help-39

[empty junco]

should be -cos(t) + e^t?

theres no integrals btw

calc 1

[odd coral]

i think so
double check by substituting

s(0)

it mentions s(0) for that one should total to 0

[cloud zephyr]

ah, antiderivatives

[empty junco]

bro wait doe

cos 0 = -1

so

[odd coral]

double check that

[empty junco]

since V(t) = -cos(t) + e^t

[cloud zephyr]

well, it's just polynomials for (b), so um, reverse the power rule

[empty junco]

and e is like 2.7 something

and -cos(0) is -1

so it cant be 0

[empty junco]

ok

[odd coral]

-cos(0) = -1, e^0 = 1

watch out with the e^t

[empty junco]

WAIT WHAT

i forgot holy

good catch

[odd coral]

yeah now you can remember it
nice

[empty junco]

how do i reverse power rule on t^4

oh

1/5 t ^4

[odd coral]

how do you remember it working for normal power rule

[empty junco]

im dumb mb

[empty junco]

i know the power is automatically +1

[odd coral]

im asking abt derivative power rule rn btw

imagine smth like an expression 3x^2
what would the derivative be through power rule

[empty junco]

uhh

6x

[odd coral]

right

[odd coral]

how would you go back to 3x^2 from there

think about the power of x in "6x"

[empty junco]

yeah i got it, since power is implicit ^1, add 1 to power and divide coefficient by 2

also, for b i got A(t) = 1/5t^5 - 1/2t^2 + 2x + 3

but i think its wrong

since v(0) = 0

[odd coral]

a(t)?

or v(t) or s(t)

also careful with the x you typed

try getting v(t) first again and lmk what you find

from scratch
for b)

[empty junco]

oh shi

mb

[empty junco]

ok

mane i gotta go to bed sorry guys

popped a melatonin earlier so my study time was limited

.close

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[wild fable]

hello i want to know if im on the right track

[wild fable]

this is what i have so far

[random ermine]

hmm not sure about this

because the entries of your matrix don't directly give you a coordinate chart

to start, the dimension of SL(2, R) as a manifold is 3

[rustic gate]

the lie algebra of SL(2, R) is 3-dimensional

[wild fable]

[random ermine]

do u know what coordinate charts for this look like

[wild fable]

uh ion think so

[rustic gate]

this is like fine though

it's not like you have to work with SL(2, R) on its own

just take it as a subgroup of GL(2, R)

[wild fable]

the approach im using?

[rustic gate]

the tangent space will be a subspace of the one of GL(2, R) with an additional equation

[rustic gate]

uh yeah

[wild fable]

okay thanks'

[rustic gate]

you've found T_gSL(2, R)

just calculate T_idSL(2, R) and the derivative map between them

[cinder flower]

[wild fable]

alr. in my notes they also used the gradient vector and level sets for showing T_idSL(2,R). i can just do the same thing here?

[wild fable]

slayla blud

[rustic gate]

uh yeah

you could also like

just set g = id in your other calculation

[pearl pond]

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[crimson haven]

question d pls

[crimson haven]

How do I calculate period

[still hamlet]

how would you typically calculate period?

[crimson haven]

my teacher said it’s the distance

[still hamlet]

well what effect does cos(x-30°) have to the graph of cosx?

[crimson haven]

It moves right?

[still hamlet]

yeah, so does the period have any change?

[crimson haven]

idk

[dire tapir]

Well think of the period for cos(x)

What would that be?

[crimson haven]

360?

[dire tapir]

Yes

Now it shifts to the right by 30° degrees

Does the period stay the same?

[crimson haven]

yes?

[dire tapir]

Nice

So its period must be..?

[still hamlet]

there's your answer

[crimson haven]

360

[dire tapir]

That's correct

[crimson haven]

Oh yay tysm

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[crimson haven]

.reopen

[pearl pond]

✅ Original question: #help-39 message

[crimson haven]

erm may I ask how I calculate coordinates

[dire tapir]

Think about the endpoints for cos(x)

By endpoints I'm assuming they want you to find the point at which the period starts and ends

Although be careful since they have given us a domain specifically $x \in [-90^{\circ}, 360^{\circ}]$

[crimson haven]

no,for question d and f

[dire tapir]

Oh d and f then

Oh whoops

[jolly parrot]

MxRgD

[dire tapir]

Okay yeah that should be right

[crimson haven]

Ty

Bye

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[tiny egret]

hiya! im really stuck on how to proceed with finding how many elements of R/I there are.. like i just cant think of a strategy here

[tiny egret]

,cw

,rotate

[pearl pond]

@tiny egret Has your question been resolved?

[tiny egret]

<@&286206848099549185>

[timber cape]

Hm

-# out of my domain 😐

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[tiny egret]

ive reduced the problem to showing that any polynomial of the form $q_3X^3+q_4X^4+\ldots+q_nX^n$ is divisible by $X^3+2X+1$, but im struggling to show it efficiently

[jolly parrot]

lyric

[tiny egret]

tried determining the coefficients of some $A=a_0+a_1X+\ldots+a_mX^m$ by the equation $A\cdot(X^3+2X+1)=q_3X^3+\ldots+q_nX^n$ but its horrific

[jolly parrot]

lyric

[tiny egret]

so wondering if anyone can think of another way

[tropic saddle]

well thats certainly not true

and also not relevant for your question

the key point to realize is that each coset has precisely one polynomial of small degree

how could you find that?

