(20)

x = sqrt(20)/sqrt(2)

[pure pendant]

isn't that the sum of 2 side in B square

[weak shuttle]

?

[pure pendant]

kinda confused

[lean otter]

Kk

[pure pendant]

rearrange the 2 triangle in B square to A

[lean otter]

How?

[pure pendant]

told you its visual

well

anyhow

[pure pendant]

@weak shuttle

yes, i get what you are saying

countiue

[lean otter]

I know how to get it

I think

First change sqrt(20) to 2sqrt(5)

Since that's a square if we draw both diagonals to get a 90° triangle beside each square both with sides sqrt5

Sqrt(Sqrt5+sqrt5) *4

[pure pendant]

yse

[safe radish]

@lean otter Has your question been resolved?

[lean otter]

.Close

. Close

.close

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[lean otter]

@pure pendant @weak shuttle thanks guys

.close

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[smoky vortex]

https://media.discordapp.net/attachments/903463351789694986/903558352095813653/20211029_091758.jpg

[smoky vortex]

anyone know how to do this? I forgot

[oblique trench]

if divisible answer will be 0 right?

[smoky vortex]

yes

[oblique trench]

and soo

u know that remainder theorem

actually idk if its called remainder theorem

but u can sub 2 and -3

in

[smoky vortex]

all i know is if theres a remainder u just write it with the (x-2) below

[oblique trench]

and they ll will equate to 0

so like

[smoky vortex]

factor theorem u mean?

[oblique trench]

idk what its called lol

but u can just sub 2 into the equation

and equate the equation to 0

so like f(2) = ...

and then once u sub 2 in for x then u can equate all of that to 0 so now: 0 = ...

[smoky vortex]

what do indo with the p and q values

idh their values

[oblique trench]

you ll find them

when u sub the 2 and -3 in

as that will make 2 simul equations

which as we know when given 2 unknow variabesl we must have another equation to find them

[smoky vortex]

okay so i have

2^4+p2^3+q2+36
-3^4-p3^3-q3+36

[oblique trench]

i mean if u sub that in right then yes

and u can equate both of those to 0

2^4+p2^3+q2+36 = 0
-3^4-p3^3-q3+36 = 0

like this

then u can now solve

[smoky vortex]

then do i solve them as simultaneous?

[oblique trench]

yess

[smoky vortex]

lol idh my classwiz on me snh

1src

sec

[oblique trench]

huh

ookk

[smoky vortex]

my calculator usually solves simultaneous equations lol

[oblique trench]

loll nice

[safe radish]

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[sour frost]

i managed to get part i which is 8/3

but im not sure how to rotate about x axis

[quasi bison]

@sour frost do not hog multiple channels.

[sour frost]

okay sorry

i have removed it

[quasi bison]

ok

have you done solids of revolution before

[sour frost]

yes but ik the formula too

but im not sure how to approach it

This is my part i

[quasi bison]

are you more familiar/comfortable with the shell method or the washer method

[sour frost]

i think shell

im okay with both tbh

[quasi bison]

okay then shell it is

you will be integrating 2πx(3 - (sqrt(x) + 1))dx from 0 to 4

[sour frost]

oh

so its a semisphere?

[sour frost]

and this gives the volume?

[quasi bison]

no it's not a hemisphere

but yes that integral i gave you will yield the volume

[sour frost]

i cant seem to visualize the image tbh

whats the image may i know

[quasi bison]

the image?

[sour frost]

yea shldnt it form a3d shape

[quasi bison]

oh, you want to see the actual shape made by this

[sour frost]

yes

i cant seem to visualize how it will look like

[quasi bison]

...i don't have an image on hand, but it looks kind of like a disk with a spike sticking out from the bottom.

