help-10

[edgy grove]

Yup what is your doubt ma'am?

[scenic fjord]

I wanna know why lamda is equal to cp

Like which parts tell us it’s equal to cp

[edgy grove]

Actually i will say something useful for advanced concepts like this try and use the specific topic channel as you might get answer faster

[edgy grove]

Yup i will see it

[scenic fjord]

Oh, so next time I post in ODEs and PDEs channel?

[edgy grove]

Umm i mean like #calculus

Or #advanced-lounge

[scenic fjord]

But this is under PDEs

[edgy grove]

Can you say full form of pde

[scenic fjord]

Partial differential equations

[edgy grove]

Yup then try

#multivariable-calculus

[scenic fjord]

I think I’ll try in PDEs channels since this is under this topic

[edgy grove]

Umm i don't want to read the theory you sent can you please summarise it for me

I can understand wave parts

And also pde

But then why are we finding wavelength

[scenic fjord]

I’m not use, it’s a heat equation solving using Fourier series

[edgy grove]

And what is c and p

[edgy grove]

Heat?

[scenic fjord]

That’s why I’m blur

[scenic fjord]

Yea

[edgy grove]

Wave in heat?

Ok just say what c and p is

[scenic fjord]

I don’t know

This is found from the mathematical methods shown in textbook

[edgy grove]

Ok i understood

Umm here you are asking where pi/l came from?

[scenic fjord]

I’m asking why lamda is equal to cp

I don’t know where this comes from

[edgy grove]

Umm can you just highlight the place where cp is mentioned in sinosodul equation

[scenic fjord]

So why lamda is equal to cp?

[edgy grove]

There is a seperate chanel for pde and ode

Wow

Oh ok @scenic fjord

I will send a pic of what they are speaking

[scenic fjord]

Huh? I just want to know why lamda is equal to cp

[edgy grove]

Just one question ma'am can you pls say what that p is

Is it the time period

Or frequency

[scenic fjord]

I already put in the red box

[edgy grove]

Ma'am i can only see p there

I can't see what p means

[scenic fjord]

I also can’t see what p means

P is obtained from previous solution

Just sub it

[edgy grove]

Hmm

Ok Lemme check

[scenic fjord]

The point now is, I wanna know why lamda is equal to cp

After I know this, I can proceed with the textbook workings

[edgy grove]

Oh i understood finally

It is the coefficient of x

[scenic fjord]

Erm, I don’t think it is

Anyway, I just see again and I know why lamda is cp

[edgy grove]

I mean coefficient of wavelength

Umm can you understand?

[scenic fjord]

Erm

[edgy grove]

I will explain why p is npi/l

[scenic fjord]

This is wave in physics right ?

[edgy grove]

And why c is speed

Yup

[scenic fjord]

I learnt this before and well, I mislook wave equation to heat equation

Anyway, actually I have no idea why we need to solve like this

[edgy grove]

Umm you mean like the book or mine?

[scenic fjord]

The book

[edgy grove]

Actually why are you studying Fourier series

?

[scenic fjord]

The normal solving method with waves I already learnt

[scenic fjord]

Mathematical method for physics

[edgy grove]

I learnt both on physics

Sooo idk why it comes on maths

[scenic fjord]

Hmm, I learnt wave in y1 without knowing fourier series

[scenic fjord]

Because it’s a textbook for engineering mathematics

[edgy grove]

Ma'am can i DM and can you explain why we use Fourier series in wave ?

[scenic fjord]

Bye and thank you

.close

[obtuse pebble]

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[north anchor]

help with this

[north anchor]

how i can calculate the intervals a and b

[native inlet]

you mean -1 and 1?

[north anchor]

yes

how you know if a and b

[native inlet]

find where f(x)=g(x) is what I would do, but there might be a better way

[north anchor]

Always works?

[north anchor]

if you use g(x)=f(x) works?

[north anchor]

how did you determine that a -1 and b 1

someone

[obtuse pebble]

@north anchor Has your question been resolved?

[distant cairn]

khm

idk but it seems like about solving quadratic equation to get a and b

graphs are intersecting so... x^4 -4x^2+1=x^2-3 => x^4 - 5x^2 +4 =0

On Viet 1 and 4 are roots for x^2

that's about all

[distant cairn]

simple?

[north anchor]

how you know if a or b

[distant cairn]

the area is calculated under curve

here between one curve over Ox and anouther negative curve under Ox

[north anchor]

oh i see

gpt chat told me that to know if a function is superior or inferior, you had to evaluate both functions by the interval

[obtuse pebble]

@north anchor Has your question been resolved?

[left hornet]

If it's hard for you to find the intervals a and b by finding the roots of a degree-4 equation, you can call x^2 a variable like t, and solve for t like a normal degree-2 equation
Now that you have t, you can find x by calculating the square root of t

[north anchor]

I know but I don't understand why a = -1 and b = 1 and not vice versa

[left hornet]

If you calculate it as I said, you should get a = -1 and b = 1

You did it, and got something else?

[north anchor]

I didn't understand

[left hornet]

You didn't understand this^?

So why did you say "I know"?

and my advice is, try to understand it

What part of it is unclear?

[north anchor]

I mean, I don't want to find a and b. I want to identify if it is the interval a or b.

[left hornet]

Cause any other root of the equation would be out of it

These two, the only other-roots of the equation, are out of the interval that you want

[north anchor]

The largest number is b nd the smallest number is a?

[left hornet]

The largest number?