[tiny egret]

ah, well i found using sage that the ring is just the polynomials of degree <=2, so wondered if theyre equivalent by making it a cubic

guess not

[tiny egret]

really no idea

[tropic saddle]

lets say we are in Z

and are doing Z/(13)

aka mod 13

what is going on there

[tiny egret]

well two integers a,b are equivalent if a-b = 13k for some k

[tropic saddle]

ok so the cosets have the form {a+13k}

lets say I have the number 1073

its in some coset

how would you find a smaller representative?

[tiny egret]

taking its remainder when dividing by 13

[tropic saddle]

good

have you heard of euclidean rings/domains?

[tiny egret]

i havent nah

[tropic saddle]

ok then ignore that

good so for numbers we take the remainder

can we do something similar for polynomials?

[tiny egret]

polynomial long division?

[tropic saddle]

yes

[tiny egret]

sounds exhausting

but i can't think of another way so ill try it

[tropic saddle]

you dont actually have to do it

the point is that you could

did you actually do the division 1073/13? no

you just saw that you could

we know that we can get every number 0-12 as a possible remainder

and they are all in distinct cosets

what possible remainders can we get for polynomials?

are they all in distinct cosets?

[tiny egret]

oh shittt

okay yeah ive got you

[pearl pond]

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[zealous moon]

wanted to ask if x limits after changing the order will be from 0 to y^2/a

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[zealous moon]

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[fading nexus]

How do I know which side is b and which side is c ?

[proper nova]

,rccw

[proper nova]

people usually label the sides that are opposite to the angle

like $b$ is the opposite of the B angle

[jolly parrot]

1 divided by 0 equals Infinity

[fading nexus]

Ah ok

Does it technically matter or no?

[proper nova]

that's how people usually label it

you should follow that convention so it matches the rule

[livid goblet]

yes

at least I think so

[fading nexus]

I’m not sure why I got a different answer even tho I did it the same

Now I just put it fully in my calculator and got another answer that’s 7.4

Ok I originally did it wrong or put It in wrong but my corrected version is still different

[glass meadow]

Your calculator is in radian mode

Do 10 sin(40*pi/180) / sin(60*pi/180)

(or use degree mode but I personally would stick to one mode so I don't forget to switch every time)

Wait no it's in degree mode

Are you doing 60º or 80º

[fading nexus]

Wdym

[glass meadow]

Yeah you're using 60º

[fading nexus]

Yea ?

is that wrong?

[glass meadow]

a/sin(A) = c/sin(C)

[fading nexus]

But I’m using angle b

or should I use x

C

[glass meadow]

But you don't have length b

You are doing a/sin(A) = c/sin(B)

That's not the law of sines

[fading nexus]

Ah ok

[glass meadow]

Also it's "law of sines", not "law of signs"

sine for sine function, aka sin(...)

[fading nexus]

Also why is my teacher sometimes using squiggly =

[glass meadow]

That means "approximately equal"

[fading nexus]

Whoops must of misheard it. My teacher told me to use the law of sines like a minute before the bell so I quickly wrote it down

[glass meadow]

,, \approx

[jolly parrot]

Nel

[glass meadow]

This sign

[glass meadow]

As for the labeling, no it does not matter, but in general it's a good rule to follow to better communicate

It also just makes the law of sines easy to remember:
a/sin(A) = b/sin(B) = c/sin(C)

[pearl pond]

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[quick venture]

i dont understand how they have applied the boundary conditions

[dense jasper]

$y(0)=0$ means that $y=0$ when $x=0$. Setting $x=0$ in $y=Ae^x+Be^{-x}+xe^x$ gives
$$y(0)=Ae^0+Be^{-0}+0e^0=A+B$$
and so $A+B=0$. Similar for $y'(0)$.

[jolly parrot]

Civil Service Pigeon

[quick venture]

OH

okay i see

[dense jasper]

[pearl pond]

If you are done with this channel, please mark your problem as solved by typing .close

[quick venture]

.close

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[quick venture]

im abit confused with this area question

is it becuase these are the intervals

and i've calculated the intercepts?

but we ignore the x=-1

becuase it isnt in the interval range?

[sharp vigil]

from the problem statement we are asked to calculate $\int_0^2 \abs{x-x^3} \dd x$ (would be different if the interval was not specified), so determining the intersection point is only relevant to finding where the expression inside absolute value changes sign

[jolly parrot]

cloud ☁

[quick venture]

ah okay i see

thanks

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[lethal ruin]

Fifty students were polled as to what courses they are currently taking. The results are shown below:

24 take Data Management, 17 take Calculus, 22 take Functions.

5 take both Data and Calculus, 7 take both Functions and Calculus, and 13 take both Data and Functions.

3 students are taking all three courses.

How many students are ONLY taking Calculus?

[flat hornet]

can you draw a venn diagram and post it?

[lethal ruin]

Yeah an incomplete one give me a second

[flat hornet]

np

[flat hornet]

perfect

so how many people take only calculus and funcation but not Data management

like this area

[lethal ruin]

7

[flat hornet]

7 is for this area

because that would be C and F

[lethal ruin]

I'm not sure actually. I have trouble understanding what to subtract. Could you give me a hint?

[flat hornet]

the area here contains C and F, but also people that take all 3 courses. We would only like the people that take C and F but NOT D

[lethal ruin]

Yes I know but I'm having trouble figuring that out

[flat hornet]

so we would subtract the amount of people that take all 3 courses froom the number of people that take C and F

[lethal ruin]

Wait is my venn diagram inaccurate then?