[sour frost]

ah okay thank you isit possible to ask another qn

[quasi bison]

well i guess while we're still here and the channel's still occupied by you

go ahead

[sour frost]

for part ii and iii also

[quasi bison]

okay so these are solids of revolution again

one will require shell and the other will require washer

[sour frost]

part iii is washer rite

[quasi bison]

no, it's the other way around afaict

part iii will require shell and part ii will require washer

assuming you integrate along the x axis

[sour frost]

oh okay so what r the steps

[quasi bison]

look up "solids of revolution"

[sour frost]

uh yes i just did

but how do i form the eqn

i get how to form the earlier qn now

because its just 1

but theres 3 eqns now

[sour frost]

help pls?

[quasi bison]

...

sorry, i've run out of energy for the time being

you'll have to look things up yourself

[sour frost]

can someone else help instead?

.close

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@sullen thunder Has your question been resolved?

[safe radish]

@sullen thunder Has your question been resolved?

[lean otter]

every question you've asked after 3:50 am pst has been something that could be googled so I'm not really sure how I can offer any type of substantial help

[tame raft]

yes

sometimes it's easier to learn when actual ppl are talking to you than from google

this isn't one of those times

[lean otter]

some questions can't be googled, like if you're stuck at a very particular point in a problem

[tame raft]

no one can be bothered to teach you entire topics when you can just google online videos

if you have specific queries that aren't basic or easy to find on google

we'd be more inclined to help

but if it's too big, ppl won't want to help

[lean otter]

this doesn't just apply to this server, this applies to any type of office hours

[tame raft]

https://www.youtube.com/results?search_query=how+to+calculate+canonical+correlation

videos, have fun

no one can be bothered to teach you entire topics when you can just google online videos
if you have specific queries that aren't basic or easy to find on google
we'd be more inclined to help
but if it's too big, ppl won't want to help

[lean otter]

your definition of "help" is tell me how to do this problem

for us to do that, we'd basically have to write an entire essay's worth of writing to explain everything, which would be the same amount of time spent reading a Wikipedia article, reading someone's webpage about the concept, reading a pdf resource on the topic of a university resource, or watching a video on it. And at that point, just skip the middle man (us) and look it up yourself. If it wasn't a basic "how" or "what", but instead a "why" question, we'd be happy to help.

[past cypress]

bro just help him

its not that hard

[lean otter]

JUST FUCKING HELP HIM

[tame raft]

bruh

[lean otter]

i got to go

rio

[tame raft]

you all feel so entitled

you feel like you're entitled to help

but you're not

[lean otter]

does anyone read the rules here

[tame raft]

a certain type of help

this is dumb

[past cypress]

damn.

wiat kaisheng i think i saw u in my cs lecture at MIT yesterday

r u the asian kid with glasses

[warped roost]

.close

[safe radish]

Channel closed

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[warped roost]

troll banned

[safe radish]

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[past cypress]

theyre asking if they can be unbanned

[past cypress]

they wanna apologize bw

i think

[warped roost]

lmao

no

[past cypress]

its up to yall ofc tho

idrc

[lean otter]

can i ask a question here?

[warped roost]

we dont tolerate intentional trolling and we especially dont tolerate false reports

they arent getting unbanned

[warped roost]

please use an opened channel

like #help-5

[lean otter]

it says i dont have permission to text

oh i found the others

.close

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[gloomy thicket]

typename BaseVertexShader::Output operator()( const Vertex& v ) const
        {
            // calculate some triggy bois
            const auto angle = wrap_angle( v.pos.x * freq + t * wavelength );
            const auto cosx = std::cos( angle );
            const auto sinx = std::sin( angle );
            // sine wave amplitude from position w/ time variant phase animation
            const auto dz = amplitude * cosx;
            const auto pos = Vec4{ v.pos.x,v.pos.y,v.pos.z + dz,1.0f };
            // normal derived base on cross product of partial dx x dy
            auto n = Vec4{
                -freq * amplitude * sinx,
                0.0f,
                -1.0f,
                0.0f
            };
            n.Normalize();

            return { pos * worldViewProj,n * worldView,pos * worldView,v.t };
        }

[gloomy thicket]

Ay so basically im confused as to how they are calculating the normal direction based on the partial derivative

this code basically turns a plane into a wiggling surface

and they are calculating the normals that move up and down on it but im not exactly sure how

freq = 45.0f, wavelength = pi, amp = 0.02

so they get the normals mapped out here on this surface which changes with time

[inland ivy]

Is that the eye of sauron

[gloomy thicket]

yes

they mention the cross product to calculate the normals direction but not sure what they mean

cross product of the partial deriative

[inland ivy]

Cross product is just like getting a perpendicular vector

That should make sense conceptually

[gloomy thicket]

yeah i get that

[inland ivy]

And how the normal is perpendicular to the vector representing the derivative?