[north anchor]

yes a is negative , -1 and b is 1

[left hornet]

In the set of the common values of the area of the first equation and the second equation, yes

[north anchor]

will it always be like this?

[left hornet]

What do you mean by always?

By changing what object?

[north anchor]

a must be less than b and b must be greater than a

[left hornet]

If it swaps, the answer would just get negative

integral f from a to b = -1 * integral f from b to a

[north anchor]

yes

[obtuse pebble]

@north anchor Has your question been resolved?

[obtuse pebble]

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[silver plover]

could someone explain plz how they drew that graph on the uv axis

like i dont get what they based that on

the sketch

[haughty coyote]

R is the region represented by that rectangle on the u-v basis of R^2

[silver plover]

how does it give a length of 3 tho

it doesnt look it

from the xy plane

[haughty coyote]

v goes from 0 to 3 by definition of R

"2x-y = 0 to 2x-y=3"

[silver plover]

sorry im still confused on how these two regions r the same

[haughty coyote]

Change of basis

It's linear algebra

Of course they're not the same geometrically

But they describe the same region of R^2

[haughty coyote]

And with all the jacobian stuff you can see how the integral is transferred from one basis to the other

[fierce lagoon]

You take one region from one "paper" and you "re-map" it onto a different "paper"

That's how I think of Jacobians

[obtuse pebble]

@silver plover Has your question been resolved?

[silver plover]

makes sense

thanks

.close

[obtuse pebble]

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[obtuse pebble]

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[exotic nebula]

.

[fathom flicker]

surely it would be easy to substitute it in and try

if y=-7e^(2x)

find y'

and then find y''

plug all of those in

is it equal to 0?

if so, true. If not, false

[exotic nebula]

lemme pull up my calculator

d/dx is -14e^(2x)

[fathom flicker]

dy/dx

[exotic nebula]

missed a letter but yeah

y''=-28e^(2x)

how do i solve the DE after plugging it all in

i plugged it in, i'm just not sure what to do now

[brisk matrix]

simplify

if the left hand side is equal to the right hand side, then that is a solution

[exotic nebula]

Oh ok gotcha

so it's not equal

so false

[brisk matrix]

then it’s not a solution

[exotic nebula]

yep

what about this one?

This one can be done using separation i'm pretty sure

Just by looking at it, it's probably false. But how would I show that?

[brisk matrix]

plug it in again

[exotic nebula]

I got y=e^(x+((x^2)/2)))

[brisk matrix]

you don’t need to solve the ode

[exotic nebula]

oh

[brisk matrix]

you can just plug in the y

[exotic nebula]

ohhh

I got some weird number

like you mean plug in y for y'?

[brisk matrix]

x^2e^x for y

and then the derivative of that for y’

[exotic nebula]

yeah thats what i mean

oh ok

y'=2xe^(x)+4x^(2)e^(x)+x^(3)e^(x)

[brisk matrix]

where did that come from

[exotic nebula]

finding the derivative of y=(x^(2)e^(x))(1+x)

[brisk matrix]

that’s not y

y is x^2e^x

[exotic nebula]

Oh i did the whole thing, not just y

ok y' is 2xe^(x)+x^(2)e^(x)

[brisk matrix]

then plug them into the equation, and see if the lhs = rhs

[exotic nebula]

It does

it's true

[brisk matrix]

it does?

[exotic nebula]

Yes?

Does it not?

[brisk matrix]

can i see your work?

[exotic nebula]

Oh i see where i went wrong

Yeah it's false too

Not sure about this one

nevermind i got it

[obtuse pebble]

@exotic nebula Has your question been resolved?

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[tight pike]

I have two lines y=-300r+9000 (IS) and y= 20r + 1500. (LM) My problem asks which policy (fiscal or monetary) will be most effective. Based on our notes it says that depending on which slope is steeper will decide which policy is better. Here I thought the first slope was steeper but according to my prof. and everyone else it's not, so I'm confused about how the first slope isn't steeper.

[obtuse pebble]

@tight pike Has your question been resolved?

[obtuse pebble]

@tight pike Has your question been resolved?

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[obtuse pebble]

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[foggy pumice]

I am working on a problem similar to f(x,y)=xye^x^2 D=[-2,2]x[-2,2]
I have a question asking me to do the following :⦁ Find the absolute maxima and minimum by evaluation f on the boundary of D
what does that mean

[rigid pine]

The boundary of the domain is made up of four line segments, for example, along the straight line connecting (-2,-2) and (-2, 2). Consider the function (on this line segment) f(-2, -2) = 4e^4, -2 <= y <= 2.

[foggy pumice]

While I appreciate the rewording, I still don't know what that is asking me to do

[rigid pine]

I wouldn't describe it as a rewording.

[foggy pumice]

ok i agree with you

[rigid pine]

You are trying to apply this theorem. If the extrema occur within the boundary (not on it) they will occur at the critical points. But they may exist on the boundary curve. You effectively need to consider four functions as suggested above to be able to determine the absolute maxima and minima.

Note. For clarity hypothesis 2 means some f(a, b) inside the boundary is higher than f(x, y) along the whole of the boundary. It might be clear but I, for some reason, got confused by it. Of course, a analogous theorem exists for minima.

[obtuse pebble]

@foggy pumice Has your question been resolved?

[foggy pumice]

ok so ive read this 15 times now. how would I apply this? theres a lot of symbols here I have never seen before

whats the name of this theorem maybe? that way I could look it up to see its application?