[flat hornet]

slightly

but its ok

it works

[lethal ruin]

so $C \cap F = 7-3=4$

[jolly parrot]

Elliot Pixel

[flat hornet]

basically

but notice that it should be CnF/D

/ for not including

n for intersetion

but the number you got is right

[lethal ruin]

so $C \cap F \cap D'= 7-3=4$

[jolly parrot]

Elliot Pixel

[lethal ruin]

I think that's right

[flat hornet]

so D' is not including D right? to be sure?

[lethal ruin]

@flat hornet

[lethal ruin]

yes

[flat hornet]

nice

that perfect

so do u think you can finish the problem now?

[lethal ruin]

Yes I think it's

$17-2-3-4=8$

[jolly parrot]

Elliot Pixel

[flat hornet]

yes that right

perfecto

[lethal ruin]

thanks a lot

[odd coral]

nice

[lethal ruin]

AI was not able to help with that lol

cuz it's easier with a venn diagram

[odd coral]

that taught me another approach too

[lethal ruin]

what would you have done?

[odd coral]

i just drew rows of numbers with letters

[flat hornet]

np

[lethal ruin]

elaborate pls

[odd coral]

i do subtracting from certain sections in order

the drawing skips a lot of the details though

[flat hornet]

huh ive never seen this before

its pretty cool

[odd coral]

i started getting rid of the 3 students that share all 3 courses

[lethal ruin]

my mind is blown

[odd coral]

and affect all other groups

and apply smth similar in stages

if you can call them stages

separating different steps in the process with those horizontal lines i drew

i like the venn diagram approach it's pretty swag

[lethal ruin]

yeah fr

[flat hornet]

venn diagram is esier to understand, but this is quicker

its rly cool

[flat hornet]

if you have no furthur questions, feel free to close the channel

type ".close"

have a nice day!

[lethal ruin]

I'm working on the same question rn and I may close it if I figure this out

ty

thanks for your help

[flat hornet]

ok np

ig u could just leave it open for a bit then

[lethal ruin]

@flat hornet @odd coral

Yea I got it awesome

thanks again

.close

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[crimson haven]

can I get help with calculating coordinates

[proper nova]

,rccw

[proper nova]

Only the endpoints tho

So that's just the coordinates where x is at the bounds

[proper nova]

Do you get what I'm saying?

[crimson haven]

No

[proper nova]

Your exercise asked for the coordinates of the endpoints in these graphs

[crimson haven]

how do Ik which to do

[proper nova]

They specified what are the ranges of $x$ in each of the graphs

[jolly parrot]

1 divided by 0 equals Infinity

[crimson haven]

So we use the range

[proper nova]

Yea

The endpoints usually lie on the boundaries

[crimson haven]

what’s boundaries

[proper nova]

For instance $[-90^\circ; 360^\circ]$ then the boundaries are at $-90^\circ$ and $360^\circ$

[jolly parrot]

1 divided by 0 equals Infinity

[crimson haven]

Wait could u help me

[proper nova]

?

I don't really do trig much but yours is a common sense question so I might be able to help

[crimson haven]

How do I do this coordinate

Actually how do I do them in general

[proper nova]

-# derivatives

I think that's called a turning point

[crimson haven]

Huh

Oh

[proper nova]

With calculus, you could do this using derivatives

[crimson haven]

Will it be (45,0.9)?

[proper nova]

But without calculus

You should remember the ranges for each function

[crimson haven]

Huh

[proper nova]

So what graph is this first

[proper nova]

And are you asking for the points that curve the direction?

[crimson haven]

Cos

[proper nova]

cos(x - 30) or smth?

[crimson haven]

I’m asking for how I do and know which parts to find the coordinates

Yes

[proper nova]

Yes, so that point would be (45, 0.9)

I thought you asked for the maximum of the function

[crimson haven]

The book and what ur saying is confusing

It’s d

[proper nova]

,rccw

[pearl pond]

@crimson haven Has your question been resolved?

[pearl pond]

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[south ice]

Hello, can someone walk me through this problem? I’m new to vector function derivative btw

[unborn abyss]

velocity is the derivative of position (which is r)

[south ice]

What do they mean by show that the velocity is orthogonal to the position

[unborn abyss]

well, velocity (at a given t) is a vector and so is position

take your position vector and differentiate it with respect to t

[south ice]

Sure but the general velocity is just a curve no?

How is a curve orthogonal

[unborn abyss]

at a given time, the velocity can be expressed as a vector

direction, magnitude

[south ice]

Yes but isn’t that when we instantiate a specific time

[unborn abyss]

well... yes, but you can talk about it in terms of t

and youll find that no matter what you choose for t, the position vector is always orthogonal to the velocity vector

[south ice]

I see your point, but how are you supposed to show that the entire velocity is orthogonal to the position ? And how does that look like

[unborn abyss]

well, how would you normally show that two vectors are orthogonal?

[south ice]

Dot product = 0

[unborn abyss]

great

so lets do that here

lets find the velocity vector in terms of t

[south ice]

Okay

[unborn abyss]

so go ahead and do that :)

its the derivative of the position vector

which is this

[south ice]

So is that

$$ bwsin(wt)i - bwcos(wt)j + C = v(t) $$

[unborn abyss]

signs messed up and youre missing a j

i think you attempted to integrate it?

where did that C come from

[jolly parrot]

warbeast002

[south ice]

Yes

Ooops

Oh no that’s supposed to be the derivative

[unborn abyss]

yes

[south ice]

One sec

[unborn abyss]

you can use \vec i if you want

or dont bother

[sharp vigil]

or \vb i

[south ice]

Alright

$$ v(t) = dr/dt = -bw * sin(wt) \vec i + bw * cos(wt) \vec j $$

[jolly parrot]

warbeast002

[south ice]

There we go

[unborn abyss]

ok great

so now can you find v(t) dot r(t)?