[gloomy thicket]

yeah its more to do with how the code is portraying that process tbf

it takes in some vertex in a 3D space as input

then takes the cos (x_pos * wl + t *freq) and sin of same thing

the coordinates of my world are right handed i believe

[inland ivy]

Discord formatting for code on phones is so bad

What language is that C++?

[gloomy thicket]

yeah

[inland ivy]

Thanks

[gloomy thicket]

im just confused with overall logic of whats going on here to make this plane has surface normals

i understand the concept though

wrap angle keeps the angle between 0 and 2pi also

$cos(45x + (freq \cdot time))$

so we do this for sin also ofc

my idea is that this moves the x coordinate 45 units every time shift based on frequency yeah

then they calculate dz being this multiplied by the amplitude?

why is that

[flat frigate]

shriller44

[inland ivy]

z is up right?

In your progrma

[gloomy thicket]

yeah so its a right handed space

so actually z is into the screen

[inland ivy]

Well I mean that makes sense

Cause you're taking a flat plane

And making it go up and down based on a cosine wave

[gloomy thicket]

but it makes the y go up by dz?

z*

that doesnt make sense conceptually

[inland ivy]

It's just like if you had a flat line

[inland ivy]

y=4 or something in 2d space

Then you added cosine*amplitude onto 4

It'd make a cosine wave at that height

4

[gloomy thicket]

but shouldnt they increase the y value to make it go up?

[inland ivy]

For my example?

[gloomy thicket]

https://youtu.be/PajM-QC5OUM?t=1293

look here in this animation

each y pos is going up and down no?

unless ive got myself confused

[inland ivy]

Wait you said z was up and down right?

[gloomy thicket]

well z should be into the screen technically

y is up and down

[inland ivy]

Hmm

[gloomy thicket]

acc i think whats happened is

this object of the surface

started initially as a square right

facing like on the screen

i think ive been dumb as he has rotated it downards

so its local coordinate system is actually different

so z is the pulsating up and down now

[inland ivy]

Oh yeah probably

[gloomy thicket]

cause thats what 3d shit is

objects have there own perspective against the base canonical system

[inland ivy]

Useful but also confusing

[gloomy thicket]

so we technically have as that frame i think

y forwards, x, right and z upwards

[dense igloo]

Hi, the real numbers in between 1 and 2 are infinite right?

I am learning about open set

[gloomy thicket]

@dense igloo yes

[inland ivy]

Probably

[dense igloo]

@inland ivy probably?

[inland ivy]

I mean if you think about it

You can have infinite decimals after a number

[dense igloo]

@inland ivy yes anyway thank you a lot

[inland ivy]

Np

[gloomy thicket]

yeah why is it amplitude times cos(45x + ft ) again

for change in the z

cos is the direction of the x axis i gues

so in the cross product part of the normal derivation

they first multiply the wavelength to move the normal along the ripple

and then multiply this by amp * sin(x) as to scale the size of the normal relative to upwards direction

thats my logic on it

so yeah you see saurons object has been rotaed by pi /2

which is 180 ??

wait is it no

[inland ivy]

90

[gloomy thicket]

90

[inland ivy]

Yes

[gloomy thicket]

so yeah its rotated from facing up to being plopped down as its rotated on X axis by 90 degrees

what is the signifance of cos(x) and sin(x) for directions again

of the same angle

sin i just see as up and cos right

[inland ivy]

-sin(x) is the direction of cos(x)

[gloomy thicket]

is that just to do with the derivative

[inland ivy]

Ye

For vectors

It'd be just like he had it

(1, -sin(x)) normalized is the direction of cos(x)