[obtuse pebble]

@foggy pumice Has your question been resolved?

[obtuse pebble]

@foggy pumice Has your question been resolved?

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[fathom flicker]

I am trying to find all the critical points of $\newline$ $\dot{x}=(2-x-y)x$ $\newline$ $\dot{y}=(x-1)y$

[warm shale]

AustinU

[fathom flicker]

I think that they are, (0, 0), (1, 1), and (2, 0)

Could anyone confirm, or point out if I am missing something?

[gilded needle]

what does the dot notation mean?

[warm shale]

AustinU

[forest sinew]

physics derivative

[fathom flicker]

it means derivative

[gilded needle]

isn't that with respect to some other variable?

like time

[fathom flicker]

yes

this system is a system x(t) and y(t)

[gilded needle]

got it, thx

[forest sinew]

you can use ZFT right

youll get two lines from each

[fathom flicker]

I don't know what ZFT is

[forest sinew]

zero factor theorem

in the x null cline, (2-x-y)=0 or x=0

[fathom flicker]

Would you mind elaborating?

right

[forest sinew]

so two lines

[fathom flicker]

so either x=0, or 2=x+y

[forest sinew]

yea

intercepts (2,0) and (0,2)

and also the yaxis

do the same for the y nullcline

[fathom flicker]

if x=0, the only critical point that comes from that is if y=0

so (0, 0)

if 2=x+y

[forest sinew]

doing too much casework i think

[fathom flicker]

we can get one with y=0 x=2

or with y=1 x=1

[forest sinew]

just extract the lines

well you do you i guess

[fathom flicker]

how would I do it with the lines instead?

[forest sinew]

you have two lines so far

[fathom flicker]

sorry if I brushed over that

[forest sinew]

one is the line connecting (2,0) and (0,2)

and the entire y-axis

then from the y nullcline, you get either y=0, or x=1

so you have the origin as stationary

then, the line connecting (2,0) and (0,2) crosses the y=0 line

then, the x=1 line crosses the line connecting (2,0) and (0,2)

3 stationary points

i think?

you may need to graph

https://www.desmos.com/calculator/d5gbnabxsc

here

anywhere a blue line crosses a green line is stationary

since it produces a zero factor in each equation

but you could just as easily create this plot by hand too

[fathom flicker]

oh that is cool, I didn't ever see that you could do it that way before

[forest sinew]

usually they arent nice lines lol

[fathom flicker]

produces the same three points (0,0), (1,1), and (2,0) that I got

[forest sinew]

yea

oftentimes you have to do this for ilke

grossly nonlinear messy things

[fathom flicker]

Okay cool that is good to know

thank you jan

[forest sinew]

[fathom flicker]

.close

[obtuse pebble]

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Remember:
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• Show your work, and if possible, explain where you are stuck.
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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[fathom flicker]

I'm reclaiming this from the same question as above

[fathom flicker]

So I found the critical points of the system

I found them to be: (0,0) (1, 1) (2,0)

and now I want to linearize the system near these critical points

Starting with the (0, 0) critical point

[forest sinew]

what class is this

[fathom flicker]

could I just linearize it as x'=0 y'=0

[fathom flicker]

I'm just self-studying some differential equations

[forest sinew]

so im gonna bastardize is

isnt the linearization like

youll take the partial of xdot wrt x

[fathom flicker]

ohhhh

right

[forest sinew]

wel li can write it

[fathom flicker]

I forgot about that

you do the Jacobian

[forest sinew]

yea

then you evaluate it

you wind up with a 2x2 matrix

you use the poincare diagram

to make your life easy

[fathom flicker]

poincare diagram?

[fathom flicker]

One second, I am going to look at my notes on the jacobian and see if I can't figure out how to linearize near the critical points

[forest sinew]

classification of 2x2 linear systems

[fathom flicker]

oh I know that

determinant trace plan

[forest sinew]

yea

[fathom flicker]

didn't know it was called poincare diagram

another good thing to know

lol

[forest sinew]

there are cooler things also called poincare diagrams

there are cooler things also called poincare diagrams

poincare maps

[fathom flicker]

I saw someone in diffeq channel earlier with one of those I think

something about it being a cross section model of a system in higher than 3D right?

[forest sinew]

im not that smart yet

[fathom flicker]

what are you taking right now?

[forest sinew]

last semester classes

about to graduate

[fathom flicker]

that's awesome!

math major?

[forest sinew]

i just took a handful of grad ODE classes so i know a tiny bit

yea

[fathom flicker]

just a tiny bit 😉

lol

well congrats that is super cool

[forest sinew]

eh u forget most of it

ah u mean terrifyign

[fathom flicker]

What was your favorite part / favorite class?

or least favorite

if you particularly disliked it

XD

[forest sinew]

i hated analysis

i like making graphics

so all the classes i got to make cool animations for

dynamics was a big one

also math biology

and neuroscience

made this one but i dont think we will use it

for my last class

[fathom flicker]

does that one have a short explanation or is it too above my grasp probably

what's it mean?

[forest sinew]

its a slowed down bird song, we are like

summing up the frequencies that are present in a way

i cant really explain it either

[fathom flicker]

aha no worries. Still cool

one second I am going to try to do the problem I started this with

[forest sinew]

its using a bunch of FFT's to pluck out the dominant frequencies

[fathom flicker]

then I'll be back

[fathom flicker]

see this part I can understand a bit

[forest sinew]

i made it run on every 100 or so samples of the audio so you can see the dominant frequencies appear alongside when you hear them

theyre distinct enough in this recording, not all layered

[fathom flicker]

is that also about the bird songs?