[south ice]

Uh

Yeah idk how that should work

[unborn abyss]

well... how would you take the dot product of two vectors normally?

like (4i + 3j) dot (i + 8j)

[south ice]

Oh yeah I see

Should I not think of it as a curve for this instance

Like should I just think of all possible vectors at every t

Like infinite velocity vectors orthogonal to infinite position vectors

[unborn abyss]

i think that would be useful yes

[icy bluff]

Is the question solved?

[south ice]

Yessir

Thanks a lot, I wanna keep the channel open tho incase I need more help

[icy bluff]

So i can share my solution

[south ice]

Oh yeah if you wanna share anything go ahead man

[icy bluff]

Ignore my handwriting

[south ice]

My answer was up to the dot product = 0. What was it that you did afterward with the square root? Are you taking the magnitude of all velocity vectors ?

I can’t tell

[icy bluff]

That's for question b

Speed of particle means magnitude of velocity

And it is proportional to omega

[south ice]

Oh wait can you explain what part b is asking for? Because I don’t even get what they’re asking

So i know speed is absolute value of velocity

But I don’t get the rest

[icy bluff]

Part b is asking us to show that the speed is proportional to angular velocity

That's it

[south ice]

What does proportional mean

And what’s annular velocity

Angular *

[icy bluff]

Angular velocity means omega

Which is dtheta/dt

[icy bluff]

When one quantity is equal to some constant × another quantity

Then both are directly proportional

If one quantity = constant/another quantity

[south ice]

I still don’t understand this thought

[icy bluff]

Then both are inversely proportional

[south ice]

What is that supposed to be conceptually

The angular velocity

[icy bluff]

Its basically the rate at which an object rotates or revolves around an axis measuring how fast the angle changes over time

[south ice]

I don’t understand bro

[icy bluff]

In which grade are you in

[south ice]

Undergrad

[icy bluff]

Ok wait

[south ice]

Alright

[icy bluff]

https://youtu.be/lV3h2kpJzp8?si=c9jvYpZjLHA2qPbT

Watch this

[south ice]

Alright thanks

[pearl pond]

@south ice Has your question been resolved?

[pearl pond]

@south ice Has your question been resolved?

[pearl pond]

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[pearl pond]

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[quick venture]

how do we know the limits are 3 and 1?

[timber cape]

Okay suppose if the circle was centred at origin, what would the limits be?

[quick venture]

(1,-1)?

[timber cape]

Exactly!

So

If

You shift the circle

By 2 units

What would the limits be?

[quick venture]

3 and 1

i think im just finding it hard to visualise

[timber cape]

Best way to counter this is shifting of origin

Let some big X+2 = x then everything shifts back like if it was at origin

And in the end just put the value of X instead of the ones of small x

[quick venture]

so x is the radius here

[timber cape]

Nono

x is just coordinates

[quick venture]

thanks

.close

[pearl pond]

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[pearl pond]

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[thorny radish]

use \setminus

[prime stream]

I started as follows. Let $\alpha \in F(c)\setminus F$. Then we can write $\alpha = \dfrac{f(c)}{g(c)}$ for some $f(x), g(x) \in F[x]$. Suppose, there exists $p(x)\in F[x]$ such that $p(\alpha) = 0$. Then $p(f(c)/g(c)) = 0$, implying that $c$ is a root of some polynomial, which is a contradiction.

[jolly parrot]

Dedekind

[prime stream]

However, in order for this to work, I believe, I need to show that $p(f(x)/g(x))$ is not the zero polynomial. And also, it seems that the fact that $\alpha \notin F$ is not used in the proof, so probably I am on a wrong way. Any suggestions?

[jolly parrot]

Dedekind

[thorny radish]

so assume f and g are coprime, then clear the denominator and see if you can conclude f and g are both degree 0

but you're on the right track

[prime stream]

what does it mean clear the denominator?

[thorny radish]

like 1/4 + 3/2 + 2 = 0 implies 1 + 3*2 + 2*4 = 0

[prime stream]

ah, I see, thanks

[pearl pond]

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[prime stream]

.reopen

[prime stream]

So I have $g(c)\alpha - f(c) = 0$, but unfortunately I failed to see why this implies that both f and g are constant. Would you mind to give a further hint?

[jolly parrot]

Dedekind

[cursive wraith]

wait, did you transform p(f(c)/g(c)) = p(alpha) into f(c)/g(c) = alpha? That assumes p is injective, which is false

[prime stream]

no, the other way around. We know that f(c)/g(c) = alpha and then substitute into p(alpha)

[cursive wraith]

ok yes sorry

but I don't think showing things about f and g is necessary

just show there is a polynomial that cancels c as you were trying to do earlier

write p(x) = ax^d + ...

and transform p(f(c)/g(c)) = 0 into (polynomial in c = 0)

by multiplying by g(c)^d (which takes care of every denominator)

[prime stream]

ok, thanks, let me try

[cursive wraith]

If you need to argue why the resulting polynomial isn't 0, just compare the degrees of each term

||one of the terms should have a degree higher than any other||

[thorny radish]

i think this is what i said lol :P

[thorny radish]

i'm not sure if this argument works since f and g could have the same degree

i suggested assuming f and g coprime earlier

[prime stream]

So let $p(x) = a_0 + a_1 x + ... + a_nx^n$. Then $p(f(c)/g(c)) = a_0 + a_1\dfrac{f(x)}{g(x)} + ... + a_n \dfrac{f^n(x)}{g^n(x)} = \dfrac{1}{g^n(x)}[a_0g^n(x) + a_1 f(x) g^{n-1}(x) + ... + a_n f^n(x)]$.

