help-23

[golden nebula]

Wait really?

[golden nebula]

Then my bad

[spark leaf]

Ohh for tan that was a mistake

[golden nebula]

Oh ok

[spark leaf]

But its still wrong

[vague phoenix]

sqrt 2 / 4 is basically 1/2sqrt2

No you are correct

[spark leaf]

How can I make it that form

[vague phoenix]

You irrationalise it

[spark leaf]

How

[vague phoenix]

1/2sqrt2 is actually a bad format, write in sqrt2/4 is better

[spark leaf]

Can u use latex pls

[vague phoenix]

1 sec

[spark leaf]

Ok

[vague phoenix]

$\frac{1}{2\sqrt{2}}$ is a bad formatting, since you have the square root in the denominator, $\frac{\sqrt{2}}{4}$ is better

[flat frigate]

Nefer

[spark leaf]

Yes but how would ik if im right or wrong if I don't have the book answer

[vague phoenix]

You can just check with theta though

use inverse

[spark leaf]

How

What u mean inverse

[vague phoenix]

tan^-1, in calculator

[golden nebula]

.

[spark leaf]

And what am I supposed to get

[errant rapids]

$\frac{1}{2\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}$

[spark leaf]

Ik whats that

[flat frigate]

flynger

[vague phoenix]

angle theta

then chuck it back in sin to see if you get 1/3

[errant rapids]

You can always check if your answer matches in a case like this by rationalizing fractions (removing the radicals in the denominator)

Multiply the top and bottom by the radical and simplify

since you are multiplying by 1 it is equal

[spark leaf]

There is no radical in tye denominator

Whats a radical

Idk

I assumed its a sqrt number

[errant rapids]

yes its a number under a root, usually square root

$\frac{1}{2\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2\cdot 2}=\frac{\sqrt{2}}{4}$

[flat frigate]

flynger

[errant rapids]

for the question asker

[vague phoenix]

Then add 180 to move it to the second quadrant

[spark leaf]

Explain it pls

Why do I do 2x2

Is it cause its the sqrt number?

[errant rapids]

if you want to remove the $\sqrt{2}$ from the denominator, we know multiplying by itself will get rid of the radical (root). So if you multiply both the top and bottom, you can get rid of it, while keeping the value of the fraction the same

[flat frigate]

flynger

[spark leaf]

Ok

Thanks

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[azure harbor]

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[hardy lion]

this book is riddled with errors, what the hell is this pfd

[hardy lion]

all of a sudden it gets fixed by the system of equations

are they even right?

[cloud hound]

It does appear to be correct

[hardy lion]

preesh

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[cloud hound]

but yes those errors are egregious

feels schizophrenic

[hardy lion]

eh you know what they say, get good at calculus to fail at algebra

[noble mango]

what

[opaque fern]

welcome back

long time no see

[hardy lion]

heyo

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[opaque fern]

Not sure if control theory/dynamic systems are applicable here, but i will try my shot:

\medskip
Im working with a linear dynamic system in $\vb R^2$ governed by the two ODES:
\e{align*}{
\dot x_1(t) &= -x_1(t) +2u(t) \
\dot x_2(t) &=2x_2(t) + 3u(t)
}
where $u(t)$ is a scalar control input. We care about $y(t) = x_1(t) - x_2(t)$.

\medskip
Assume the system starts at rest where $x_1(0) = x_2(0) = 0$. I want to design some sort of feedback control function $u(t) = k(x_1, x_2)$ that drives the output to a strictly positive constant $y^* > 0$ but with $u(t) \in [-1,1]$

[errant rapids]

$u(t)=k(x_1(t),x_2(t))$ ?

[flat frigate]

flynger

[opaque fern]

If you analyse the gain of the system you get[
\4{Y(s)}{U(s)} = \4{1-s}{(s+1)(s+2)}
]
but the issue thats giving me a headache is the zero in the numerator. Any controller designed to push $y(t)$ toward a positive value will guarantee that $y(t)$ moves in the negative direction, at least initially

[opaque fern]

yeah

[opaque fern]

but then if you try to aggressively overcome this wrong movement you need to demand a huge spike in u(t)

which violates the [-1. 1] boundary

basically: i want help figuring out how to handle the initial reponse without saturating the controller

i also asked in #dynamical-systems fwiw

[safe radish]

@opaque fern Has your question been resolved?

[safe radish]

@opaque fern Has your question been resolved?

[opaque fern]

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[opaque hamlet]

Did i do anything wrong?

The answer seems pretty long to me.

[burnt notch]

Yes 😅

The points you've been given are written in polar forms

[opaque hamlet]

oh..