[forest sinew]

nah this is bifurcation stuff

u just got me going through old gifs

lol

[fathom flicker]

lol

yeah I wish I could understand more

but I haven't started school yet

[forest sinew]

you must be on bifurcations

o wow

not even a freshman?

[fathom flicker]

I'll be a freshman next year

in college

if that's what you mean

[forest sinew]

you should take a peak at bifurcation stuff

if you havent yet

[fathom flicker]

what topic is that

[forest sinew]

youre in 2d nonlinear systems

[fathom flicker]

or is it just bifurcation

[forest sinew]

ODEs

its dynamical systems

you are already looking at more interesting 2d systems apparently

you could look at like hopf

or study the 1d ones

theyre just cool is all

[fathom flicker]

yeah this is just kind of the tip end of my ODEs studying for now

my last topic was like

structural stability, borderline sketches and volterras equation

[forest sinew]

volterra is fun

idk what borderline sketches is

structural stability sounds like bifn lol

[fathom flicker]

I think that it is

I googled bifurcation

lol

[forest sinew]

its just a change in the quantity or stability of stationary points along variation of a parameter

[fathom flicker]

yeah

[forest sinew]

if you remember grist

who got banned

[fathom flicker]

I don't

[forest sinew]

ah

[fathom flicker]

sorry

[forest sinew]

he studies them in neurological models IIRC

for his PHD

[fathom flicker]

I haven't been around here long

[forest sinew]

ah

[fathom flicker]

he had a PhD and got banned?

[forest sinew]

he is persnickity

not stubborn

idk the word

[fathom flicker]

I feel like a lot of people are here

avid....

lol

[forest sinew]

avidrunner?

[fathom flicker]

yes

is 'persnickity'

I'd say

[forest sinew]

ah, resolute i think is the word im looking for

resolute describes avid pretty well

[fathom flicker]

true

How did that get him banned?

[forest sinew]

i have no idea lol

he has really strong opinions on authors too

thatd be my guess

sorry did i totally distract you

[fathom flicker]

It's fine, I should do some work but I enjoy talking too

I don't really have any work to do

since I'm not in school

lol

[forest sinew]

im up playing a game of see if i can not get entangled in a help channel while my sleeping pills kick in

[fathom flicker]

that's a dangerous game

idk what it is about this server that makes me stay awake hours past when I should be helping random people with random things

but it happens often

[forest sinew]

i usually feel like i know nothing

its nice to know something

wish it paid

[fathom flicker]

:/

the nice thing for me is right now I actually think that I want to become a professor

so that's like the 1 job I could get when I am done anyways

Are you going to go to graduate school?

[forest sinew]

idk that i can afford it

i only got accepted into a non thesis

i feel salty about taking out so much in loans

at the same time im scared about entering the job market

[fathom flicker]

are you in the US?

[forest sinew]

yea

[fathom flicker]

what is a non thesis?

[forest sinew]

like, you pay

you dont do research under a professor

you just take another year of more specialized classes

its half a degree for suckers

[fathom flicker]

If you don't mind me asking, how many programs did you apply to?

[forest sinew]

one

[fathom flicker]

dang

[forest sinew]

life on rails

[fathom flicker]

if you want to feel better, I have a question about my question that you probably know how to do

[forest sinew]

https://tenor.com/view/cat-kitten-skateboard-box-box-cart-gif-19024695

sure

[fathom flicker]

So, I did the jacobian stuff

at (0,0) I get tr=1 det=-2 so a saddle

at (2,0) I get tr=-1 det=-2 so a saddle

but

at (1, 1) I get tr=-1 and det=1

and I'm not sure if this is a nodal source or spiral source

Idk how to use the poincare map like with non obvious points

how would I find out which it is?

[forest sinew]

usually youd use the line

well its not a line

theres a parabola crossing through the upper half

[fathom flicker]

what would the parabola be in my case

it's det=tr^2/4

but

1=1/4 doesn't make a parabola

so I must be missing something

[forest sinew]

damn lets see if my sleeping pilled brain can do this in my head

[fathom flicker]

do I just add a y and an x to that

y=x^2 / 4

[forest sinew]

yyea

so in order for these to be equal

what x would have to be a lot smaller?

holding y the same, youd have to decrease x a lot to get to 1/4

which is the parabola

[fathom flicker]

wait I'm not understanding

[fathom flicker]

for what to be equal?

[forest sinew]

Im just trying to figure where we are in relation to the parabola

say right now we have x=1/4 y^2

this parabola

sitting somewhere in relation to us

[fathom flicker]

oh, well at tr=-1 the parabola has point (-1, 1/4)
and our point is (-1, 1) which is higher

[forest sinew]

our 'equation' right now looks like 1=1/4

[fathom flicker]

so we have a spiral sink

I think

[forest sinew]

at (1,1)?

[fathom flicker]

(-1, 1)

for trace = -1 det = 1

but it is the point (1, 1)

x= 1 y = 1

if you get what I'm saying

[forest sinew]

oh i get what you mean

damn my sleeping pills are kicking in hard

[fathom flicker]

yah so our trace is -1, our det is 1

if our parabola is y=x^2/4

(-1, 1) is above the parabola

to the left of the y-axis

so I'd conclude that we have a spiral sink

do you approve?