Hence have $a_0g^n(c) + a_1 f(c) g^{n-1}(c) + ... + a_n f^n(c) = 0$. Now, it looks like, for example, $a_0g^n(c)$ will be the only one that have the highest order, but only if we assume that the degree of f is less than the degree of g. It's not the same as to assume that f and g coprime, or I am wrong?

[jolly parrot]

Dedekind

[cursive wraith]

Yeah I made a mistake in the degree argument which doesn't always work

the point is backwards that this polynomial, p(f(x)/g(x)) * g^n(x), has to be 0 (as only the 0 polynomial cancels c), and so p(f(x)/g(x)) is also the 0 fraction, as g^n(x) is not 0 and F(x) is integral

[prime stream]

ah, I see now. The only question remains: where we use the $\alpha \notin F$ here?

[jolly parrot]

Dedekind

[prime stream]

no, wait. If p(f(x)/g(x)) is zero, does it follow that p(x) is zero?

I mean, apparently it does, but I don't see how

[thorny radish]

no, for example f could be 0

[prime stream]

this I believe we can rule out, since $\alpha \notin F$

[jolly parrot]

Dedekind

[cursive wraith]

f/g can also be constant

and p = x - that constant

the point is that you need to show those are the only cases

which is not trivial

[cursive wraith]

maybe the degree idea can work when f and g have different degrees
when deg(f)= deg(g) you need to use their coprimeness

[thorny radish]

f and g coprime just solves for all cases though

[prime stream]

Let me try this. Since $a_0g^n(x) + a_1 f(x) g^{n-1}(x) + ... + a_n f^n(x) = 0$ (as polynomial), this implies that $g^n(x)$ is divisible by $f(x)$ and $f^n(x)$ is divisible by $g(x)$. But since $f$ and $g$ have no common factors, this is impossible, unless $f$ and $g$ are constant. But then $\alpha \in F$, a contradiction

[jolly parrot]

Dedekind

[thorny radish]

that's right

[prime stream]

@thorny radish @cursive wraith thank you!

.close

[pearl pond]

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[cursive wraith]

out of the two you really have only f^n divisible by g(x) (as only a_n is assuredly not 0)

for the other way around, consider the first coefficient a_k that is non zero

and you can obtain a similar reasoning

[rustic gate]

the minimal polynomial is irreducible

[cursive wraith]

p isn't supposed to be the minimal polynomial

but that's another way

[rustic gate]

assume it is then ig

[cursive wraith]

yes, that's what I agree with

[prime stream]

good catch, thanks!

[rustic gate]

i think it was rafilou that caught it

[pearl pond]

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[sharp smelt]

(a) $\sigma^2,\sigma^2, \sigma^2, \frac{4}{3} \sigma^2$
\
SO
\ Y,Y,Y,N

[jolly parrot]

wai

[sharp smelt]

the first 3 have 0 MSE

How woul T_5 be unbiased for any multiple though

the MSE is sigma^2

[pearl pond]

@sharp smelt Has your question been resolved?

[brisk scarab]

It's been a while since I did stats, but I'm pretty sure T4 is unbiased

those MSEs look very wrong as well

(on average, they're bang on? I think not)

[pearl pond]

@sharp smelt Has your question been resolved?

[pearl pond]

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[south ice]

Hello need help on this

[eager jewel]

b part?

[dusty jungle]

use the fact that the velocity function is the time derivative of the position

[south ice]

Part b yes

I don’t understand angular velocity

[eager jewel]

What did u get as velocity

[south ice]

Oh I’ve already solved that one

I got it right

[eager jewel]

Yea just tell

[south ice]

$$ v(t) = dr/dt = -bw * sin(wt) \vec i + bw * cos(wt) \vec j $$

[jolly parrot]

warbeast002

[eager jewel]

Yea correct

Now speed is absolute value of velocity

[south ice]

Yes

[eager jewel]

And u can take w common from this expression

[south ice]

$$ speed = | v(t)| = | - bw * sin(wt) \vec i + bw * cos(wt) \vec j | $$

[eager jewel]

No

U can't remove the - sign for when the expression is added

[eager jewel]

So put the modulus around this itself

And take w common

[south ice]

Wdym

I don’t understand what you’re saying tbh

[eager jewel]

Ia-bI is not equal to a+b

[eager jewel]

What exactly did u not understand

[south ice]

“Take with common “ “put modulus around itself”

[jolly parrot]

warbeast002

[eager jewel]

w is not with ,w is angular velocity

[south ice]

You mean this

[eager jewel]

Yes

[south ice]

,taken

Whatever the command is

[eager jewel]

!occupied

[pearl pond]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[eager jewel]

@south ice

[south ice]

Sorry but I don’t get what you’re saying here

w is not with ?