[burnt notch]

(as the π is suggesting)

[opaque hamlet]

bruh

my reading comprehension skill strike again

welp

rip

anyways, thanks a lot

appreciate it

[burnt notch]

Happens, don't worry 🤗

[opaque hamlet]

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[golden hazel]

can someone help me on these problems please?

[spice furnace]

!show

[safe radish]

Show your work, and if possible, explain where you are stuck.

[golden hazel]

im just confused on where to start

do i start with factoring the trinomials or do i multiply first?

[spice furnace]

So for the steps is determining the theorems that you can use.

[open wedge]

What problem first?

[spice furnace]

For Q23 You can use a/b * c/d = (ac)/(bd)

You need also make use of equivalent fractions. i.e. (ka)/(kb) = a/b

[golden hazel]

oh i don’t need help on 23

i need help on 24 and 25

sorry i didn’t clarify that

[open wedge]

How about we start from 24

[golden hazel]

yes

[open wedge]

Golden rule: factor if you see a mess on fractions like 24

[golden hazel]

mhm

[azure harbor]

I mean onyl the 2nd one looks factorable

[golden hazel]

i can’t factor the first trinomial?

[open wedge]

These guys have to be factorizable right?

[azure harbor]

No, due to -6

[open wedge]

Hm

[open wedge]

No that's possible

[golden hazel]

wouldn’t it js be (x + 2) (x - 3)

[open wedge]

Yeah

[azure harbor]

Fuh ig

[golden hazel]

k i thought so

[azure harbor]

,w factor xx-x-6

Yup.

[golden hazel]

so if i factor both of them it would be

(x + 2) (x - 3)/4x^3 *
2x^2 + 2x/(x + 2) (x + 3) ??

[azure harbor]

Yea

[golden hazel]

ok so what now

[azure harbor]

I mean the numerator in the second one can be factored further into 2x(x+1)

[azure harbor]

Multiply both together then simplify i guess

[golden hazel]

fym i guess? 😭😭

im in desperate need of help

[azure harbor]

Now that its factored it would be easier to multiply

[golden hazel]

my final is next class period

[golden hazel]

uh huh

i need help w that too

[azure harbor]

What

a/b × c/d = ac/bd

Simple as that

[golden hazel]

that made me more confused

[azure harbor]

I mean read this i guess https://www.mathsisfun.com/fractions_multiplication.html

Note: instead of numbers you can also multiply algebraic expressions involving variables roo

[golden hazel]

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[azure harbor]

Mhm.

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[fleet frost]

Help with trig study guide

[azure harbor]

!da2a

[safe radish]

Asking the actual question right away is more likely to get responses.

Asking "Can I ask...?" or "Does anyone know about...?" doesn't give people enough information to decide whether they can help, and answering can feel like a promise to help with the actual question, which they might find themselves unable to.

[fleet frost]

I need help solving a few trig questions on a study guide

[fleet frost]

?

[azure harbor]

"Dont ask to ask"

[fleet frost]

Ok so does this work or would I still need to rephrase it

.close

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[fleet frost]

.reopen

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✅ Original question: #help-23 message

[fleet frost]

Help with trig study guide

[safe radish]

@fleet frost Has your question been resolved?

[golden nebula]

Do you have a specific question you are stuck on?

[fleet frost]

What it is there are a few groups of questions that ask about the same thing that i am stuck on those pictures so those questions and the ones that I have already done

In total there were about 22 that I haven’t answered because I didn’t know how to

[zenith maple]

Do you not know the ASTC rule

[fleet frost]

No

[zenith maple]

If you knew you can solve 7 8 questions

[quaint pier]

All Silver Tea Cups.

[fleet frost]

Ohhh I know what you are talking about now my teacher calls it all students take classes but how would I use it in those questions

[quaint pier]

So for sin <0 , it needs to be negative

Check where it is negetive

and tan <0 is only in quadrant 2 and 4

for sin <0 it's only in quadrant 3 and 4

so for both to be negative, what do you think is the answer

[fleet frost]

So it would be quadrant 4

[quaint pier]

That's right

[zenith maple]

@fleet frost uk that cscx=1/sinx

[fleet frost]

Ya it’s starting to click I little bit again

[zenith maple]

Then answer of 10 should be clear

The values should be reciprocals of each other

[fleet frost]

And doesn’t that mean the same

[quaint pier]

So
Quadratic 1: All are positive

Quadrant 2 : Silver - Sin and cosec (since cosec is inverse of sin)

Quadrant 3 : Tea - tan and cot (tan is inverse of cot)

Quadrant 4 : Cups - cos and sec (since cos is inverse of sec)

Hence all silver tea cups or ASTC is the code word

[fleet frost]

Ok ok I understand it quit a bit better now

[zenith maple]

Yeah so in what world 0.2 and -0.2 reciprocals of eachother

[fleet frost]

They aren’t so then that would be it’s impossible

[zenith maple]

Correct

[fleet frost]

Thank you that helps me understand that a lot better it just wasn’t clicking in my head for a bit

Are you able to help me with a few more or not if not that’s ok

[quaint pier]

yeah let us know which ones

Can you type the numbers here

[fleet frost]

So the next ones are 11, 12, 13, 16, and 17 those are the ones on the next page

I feel like 11 is easy I just am missing something

[quaint pier]

Did you find the coterminal angles? If what are they

Or do you know what are co terminal angles?