[gloomy thicket]

why is he mentioning cross product when they are all scalar values ?

acc wait

the cross is liek

is that what the sin in the code means then

a = wavelength and b = amplitude

[inland ivy]

I don't think so

It could be

[gloomy thicket]

mult by sin

well - sin as you said as its deriative

so in the z it has one as we want our normals to be one siez

im just trying to visualise specifically whats going on it does add confusion with teh shifted axis

[inland ivy]

If it weren't for the wavelength I'd say n is the derivative vector

[gloomy thicket]

thats the purpose of it

its supposed to the derivative vector

[inland ivy]

Oh wait

[gloomy thicket]

in same direction

[inland ivy]

I bet the wavelength comes out cause it's attached to the variable of differentiation

[gloomy thicket]

so the bottom is the axes btw

we are working on

z acc pointnig dowin

you saying we dont need the wavelength

[inland ivy]

I think it just comes out because that's the derivative

So that makes sense

[gloomy thicket]

i think the confusing thing is its creating a normal in the normal x y z location

but basing it off the rotated values

[inland ivy]

Ye maybe

[inland ivy]

But it's basically just this I think

With extra stuff to fix the problem specifically

[gloomy thicket]

so going through the code , the logic is we

first find what angle we have from 0 to 2pi

we then evaluate at sin and cos

where cos is the direction of the wave in the right direction, and sin the upwards direction

[inland ivy]

I think cos is just the height of the wave

[gloomy thicket]

huh

cos is used as the offset to increase and decrease the z ripple amplitude acc thinking about it

to create the wave of course over time

[inland ivy]

Yes

Well over x

Modified by time

[gloomy thicket]

yeah each time unit + wavelength mult by whatever x coordinate we are looking at butt

so we have this ripple now and we want the surface normals at each point

and the derivative of cos is -sin

so this becomes the direction of the surface normal vector right

[inland ivy]

Yes

I imagine he didn't actually use cross product

But just flipped the x and y

And made it negative

[gloomy thicket]

yeah im trying to think bout that

it doesnt seem like he did

[inland ivy]

Well the 1 should be positive

So he probably flipped it from

Slope = -sin x/1

To

Perp slope= -1/-sin x

[gloomy thicket]

yeah the code seems to work with just this]

[inland ivy]

Epic

[inland ivy]

Cause yeah z is up

So sin if it was the normal slope

Would be in the z

So yeah that's 100% what he did

[gloomy thicket]

but he multiplied this by the amp and wavelength to acc give this direction some length

i think

[inland ivy]

He just did that because the derivative I think

Because chain rule

When you have the derivative of let's say

cos(5x)

It's not just -sin(5x)

[gloomy thicket]

the code is calculating the lighting correct is what i say, but maybe not perfect i dunno

[inland ivy]

It's -5sin(5x)

[gloomy thicket]

cause lighting is based on the surface normals

the idea of the demo is to show how the plane sort of has a shadow move across it as the normal direction changes

meaning light hits it differently

[inland ivy]

Ye

[gloomy thicket]

mine seems to be okay in that regards

why do u think he mentioned idea of cross product its confused tf out of me

[inland ivy]

I mean it shows that it's the perpendicular of the derivative

It's how he derived and defined it

I imagine

[gloomy thicket]

yeah so he got the derivative -sinx as the direction

from this single direction you can get cross product by finding something else to multiply it with

which is not collinear or something

so if its pointing upwards and we multiply by wavelength * amp this is kinda whats going on

[inland ivy]

Well the multiplications is just how derivatives work

I don't think it has any specific meaning

[gloomy thicket]

yeahh

but why does it work without amp / wl

well at least i think its doing it

[inland ivy]

I mean it's probably close

But not perfect

Especially for higher values of amplitude and wavelength

[gloomy thicket]

yeah true its precise on it

right so amp part is just to push the normal up and down

but in the direction of the surfce normal

and wavelength is just to push it along right

like not push it acc

cause it constant

[inland ivy]

For wot?