[forest sinew]

i dont think so

but im sleeping pilled

to me i think (-1, 1) is a point on y=x^2 right

so the 1/4 only flattens it?

then our point should be on the inside

between the parabola and the axis

[fathom flicker]

yeah that's what I said

[forest sinew]

oh, are we saying the same thing

[fathom flicker]

[forest sinew]

okay

i gotta go to bed

good luck with your study

[fathom flicker]

Goodnight!

[forest sinew]

[fathom flicker]

thanks for your help

$\bar{x}=(2-x-y)x$ $\newline$ $\bar{y}=(x-1)y$ $\newline$ this system has critical points (0, 0) (1, 1) (2, 0). When I linearize at each of these points I get the jacobian matrix evaluated at each point to be (in order as listed before) $\newline$ $\begin{bmatrix} 2 & 0 \ 0 & -1 \end{bmatrix} , \begin{bmatrix} -1 & -1 \ 1 & 0 \end{bmatrix} , \begin{bmatrix} -2 & -2 \ 0 & 1 \end{bmatrix}$ $\newline$ these have in order (trace, determinant) being $\newline$ $(1, -2) , (-1, 1) , (-1, -2)$ $\newline$ So, I conclude that in order the critical points have these trajectories near them: saddle, spiral sink, saddle. $\newline$ $\newline$ Can anyone verify?

[warm shale]

AustinU

[forest sinew]

https://www.cs.unm.edu/~joel/dfield/

i cant recall if this version has all the features

the full one is in matlab

but this will find fixed point and classify stability

[fathom flicker]

https://tenor.com/view/monsters-inc-sully-you-go-to-sleep-go-to-sleep-go-to-bed-gif-22295511

[forest sinew]

da mnit

[fathom flicker]

XD

thank you

do I want the direction field or phase plane one?

Sorry to dissapoint Jan but I can't figure out how to download/use that

So this is my question if anyone sees ^^

[obtuse pebble]

@fathom flicker Has your question been resolved?

[fathom flicker]

.close

[obtuse pebble]

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[obtuse pebble]

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[hollow sonnet]

Need help here

[hollow sonnet]

So I’m thinking V = lwh

We’ll really V = x^2 * y

[brisk matrix]

you know what the volume should be

[worn yoke]

we need to solve for y but we can’t use this equation for it

[brisk matrix]

oh never mind

we don’t have a volume

[worn yoke]

so what other physical property of the box can we make an equation for?

[obtuse pebble]

@hollow sonnet Has your question been resolved?

[hollow sonnet]

My apologies. We could do area?

Area of the square portion

[worn yoke]

we can do the surface area of the whole box

[hollow sonnet]

Ok

Hmmm

A = x^2?

[worn yoke]

that’s the area for the bottom portion, so what’s the area of the side portion?

[hollow sonnet]

A = X * y

[hollow sonnet]

Is this right?^

[worn yoke]

that’s correct for 1 side portion. How many of those are there on the box?

[hollow sonnet]

4

Oh

That’s just the bottom

But there’s 4x y

For the sides

[worn yoke]

so if we add the area of the bottom and the 4 sides we get the total area of the box

[hollow sonnet]

Ok

So it’s A = x^2 + 4xy?

[worn yoke]

yes

[hollow sonnet]

Ok

[worn yoke]

note how our constraint is the total cost

[hollow sonnet]

Yes

[worn yoke]

which is a function of the bottom and side areas

[hollow sonnet]

Right

[worn yoke]

so let’s express the total cost in terms of the area formulas we already have

[hollow sonnet]

Ok

Hmmm

I’m not sure here, do I set 160 = x^2 + 4xy

[worn yoke]

we need to multiply the areas by their respective cost per square foot

[hollow sonnet]

Ok

A = 36x^2 + 10(4xy)?

[worn yoke]

now that we have an equation we can use it to solve for one of the variables

[hollow sonnet]

Is that right?

[worn yoke]

in terms of the other variable

[hollow sonnet]

Ok

I think I might have it

[hollow sonnet]

Is that it?

[worn yoke]

look at how you solved for y again

[hollow sonnet]

Oh whoops

But I solve for y and then throw it into the V equation, correct?

[worn yoke]

yes

[hollow sonnet]

Ok so fundamentals are fine but the math isn’t right

[worn yoke]

just redo the algebra and you’ll be good to go

[hollow sonnet]

Ok and the 160 is in the right spot correct?

[worn yoke]

not sure what you mean by ‘right spot’

[hollow sonnet]

In for A

[worn yoke]

160 isn't really area, it's cost

once you multiplied by the cost of each portion of the area it's a cost equation

but it is on the correct side of the equation

[hollow sonnet]

Ok thank you for your help

.close

[obtuse pebble]

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[obtuse pebble]

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[sage dagger]

So uh I just took my eyes off of the screen for a second and see this? The lesson is about quadratics

[royal basin]

sounds like they were finding the eq of a quadratic going through 3 known points by means of a linear system

[sage dagger]

but like

what is that

[royal basin]

computational work for gaussian elimination

[sage dagger]

what the hell

ok

never heard of it

[royal basin]

in principle if you know how to solve linear systems then gaussian elimination will end up not much more than a change of notation

[sage dagger]

as in ive never heard or seen of gaussian elimination or that bracket notation before

[royal basin]

those would be matrices

[sage dagger]

University math?

.close

[obtuse pebble]

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[spiral wraith]

How do I go about beginning to solve this?