[midnight haven]

i was suggesting that this is how you find the magnitude and not asking a question but nvm

[south ice]

With what

[eager jewel]

Bro it's not w it's omega

Take omega common from that expression

[south ice]

Look bro I don’t know this wording you’re using

What’s omega common

😭

[eager jewel]

Bruh ka + kb = k( a+b )

[south ice]

Oh just factor out omega

[eager jewel]

Just do this with omega in that expression

Yes 😭

[south ice]

😭😂

[midnight haven]

$\omega(-b * sin(\omega ,t) \vec i + b * cos(\omega ,t) \vec j)$

[jolly parrot]

atlanta

[south ice]

$$ speed = | v(t)| = | w(- b * sin(wt) \vec i + b* cos(wt) \vec j) | $$

[jolly parrot]

warbeast002

[eager jewel]

Yes now do u see that we proved speed directly proportional to omega

[south ice]

No bro because I don’t understand what angular velocity even is conceptually and that’s what I’ve been trying to understand

[eager jewel]

Angular velocity is the rate at which an object rotates or revolves around an axis, measured as the change in angle per unit of time

[toxic goblet]

wsp guy

wsp guys

typo

im new

[south ice]

And what’s the formula for that

[eager jewel]

Omega=d(theta)/dt

Just like velocity=dr/dt

[south ice]

But lik what’s the θ function here

For example position has a function that you differentiate

Or does it not have one

Hope I’m not asking a stupid question

[eager jewel]

For this question yes

Position may not always be a function of something

[eager jewel]

For this question theta =wt because omega is constant

[south ice]

I don’t understand

[south ice]

What would it be?

[eager jewel]

its given in the question..

[south ice]

Θ = ?

[eager jewel]

im talking about position not theta here

[south ice]

Oh ok, also how can you not have a position function ?

[eager jewel]

no we will always have position function it just may not depend on time like the one given in the question

for example we may just have r(t) = 7

[south ice]

I mean sure but it’s just a constant for all time

So time isn’t really completely absent here

But man I’m still not understanding the speed being proportional to the angular velocity

[eager jewel]

why not..we proved it mathematically..

[south ice]

No I mean graphically

Because essentially that’s what we’re quantifying with algebra

[eager jewel]

graphically? how do u understand proportionality graphically

[south ice]

Shouldn’t that be the case ?

I don’t know broooooo

[eager jewel]

well i dont know then

[south ice]

is w angular velocity ?

[eager jewel]

yes

as given in the question

[south ice]

I’m having trouble seeing where on that graph w is expressed

[eager jewel]

its not expressed on that graph..the particle is rotating in a circle with angular velocity omega

[pearl pond]

@south ice Has your question been resolved?

[south ice]

w is a function?

[eager jewel]

no its constant with time

[south ice]

but w is angular velocity ?

Is w angular velocity ?

[eager jewel]

yes bro i already confirmed this..

why are u asking the same questions

[south ice]

Okay so shouldn’t it be a function

[south ice]

I know the answer I’m asking so that we syllogize the concept

w ⇒ angular velocity

angular velocity ⇒ function

∴ w ⇒ function

<@&286206848099549185>

[pearl pond]

@south ice Has your question been resolved?

[pearl pond]

@south ice Has your question been resolved?

[south ice]

<@&286206848099549185> can anyone help me out with this ?

[lavish bay]

ya whats up

[west ether]

What's up

[south ice]

Thank you, I’ve been trying to understand part b conceptually for a while now, can you please help me out

[lavish bay]

sure no problem

[south ice]

What does it mean to have angular velocity and how does that even look like in relationship to the position function we started with and the graph???

[south ice]

Thanks a lot

[lavish bay]

angular velocity essentially tells you how fast the angle is changing

in this problem its within the the coswt and sinwt meaning it rotates around the circle at rate w

so larger w maskes it go around faaster

[south ice]

Is that always gonna be Relative to the origin ?

[lavish bay]

yes

its always measured relative to a chosen center of rotation

[south ice]

can you please explain how this works

So w is a constant right ?

[lavish bay]

yes w is a constant

[lavish bay]

if w is a large number, the particle completes more rotations per second (faster motion)and if w is smaller, the particle moves more slowly around the circle

[south ice]

Well the thing is whenever I think of velocity I think of it in terms of a velocity function at some point v(t)

How does this differ here

[lavish bay]

thats a great point you make

i think ur thinking about velocity in terms of linear velocity

or the actual speed of an object along a path

in the case of circular motion you still have a velocity function but it's a bit different due to the motion being on a circle

[south ice]

But we have a constant w not a function right

[lavish bay]

yes

[south ice]

So where is our angular velocity function

[lavish bay]

the angular velocity is constant so there is no angular velocity functio

it simply describes the constant rate at which the angle ,tex θ(t)=ωt changes with time

[stone egret]

But how do we show it

[south ice]

Yeah im trying to use similar logic to linear velocity

[stone egret]

Tryna understand with u lmao

[south ice]

No worries

[south ice]

The θ(t) made it more confusing

[stone egret]

Theta(t) means angular velocity in most case

[south ice]

<@&286206848099549185> anyone?

[south ice]

.