[gloomy thicket]

well his formulae is -sin(x) * wl * amp

-sin and amp makes sense as its just pushing it up and down

then he multiplies by wavelength

becausss

[inland ivy]

I mean it's a true statement so there's prob many ways to prove it

And I'm sure that should make sense

Not necessarily for the upwards normal but to scale up and down based on the amplitude

Yeah that makes sense

[gloomy thicket]

what

yeah i get that part im just trying to make sense of wavelength part

unless my variable names are wrong and its supposed to mean frequency

oh his code tehcnicall it does but he said to switch

[inland ivy]

Shrug

I'm sure it makes sense in some aspect

[gloomy thicket]

cross product think comes into paly surely

as we have rippling wave

the direction determined by sin but thats one part of the value we need

-sin*

[inland ivy]

I mean rhe in this case is literally just swapping x and y and making it negative

[gloomy thicket]

swapping?

how has he swapped x and y

the third is value is z

acc technically you could say its y

its y pointing into the screen the third value

the top value is x pointing to the right which is where the calculated normal is placed

in that n = vec4 bit

[inland ivy]

I mean swapping x and z

Also I think it makes sense

He did in fact mislabel the wavelength

[gloomy thicket]

what me or him

[inland ivy]

It should be frequency like you said

[gloomy thicket]

nah that was mine that was wavelength

[inland ivy]

Oh

[gloomy thicket]

this is what he has

t starts at 0 and is incremented somewhere else

so the idea is to push the calculated direction left

thats what it does in the video

its just weird his labels as he does t * wavelength

time * wavelength

it does make sense surely

hes definitely sure about using the cross prodcut

[inland ivy]

Ye

[gloomy thicket]

https://youtu.be/PajM-QC5OUM?t=1233

[inland ivy]

He def does use cross product

He just doesn't show his work in the code

[gloomy thicket]

nah its not fully explained

i get it all apart from the mult by frequency

[inland ivy]

Well if there's a higher frequency the derivative would go up and down more often

Lower frequency the derivative would go up and down less often

[gloomy thicket]

so visually a higher freq would be closer normals?

[inland ivy]

Ye

[gloomy thicket]

as more ripples

[inland ivy]

Si

[gloomy thicket]

i guess that does make sense like its an extra offset

to move it right a certain degree to match the frequency of the original wave

and then multiply by the -sin and amplittude to obtain the direction in the right place

angle calculation using x * freq is less intuitive but it makes sense i guess

like hes using timestep * wavelength rather than freq

seems a bit weird to me

like look at this

oh shitt its his values i think

like

look at what values he used

oh wait thats my ones one sec

[inland ivy]

Wavelength must be the length of the wave in relation to time

Not of the cos wave itself

[gloomy thicket]

what ?

wavelength is the length of a full cycle

[inland ivy]

Like how it moves

[gloomy thicket]

so time i assume is scaling that between 0 and 1 to get where you are on the wave no?

i mean not acc 0 and 1 but

cos(x * freq + time * wavelength))

so our point x which is actually z technically

we first multiply by the frequency as to get the spacing right

like 45x in this case shifts x 45 units along every coordinate

so after that we translatea it the correct bit into this section of teh wave

which is the time we are on multiplied by how big a single wave is based on one timeframe

like multiply points are gonna be in the same sort of 'cycle

so each x incrementally pushes into its own wave with different angles from 0 to 2pi

thats how i am seeing it

freq determines how far they get pushed scaled by what exact point it is

im overthingking it a bit but yeah

[inland ivy]

Maybe

It's just a cosine wave that moves with time by some scale factor

[gloomy thicket]

i mean cos and sin become meanignless surely

[inland ivy]

That scale factor or almost speed is the variable wavelength

[gloomy thicket]

acc no its cos as we chose it to be tbf

nvm

we can switch cos and sin and adjust derivative a bit to cretae antoher wave ofc

essentially wrap angle takes all points, maps them to between 0 and 2pi and this mapping is constructed on specific wavelengths and frequency

wrap angle idea is as x gets bigger its just gonna get larger

as its periodic we just mod it to wrap around

[inland ivy]

Ye

[gloomy thicket]

so yeah i think it all makes sense now cheers

[inland ivy]

Adios

[gloomy thicket]

.close

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[hexed fog]

Help

[worthy hemlock]

,rotate

[hexed fog]

I replaced 2-x with Z

Then replaced every x

So ?