I know that the tangent line equation is the differential of the curve equation

[woeful cove]

I don't know how to solve it either

Can I still give it a go?

[spiral wraith]

feel free

[woeful cove]

So y' is the tangent to the curve y= ax²

[spiral wraith]

I believe so

[woeful cove]

y'=2ax

So the line -2x+y-b=0 is a tangent to y=ax² when -2x+y-b=-2ax

[spiral wraith]

ohhhh because they touch at x = -3

[woeful cove]

What?

[median yacht]

I think I found the solution

[spiral wraith]

because it's tangent to the curve

[woeful cove]

Well after forming that equation

You can just input -3

Then find the rest

I hope?

Try it and let me know

[median yacht]

That's not correct

[spiral wraith]

what's not correct

[median yacht]

The tangent and the curve share a point

That point can be found making an equation with both the curve and the tangent

In x = -3

$-2(-3) + a(-3)^2 = b$

[warm shale]

Rub05

[spiral wraith]

this is what I hve so far

[median yacht]

You can see this here

[spiral wraith]

I wanted to do this algebraically and use graphing to check after

but thank you

so I have -6a + 6 = b

how do I determine a or b from that equation

do I use a past equation that equals a or b?

[median yacht]

That's not the answer, I put another function in desmos just for proving by graphing what I said before

[median yacht]

9a + 6 = b

[spiral wraith]

I didn't say it was the answer, I was just saying I wanted to use graphing in general post-answering the question

2 * -3 = -6, not 9

[median yacht]

(-3)² = 9

[spiral wraith]

would we not use the y^1

instead of y

[median yacht]

Not yet

[median yacht]

This is just the curve and the tangent, there is not y' anywhere yet

Just two function sharing a point

[spiral wraith]

ok I see

[median yacht]

Having a and b here, we can know the point that's shared

And having that equation written is the first part of the excercise

[spiral wraith]

ok

[median yacht]

In the second part, you have to use y' now

But first

Do you know how to find the slope of a tangent?

[spiral wraith]

it's 2 isn't it?

+b

[median yacht]

False

[spiral wraith]

how come

[median yacht]

$$-2x + y = b$$
$$ +2x -2x + y = b + 2x$$
$$y = b + 2x$$

[warm shale]

Rub05

[median yacht]

y = 2x + b

[spiral wraith]

yes

and m, the slope, of y = mx + b

2 being m

[median yacht]

Correct

So now, we knoe that the slope of the tangent is 2

[spiral wraith]

??

what is false then

did I do something wrong

[median yacht]

You put -b instead of b

[spiral wraith]

+b

[median yacht]

Oh, I didn't see that sorry

[spiral wraith]

ur good

[median yacht]

Now, do you know how to find the slope of any function at any point?

[spiral wraith]

point-slope form?

y - y1 = m ( x - x1 )

[median yacht]

But what about the curve?

You cannot do that with the curve

[spiral wraith]

finding the slope of the tangent line of the function, or any function period?

[median yacht]

No

Using y'

Now, you need to use it

[spiral wraith]

so I differentiate y = 2x + b?

[median yacht]

Why do you want to differentiate the tangent if you know how to find its slope just seeing its equation?

[spiral wraith]

or do I differentiate y = ax^2?

[median yacht]

That one

[spiral wraith]

m being the slope, m = 2a

[median yacht]

Correct

Wait, no

Not correct

m = 2ax = y'

[spiral wraith]

m being the slope of the curve ohhhhh

[median yacht]

We know that the tangent and the curve share a point, we have the equation to find that point.
The tangent has the same slope as the curve in that point, so the second equation is...

[spiral wraith]

2ax = 2x + b?

[median yacht]

No

[spiral wraith]

because they share a point, so they are equal at that point?

[median yacht]

You are saying "The slope of the curve = The equation of the tangent"

[median yacht]

Yes, but that's the first equation we made before

[median yacht]

"Curve equation = Tangent equation" in x = -3

[spiral wraith]

a(-3)^2 = 2(-3) + b
9a + 6 = b
yes

[median yacht]

And know, you have to do "Curve slope = Tangent slope" in x = -3

[spiral wraith]

2 = 2a(-3)
2 = -6a
-1/3 = a

[median yacht]

Yes

[spiral wraith]

9(-1/3) + 6 = B
-3 + 6 = b
3 = b

[median yacht]

Yes!

[spiral wraith]

ohh my god that was so much simpler than I initially did it

thank you for helping me

are there any tips for future problems like this one?

[median yacht]

If you have parameters to find, think always about making equations

[spiral wraith]

ok thank you

.close

[obtuse pebble]

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[fallow kindle]

can anyone solve this problem

[obtuse pebble]

Please don't occupy multiple help channels.

[stone spoke]

there isn't much more to the problem than the hints that have been provided

have you made any progress so far?

or are you stuck on any particular step?

[obtuse pebble]

@fallow kindle Has your question been resolved?

[obtuse pebble]

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[earnest shell]

how can i check if my solution is valid in lu decomposition?

[obtuse pebble]

@earnest shell Has your question been resolved?

[obtuse pebble]

@earnest shell Has your question been resolved?

[kind hawk]

multiply it out again?

what are you hoping for here?

[obtuse pebble]

@earnest shell Has your question been resolved?

[obtuse pebble]

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[terse sparrow]

I just bought lottery ticket. 7 numbers I picked, out of 36 possible numbers, same number can't be picked twice.