[south ice]

<@&286206848099549185>

[warm swan]

What's the question

[south ice]

I really wanna understand what angular velocity is man

Like I know it’s how fast an angle changes of some particular relative to the origin

But I don’t understand the math behind that

Like the w and it being inside the cos and sin functions

[warm swan]

That's the equation of a circle

In parametrized form

You use that and the fact that angular velocity is the rate of change of angle

[south ice]

Yes but what is that

I’m trying to understand whew that comes from how do we obtain it what is it

[real arrow]

Bro theta = omega*t

They just replaced it with the angular velocity formula

Angle theta = w*t

So instead of theta, they write wt

If ur asking why they do thay

I'll just use 0 as theta

Its because the angle is changing with time

So instead of just 0

U have 0(t)

And 0(t) = wt

[proper nova]

!15min

[pearl pond]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[proper nova]

don't ping twice

[ashen mauve]

omega = theta/t

[south ice]

yoooooo

[ashen mauve]

thats the only relationship you need

[south ice]

can you please help me out on this

so im trying to think of it the same way you think of normal velocity

for example

i know velocity of some position function is the limit of the difference quotient as h approaches 0

So i know the logic that makes up this velocity

[ashen mauve]

same thing for omega

[south ice]

whats the logic that makes this one up

[south ice]

wdym?

[ashen mauve]

omega can be a differentiable function

it doesn’t have to be

think angle as position

it’s a unique type of position measurement in the fact it’s dimensionless

but it’s used similarly in the calculation of angular velocity

simply the rate of change of an angle

if i turn 1pi radians every second

then my angular velocity is just omega=pi/t

similarly you can have angular acceleration too if the velocity function is of degree 2 or higher

[south ice]

wait can we pause here

what would the positon function equivalence be

[ashen mauve]

your angle at time t

theta(t) = …

[south ice]

How would you define that in this example

Like the function

[ashen mauve]

i’m sorry wdym

the function is dependent on the scenario

if i’m not rotating then it’s constant

if i’m rotating at a constant rate then it’s linear

you could have theta(t) = t^3+t+4

or you could have theta(t)=pi

[ashen mauve]

this is the definition

[ashen mauve]

is there something more in depth you could provide

[south ice]

How do we know our equation is in what form

According to what you explained above such that it could either be linear or not

Like in this example

How do I know what my angular function is

[ashen mauve]

if it’s provided

or you measure it

or you derive it from other given functions

if you’re given an angular velocity you can integrate to get your angular position +c

c can be found if you know the starting position

ex, omega=7, i started at theta = pi

integral of that with respect to time is 7t+c

my starting position was pi, so c=pi

my position function is 7t+pi

[ashen mauve]

does that make sense

[south ice]

In our example what is our angular position function ?

They’re just saying dθ/dt

But then also w is supposedly dθ/dt

So w = dθ/dt?

[ashen mauve]

you keep saying angular function but that could refer to anything

angular position, velocity, acceleration, jolt

omega = d(theta)/dt is the definition for angular velocity

[ashen mauve]

so yes

[south ice]

I meant displacement or position function

[ashen mauve]

angular position or just position?

[ashen mauve]

which of those

[south ice]

Displacement functions aren’t the same as position functions?

[ashen mauve]

they are

refer to what i asked in my previous comment

[south ice]

Yes?

Yes the one where when you differentiate it you get your angular velocity

I’m asking so I know the root of the angular velocity and how to obtain it

[ashen mauve]

then that’s angular position

here denoted by simply theta

since its in derivative form

the function would look like theta(t) = …

[south ice]

Yes but what im asking is:

Is there anything related to angular motion (such as the position function for θ or its angular velocity?)
And where is it? They’re saying d θ/dt but where is the function itself ? Am I supposed to solve for it somehow?

[ashen mauve]

why do you need angular position

[ashen mauve]

it’s just theta

you aren’t given info about omega so you cannot solve for it here

[south ice]

Oh..

[ashen mauve]

as that’s not the intent of the problem

[south ice]

I have one more question, I see that the position function of the particular is utilizing the angular velocity of that same object in the cos and sin functions, is that true ?

[ashen mauve]

yeah

particles moving in a circle

cos tells you x coordinates sin tells you y

[south ice]

So I’m asking because angular velocity here is expressed inside the cosine and sine functions, can that differ based off of the position function itself ?

[ashen mauve]

it’s there bc it tells you how fast theta changes with respect to time

multiply it by time, you get position

that’s why it’s (omega)t inside

that just tells you what angle it’s at

thus giving position

it’s just inside the cos and sin because that’s how we can make circular motion

[south ice]

If you multiply the angular velocity by time you get position ?

[ashen mauve]

yes

think of it how you can multiply velocity by time to get position

[south ice]

Oh I’ve never knew that

[ashen mauve]

i move at 10m/s multiply that by 2 seconds i’ll get 20m

technically that’s change in position

this assuming p(0) = 0

[south ice]

Ohhh… where did the seconds unit go?

[ashen mauve]

?

what second unit

theta/t • t

t cancels

[south ice]

the seconds unit

Does it cancel out with the s in the bottom

[south ice]

But intuitively, what does it mean when you multiply this by time

I’m trying to understand why it gives you position

[ashen mauve]

i mean it just means you multiplied it by the time it takes

[south ice]

Sorry but I still don’t understand it 😂

[ashen mauve]

what s

[south ice]

Seconds

[ashen mauve]

refer to this comment

it’s just dimensional analysis

[south ice]

Yes so you are moving 10m/s

How is it that if you multiply that speed by some time , you magically get a position

[ashen mauve]

because you’re taking a distance per time

and multiplying it by the time it took

if i travel 60mph

how far do i travel in an hour

[ashen mauve]

make sense

[south ice]

60 miles

Ohhhhhhh I think I get it

@ashen mauve and in the case of this problem your angular velocity * time gives you the angle of that particular at that time? Or tue position of that particle ?

[ashen mauve]

it gives you the angle theta, then taking the sin and cos of that angle give you the x and y coordinates

[south ice]

Ohhhhhh

[south ice]

What if you have cos (t) without t being multiplied by the angular velocity ?