[hexed fog]

Make it y so the 2 do not disturb

[safe radish]

@hexed fog Has your question been resolved?

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[short rover]

I'm row-reducing a matrix, what symbol should I put between each step ? I don't think it's valid to put an equal sign since the matrices are different ?

[steel stag]

an arrow is fine -->

[safe radish]

@short rover Has your question been resolved?

[short rover]

okay thank you!

.close

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[hard prairie]

How would you turn any number to an unique number between two numbers. Like 290378487 would be turned to a number between 0-15. And if the input number is any different its going to turned into a different number between the 0-15? Its for a program but maybe there's like an equation for it

[median kestrel]

Perhaps the modulus function would do?

[safe radish]

@hard prairie Has your question been resolved?

[hard prairie]

.close

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[flat frigate]

Goose on a Moose

[frosty ridge]

This is for part (a)

[stiff inlet]

with a mean of 69 and a standard deviation of 2.5, 74 would be two standard deviations from the mean

[frosty ridge]

So two standard deviations means that we would be in the 95 area?

so it would be 5%?

[stiff inlet]

~4.55

so your answer was divided by two for some reason

[frosty ridge]

but why dont we divide by two? In your diagram, we only want to find the probability of the green area on the right (and not the one on the left).

[stiff inlet]

oh duh, you're right

[worthy hemlock]

You want the percent area of taller than 74 inches, so $$P(Z > \frac{x - \mu}{\sigma}) \implies 1 - P(Z)$$

[flat frigate]

dldh06

[stiff inlet]

Perhaps it wants the "rougher" answer of 2.5?

[worthy hemlock]

So P(Z > 2) is about 0.9972, so 1 - 0.9972

Which is 0.0028

Or 0.28%

[stiff inlet]

if it means 68-95-99.7 strictly (and not 68.27-95.45-99.73), perhaps the answer key is using 5% outside of 2 standard deviations instead of 4.45

and therefore 2.5 instead of 2.275

[frosty ridge]

I think this is most likely the solution. I haven't realized it was conveniently two standard deviations away. I just went straight to the z-score formula.

[worthy hemlock]

Z = 2 is not 0.02275

[worthy hemlock]

Negative

2

Sorry

Can't read

[frosty ridge]

okay but negative would just be 1-0.97725

so it would be 0.02275 right?

[worthy hemlock]

Yeah, should be

[frosty ridge]

okay thanks.

.done

.close

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[atomic bane]

,help

[flat frigate]

A brief description and guide on how to use me was sent to your DMs!
Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[atomic bane]

.close

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[naive comet]

Hi, I need help with these two questions. I really don't know how to go over through them. Like what do I plug in and how? I'm really in the dark here. I've been stuck on these for the past 30 mins and I can't get past it.

[stiff inlet]

for the first one, you know two things

d= 3 * x

and

d = 2 * (x + 20)

(you also know d=d)

so 3x = 2(x+20)

from there you can distribute and solve for x

for the second one, you know 94500 = x + 2x + 4x +8x + 16x + 32x

on that one, remember you don't want x (the payment in the first year), you want the payment in the fifth year

[lyric sable]

Hey I would like to know what exponent property causes the 4^2 and 9^-1 to be there

[worthy hemlock]

Don't ask in an occupied channel

Read #❓how-to-get-help

[safe radish]

@naive comet Has your question been resolved?

[safe radish]

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[zenith socket]

help

[zenith socket]

.close

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[zenith socket]

nvm

[safe radish]

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[livid blaze]

can someone explain what I'm looking at here? this thing doesn't look exactly look like what i've seen in class

for reference, what I see in class is something like:

[safe radish]

@livid blaze Has your question been resolved?