I thought that calculation for chance to win 7 correct numbers would be 36x35x34x33x32x31x30, because after each pick of a number that is one less number to choose from. This yields a chance of 1/42,072,307,200 according to calculator.

However according to our Lottery authority I bought it with the chance for 7 correct numbers should be 1/8,347,680.

So what am I doing wrong?

[obtuse pebble]

@terse sparrow Has your question been resolved?

[jolly ginkgo]

The same cases are repeated here several times

It contains the different permutation of those 7 numbers which is supposed to be as just one way

[obtuse pebble]

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[terse sparrow]

Can you explain it how that works?

[jolly ginkgo]

Yeah

Consider instead of 7 there are two correct numbers

Let the numbers be 3 and 4

So in your case you consider 1) picking up 3 and then picking up 4 , 2) picking up 4 and then picking up 3 ; as different actions

Hence you are counting then twice

While they need to be counted as one way

Same goes if we are picking three numbers

If we are picking 1,2,3 then there are 3!=6 arrangements to this
123
132
213
231
312
321
But all these arrangements are denoting same action

So we divide the ways by 3!

For your question we can get total number of ways by 36×35×34×33×32×31×30/7! , dividing by 7! So the different arrangements of those same seven numbers are considered as one

,w 42072307200/7!

[warm shale]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by yes sam#5028

[jolly ginkgo]

This is what you had in denominator of the actual answer

We are just choosing seven numbers and the arrangement of those choices doesn't matter

[obtuse pebble]

@terse sparrow Has your question been resolved?

[terse sparrow]

Oh of course, the order doesn't matter, that's what I did wrong. If the order mattered there would be 42072307200 different outcomes, but since the order doesn't matter many of the outcomes are considered the same.

[terse sparrow]

But what does 7! mean? The ! what is that

[jolly ginkgo]

7×6×5×4×3×2×1

n! = n×(n-1)×(n-2)×...×3×2×1

Factorial we call it

[terse sparrow]

Cool, I understand. Thanks much 🙂

[jolly ginkgo]

[terse sparrow]

/done

[jolly ginkgo]

Not done

[terse sparrow]

/close

sorry lol

[jolly ginkgo]

There are 7! arrangements of those 7 numbers so we divide by it

.close

[obtuse pebble]

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[vale dirge]

is it true that if ab=ba then atbt = btat?
a and b are matrices of course

[timid silo]

$A^TB^T=B^TA^T$, right?

[warm shale]

Toby

[timid silo]

the t are transposes?

[vale dirge]

yeah

[timid silo]

take transposes of both sides of AB=BA

[vale dirge]

oh, right, $(AB)^t=(BA)t-> B^TA^T=A^tB^T$

[warm shale]

marchello

[vale dirge]

forgot about it

.close

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[strong oxide]

a

[elfin ether]

?_?

[royal basin]

do you have a question?

[obtuse pebble]

@strong oxide Has your question been resolved?

[obtuse pebble]

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[north anchor]

help

[north anchor]

how i can determine f(x) or g(x)?

[high lily]

wdym

[north anchor]

How do I know if one curve is greater than the other?

[high lily]

graphing helps

[thin pawn]

rule of thumb is graph it and it's top function - bottom function

[north anchor]

last part can help me?

[thin pawn]

i don't understand, what do you need help with?

[north anchor]

I need to know what is f(x) and g(x)

I don't know if what I did is correct, so far it's working for me

[high lily]

graphing helps

[thin pawn]

you wrote them, they are sqrt(4x) and 16x^2

[north anchor]

this part

[thin pawn]

@high lily can you solve bottom function - top function and just take absolute value in the end?

[high lily]

yes

[thin pawn]

well then @north anchor just take whatever function you want and do that. No need to try and find whichever is on top

| f(x) - g(x) | = | g(x) - f(x) |

[coral glade]

ooo functions

very easy stuff thid

[high lily]

need to be careful and ensure that one is always greater or equal to the other

graphing helps
don't skimp

[coral glade]

fatneek

[north anchor]

idk how graph function

[high lily]

then go back and review

its something you should know
and one of the prerequisites before even starting any calculus

how the heck are you expecting to do stuff like analyse behaviour of function of you don't have a basic idea of how to graph them

[north anchor]

the upper function is f(x) and the lower one is g(x)?

[high lily]

are you implying you don't know how to graph stuff like
y = x^2?

[north anchor]

yes is like a u

[high lily]

and what's stopping you from graphing
y = 16x^2

[north anchor]

idk

wdym with this | f(x) - g(x) | = | g(x) - f(x) |

[high lily]

that's a property of the abs val

|k| = |-k|

don't worry too much about that

idk
try it

same idea with
y = sqrt(4x)

[north anchor]

to find that, f(x) is the upper curve and g(x) lower?

But in case both functions are the same, what do I do?

[high lily]

wdym the same

they are clearly different functions

having a visual representation of the problem as evident in the sketch they have
gives a clear indication of which function is greater

and when doing the problem yourself you should make a sketch/diagram just like they have

[wintry portal]

Help

,rotate

[wintry portal]

Help me on this one

Q8

[high lily]

#rules #❓how-to-get-help

[north anchor]

how can determine f(x) and g(x) in this

[high lily]

well for that example, it seems more convenient to integrate wrt y

,rotate

[high lily]

and observe in the interval which graph is above the other

[north anchor]

what interval

[high lily]

interval for the region

in this case from y = -1 to 1

[north anchor]

yo evaluate both function?