Or is it actually there but it’s equal to 1?

I’m guessing stuff

[ashen mauve]

then youre just taking the cos of time

assuming thats a position function

youll just move in some sinusoidal motion

[south ice]

Well then what’s the point of having w in our position function? Whats the deal with having to take the cos of the angular position of some particular , vs taking the cos of the time ? How are these two things related to the position function ?

[ashen mauve]

well bc the original function is circular motiono

while cost is a sinusoidal motion

which looks very different

[ashen mauve]

two different motions two different aproaches

[south ice]

So when it comes to circular motion you’ll always need to have your angular velocity multiplied by time in your cos and sin?

Is that because if you have a circular motion, you can utilize Θ to obtain your position

?

[ashen mauve]

yes its the way of obtainning theta in this case

since our function is parameterized by t

[south ice]

Interesting

[south ice]

So not all cos / sin functions take an angle as a parameter ..?

But also aren’t those functions specifically for angles

[ashen mauve]

sin takes a scalar value and outputs a result

those outputs correlate to angle ratios

but i dont have to use an angle if i dont want to

i can just time sine of time

and get a perfectly graphable funciton

[south ice]

Wdym by “time sin of time”

Also I’m assuming your statement is inclusive to cos as well

[ashen mauve]

*take sine of time

yes it is

[south ice]

So can I say as a conclusion that all circular functions take a θ value, but non circular functions may or may not?

Or do non circular functions not take θ at all

[ashen mauve]

Hmmmm, i’m not gonna confirm or deny that as i’m not quite sure

[pearl pond]

@south ice Has your question been resolved?

[pearl pond]

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[pearl pond]

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[zenith bramble]

How to do this???

[zenith bramble]

I tried first maximizing the sqrt term (because bigger value subtracting would result in minimum value in result)

But after that it becomes like very bad expression

[dire tapir]

,rccw

[zenith bramble]

It's in application of derivatives chapter btw if it helps lol

[dire tapir]

I'm guessing they want you to reduce the expression?

then find the derivative

[zenith bramble]

Ain't no way

It's 4th power

[dire tapir]

mind you this kinda looks like a bit like some distance formula stuff going on

hmm

[zenith bramble]

I first thought that too

But like expressions are different and stuff

Wait I think I got smth

[midnight haven]

straight differentiation would be ugly

[zenith bramble]

Holy shi

I got it

[dire tapir]

let the expression be $D^2$, where $D$ is the distance between two points $P_1(x_1,y_1)$ and $P_2(x_2,y_2)$. The distance formula is: $$D^2 = (x_1 - x_2)^2 + (y_1 - y_2)^2$$

[jolly parrot]

MxRgD

[zenith bramble]

Smth like this

Then solve it with shortest distance between parabola and point

[frank violet]

yes

[zenith bramble]

Tysm for insight @dire tapir

[tough ore]

That thing under the root seems to be equation of circle

With center (15,0) and radius 2

[frank violet]

it's the distance between 2 parabolas

I think

[zenith bramble]

Yes yes

One horizontal and one vertical

[frank violet]

exactly

[zenith bramble]

I feel like my brain just opened up lmao

[frank violet]

typical jee problem

[dire tapir]

yeah I was just gonna say that you can tell from here that $y_1 = \frac{x_1^2}{20}$ and $y_2 = \sqrt{(17-x_2)(x_2-13)}$

[jolly parrot]

MxRgD

[zenith bramble]

You damn you too?

[zenith bramble]

I gotta learn how to you texit too man

[frank violet]

no, but so many people have asked similar problems

[tough ore]

I solved this exact question last year lmao

[dire tapir]

jee is tough, they kinda want you to spot tricks like these

[zenith bramble]

Nostalgia

[midnight haven]

looks like distance between a circle and parabola

[tough ore]

Yeah

[zenith bramble]

Wait is tht circle

Oh shi

[frank violet]

not sure, just graph it tbh

[zenith bramble]

x square has -ve

[midnight haven]

if its AOD for jee always better to solve graphically 🥀

[dire tapir]

well if you think about my P_2 that i've defined seems to be an upper half of a circle when you simplify and complete the square?

[frank violet]

yeah right

[midnight haven]

hmm..

[dire tapir]

plus y ≥ 0 as well

[midnight haven]

one above the x axis

[zenith bramble]

Yea it's actually 2grapgs half circle and half parabola(the first one)

[dire tapir]

is it an upper one?

the half circle

[zenith bramble]

Cuz they said x1 belongs to positive real

[zenith bramble]

Yea

[dire tapir]

Yeah, so i'm thinking it's essentially a minimum distance between a point on a parabola and a point on a circle question

that we've reduced it down to

[zenith bramble]

That circle looks so small lmao

💀💀

[dire tapir]

😭

[midnight haven]

lol smol ass

semi circle

[zenith bramble]

Ye ty for the help guys

I think I can solve it from here

.close

.close

[pearl pond]

Channel closed

Closed by @zenith bramble

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• 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.

[vernal tinsel]

hiii guys so i have 2 questions to ask i just came out of a test and im just asking the questions i wasnt sure of to see whether i guessed them right or wrong lmao. theyre not in english so pls let me know if theres anything that ishould translate

[dire tapir]

you mind translating 29 then?

I'm guessing it's a translation question

[vernal tinsel]

the peak point of the first parabola(f(x) = ax² + bx + c) is (-4,4) and it asks for the peak of
f(-x+2)