[short falcon]

Just to notice a couple things here, the superscript p in what you see in class is in fact an exponent, in case it wasn't clear

So E_p from class is equal to the pth root of the righthand side

The 1/p from your class picture inside the sum doesn't appear to be in the image for your question

[livid blaze]

I'm also kinda confused at the wording of the thing

We didn't explicitly go over what p means which is why I'm kinda lost

like I know what the number will affect as p gets large or small but idk what p represents

[short falcon]

I'm guessing A_n is your like... list of data

And S is the optimal summary statistic, perhaps?

[livid blaze]

sounds right

sorry when you say summary statistic do you mean like the mean or mode?

or something else entirely

[short falcon]

Well it could be those potentially

Like they say

In the limit that p goes to 0, apparently S is the mode

When considering the minimization problem

[safe radish]

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[faint lotus]

I'm not sure what a summary statistic is, but
$$
E_p = \left{\sum_{n=1}^N |A_n - S|^p\right}^{1/p}
$$
is a pnorm https://en.wikipedia.org/wiki/Norm_(mathematics)#p-norm
which converges to max as p -> infinity @livid blaze

[safe radish]

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[flat frigate]

uli
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[livid blaze]

thanks il take a look at it

[short falcon]

I'm guessing that if p=2 then the S that minimizes E_p is the mean

I'd try to play around with minimizing E_p with respect to S and try to see why the S that minimizes E_p is the mode for p approaching 0, and (probably) is the mean for p=2

To gain a bit more understanding before tackling the actual meat of the question

[faint lotus]

intuitively the reason the pnorm approaches max as p->infinity is because the differences blow up when you raise them to large exponents (eg. if |A_1 - S| = 0.3 and |A_1 - S| = 0.9 then 0.3^p is exponentially smaller then 0.9^p) then taking the pth root with (1/p) brings you to the maximum. I'd expect similar logic to apply in the p -> negative infinity case

as p -> -infinity the pnorm becomes the minimum, i'm too lazy to prove it so here's some examples in python

[livid blaze]

oh

thx for example lol makes things a lot easier to read

[safe radish]

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[frigid canopy]

could someone help me with surds

[pulsar condor]

wow, what a specific question and not at all vague

answer is yes.

[frigid canopy]

bro hates me with a passion

[pulsar condor]

No, just pointing out you wont get help for vague questions

[frigid canopy]

could you help me with this or no

[pulsar condor]

help with what?

you posted numbers with no actual question

[frigid canopy]

Completely simply the following, giving your answers in the form pඥq where p and q
are positive integers, where possible.

[visual crater]

sqrt(45) = sqrt(9*5) = sqrt(9) * sqrt(5) = 3 * sqrt(5)

you have to factor the number under the radical into numbers you can find the sqrt's of. and when you can't go any further, you stick with the sqrt e.g. sqrt(7) is a stop.

[frigid canopy]

thanks

[visual crater]

or in the case of the iii) sqrt(5) is a stop because 5 is prime.

and.... sqrt(108) = sqrt(9)sqrt(12) = sqrt(9) * sqrt(4)sqrt(3) = 32sqrt(3) = 6*sqrt(3)

my * dissapeared.

oh, i know

$sqrt(108) = sqrt(9) \cdot sqrt(12) = sqrt(9) \cdot sqrt(4) \cdot sqrt(3) = 3 \cdot 2 \cdot sqrt(3) = 6 \cdot sqrt(3)$

[flat frigate]

tensor

[visual crater]

I need to work on my square root symbol

[pulsar condor]

\sqrt[]{}

[visual crater]

$sqrt{108} = 6 \cdot \sqrt(3)$

[pulsar condor]

[3] gives cube root, [4] gives 4th root, etc

[visual crater]

ooof!! 😦

I give up.

[pulsar condor]

$\sqrt{108}=6\sqrt{3}$

[flat frigate]

Mosh

[safe radish]

@frigid canopy Has your question been resolved?

[safe radish]

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[cobalt plinth]

Find the x-intercepts for
y=6x^3-18x^2+12x-20