[high lily]

wdym

[north anchor]

[-1,1] can use 0.1 to evaluate

[high lily]

wdym can use 0.1 to evaluate

where's 0.1 coming from

[north anchor]

interval

[high lily]

still doesn't explain where 0.1 is coming from

[north anchor]

of the range of the interval

[high lily]

like you said [-1,1]
that's the interval

0.1 is in that interval, but doesn't tell me why you're using it and you're expecting to get from that

[north anchor]

I hope to see when evaluating it that they tell me who is the oldest or the youngest

[high lily]

why are you so insistent on NOT graphing

if you insist on doing it purely algebraically, there are nice values to use

[north anchor]

it is graphed

[high lily]

if its graphed already

then you should be able to clearly see from the graph which function is greater

no need to do additional tests

[north anchor]

I can't see it, both functions have the same address

[high lily]

can you not clearly see from that that the purple is above the brown (except at the intersection points)

or that the purple is above the green

[north anchor]

i see it thx

[obtuse pebble]

@north anchor Has your question been resolved?

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[timid silo]

could be a mental lapse thing, but could someone help me with this

My question:
a number when divided by 7 gives 4 as remainder and when divided by 3 gives 2 as remainder. Find the possible values of the number.

My answer:
n_1 = 7q + 4
n_2 = 3q + 2

I know i could list out a bunch of numbers where q takes any integral values and see the common points

[timid silo]

But is there a way for me to get the entire family of solutions so that i can instantly find instances where n_1 = n_2

[kind hawk]

this is the chinese remainder theorem

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

?

Mhm, what about it?

it's the euclidean division algorithm i presume

[kind hawk]

no that's something else (although of course related)

the chinese remainder theorem first tells you that there are solutions to these types of equations (provided the numbers you divide by are coprime) and gives you an algorithm to calculate the solution (which is unique up to the lcm of the numbers)

btw dont ignore the bot

[timid silo]

ahh interesting

it'd given me the entire family of solutions?

how do i even know the two numbers are coprime though

[kind hawk]

I mean the 7 and 3

those have to be coprime

[timid silo]

ah you mean the number we're dividing with have to be coprime

sure

they are

[kind hawk]

if you asked the same question but with eg 8 and 6 instead of 7 and 3 then you might not find solutions

[timid silo]

"might not" meaning it's more of an undefined behavior thing?

Also just to confirm the way i setup the equation is fine using division algorithm right?

or was i supposed to use different variables for q

for some reason they used a different variable in one of the solutions but i never saw the need to do so

[kind hawk]

"might not" means that sometimes there are solutions and sometimes there are not. depending on which remainders you want

yes you should have used different variables for the q

n_1 = 7q+4

and n_1 = 3k + 2

here one solution is obviously 11

11 = 7*1 + 4

and 11 = 3*3 + 2

q and k are different

[timid silo]

right i bruteforced it and i did get 11 and 47

i could keep going but i was just curious about a generalized method

[kind hawk]

47 = 5 mod 7

[timid silo]

wait i must be bugging but why is the q variable different?

isn't q integers?

in both cases

[kind hawk]

and?

are all integers the same?

[timid silo]

Lol are they not? I'm confused wym

[kind hawk]

is 1=3?

[timid silo]

No

[kind hawk]

but here we have q=1 and k=3

so we cant use q for both

[timid silo]

So basically you define q differently because they have to be yield the same value (n_1)?

[kind hawk]

they will be different numbers

you cant use the same variable for different numbers

[timid silo]

Yeah i get that you use different values of q to get n_1 in both cases cuz ofc you require different linear combinations to get the same value

i was confused why my definition would be wrong then

I'm using the logic that q is same in both instances so i take the same integer

but ofc in the process the n_1 and n_2 would be different

no?

I guess it doesn't make practical sense for me to define it the way i did (?)

[kind hawk]

well but that's not what you want

[timid silo]

okay yeah brain's just foggy now lol you just want instances where n_1 is equal in both cases therefore q has to change

also just to clarify for the generalized remainder theorem: a = bq + r

0<= r< b and a,b, and q are integers would be the only conditions?

[kind hawk]

so the general theorem is that given integers a and b there exist unique integers q,r with a=bq+r and 0<=r<b

which you prove with eg euclidean algorithm

[timid silo]

okay sure i guess that clarifies it

any reason why they use the word "unique integers" and not just "integers"?

or is that supposed to signify the shifting nature of q (taking any integer values)

[kind hawk]

well in general there will be lots of options to write a like that. 7 = 3*2+1 = 3*1 + 4 = 3*15 + (-38) = ...

but the condition 0<=r<b actually makes this unique so that you only get one solution

[timid silo]

wait i got the first line but this has me a little confused

one solution of what? could you give an example

[kind hawk]

of writing it in that form with 0<=r<b

there is only the one way to write 7 = 3*2+1 with remainder r=1 satisfying 0<=1<3

there is no other option

which is trivial for small numbers but not so obvious for big numbers

you can't write 7 = 3*some integer + (some number between 0 and 3) in some different way other than 7=3*2+1

[timid silo]

ah you mean given a and b are integers and r is in the range right?

[timid silo]

okay so it depends on r then

isn't that just algebra or am i missing something

7 = 3*(some integer) + (take any integer from (0,3)) then find the "some integer"?

i think i'm probably missing the intricacy lol

also could we go to the chinese remainder theorem part now?

[kind hawk]

its not obvious that there is a unique solution to this