[fleet frost]

No i couldn’t even think of the right way to start it and I can’t think of what those are off the top of my head right now

[zenith maple]

Uk how we measure the angle from the x axis they are telling you to measure the given angles starting from 115

[fleet frost]

Ohhhhh ok I think I got what you are saying

[quaint pier]

So there are two types of co terminal angles

1. Positive co terminal
2. Negative co terminal

For positive co terminal you add +360° to the give angle which is 115 here

For negative co terminal , you subtract 360° to the given angle which is 115 here

[fleet frost]

Ok ok I got that now so why are there two things with the same numbers but different pictures

[quaint pier]

And you might ask , then what will be the answer A or B?

[zenith maple]

Or basically rotate in clockwise direction for -ve angles

[quaint pier]

you know why?

[fleet frost]

No not really

[quaint pier]

So 115° in figure A shows the correct slanting of the arc whereas in option B , the arc angle is less than 90°

so which is the correct answer?

[fleet frost]

It would be A then

[quaint pier]

that's right

[fleet frost]

Ok I understand now how to choice the right picture

So how would you do 12 and 13 then I have never been good with those ones

[quaint pier]

so

3,4,5 and triplets

uk that?

[fleet frost]

Huh

[zenith maple]

Are you familiar with the unit circle

[fleet frost]

Ya some what

[quaint pier]

I'll let @zenith maple answer cuz I believe I'm intruding 💀sorry my guy

[fleet frost]

Your all good more help the better

[zenith maple]

Ok what you do is plot the point

Join it to the origin

And drop a perpendicular onto the x axis

[fleet frost]

Ok ok I got that

[quaint pier]

im not interrupting, watch this and I'll tell you a trick my teacher taught back when I was in school https://youtu.be/PUB0TaZ7bhA?si=KWzGm3aCCCRBgEKk

[fleet frost]

Ok I will lyk when I am done watching it

[zenith maple]

You get a right triangle so use trig ratio concept

[fleet frost]

Ok ok sorry for the confusion what I am seeing now isn’t that I don’t know what to do on this problem it’s that I don’t really know what it’s asking for I have a hard time with word problems like this I am better with just numbers and no words

[zenith maple]

See they are asking for a trig func of an angle corresponding to that point

[fleet frost]

So there basically just asking for it to be one of the functions things but only from the point they pick

[zenith maple]

Wdym?

[fleet frost]

What I mean is that after you make the triangle they just want you to put it in like sin, cos, tan like those type of things from the part of the triangle they pick

[zenith maple]

Yeah exactly

[fleet frost]

Ok ok

That’s Makes it work better in my head

[fleet frost]

Ok so for 16 I know that the evaluate is you make it either like a 1 or 2 either pos or neg but how do I know which number and if it’s pos or neg

[fleet frost]

I just have the sohcahtoa thing in my head I always just write it down on my paper but thanks for a different way to think about it

[quaint pier]

so that is what we use in the 12th question

[fleet frost]

Ok I understand it now just totally forgot how to do it just like all the ones I have questions on

Been a long semester

[zenith maple]

Wdym? It is just (cot90)²

[fleet frost]

But for it the answers are 0, 1 , -1 , 2

[zenith maple]

Yeah put the value of cot90 like how you did with sec180

Btw cot90=0

[fleet frost]

About that I had no idea what to do on that I looked through some of my notes and saw it turn into a one but don’t know how it did or what do to to do that

[zenith maple]

You get 0²-(-1) which is 1

[fleet frost]

Ya so how do those turn into a 0 and a 1 I don’t understand that part

[zenith maple]

See sec(180) is just 1/cos(180) and cot90 is cos90/sin90

[fleet frost]

So why is the 180 a 1 how did you get that and same with the other one I understand like the changing it to sin and cos I just don’t understand the getting it to 1 and 0 part

[zenith maple]

See you need to use the unit circle

[fleet frost]

I understand most of it but not all of it because we got thought a different way that we didn’t really use the picture

[zenith maple]

And the fact that if there an odd multiple of 90 the function changes into it's cofunction

[zenith maple]

Ok then wait a minute

[fleet frost]

What

[zenith maple]

Then are you familiar with these

[fleet frost]

No what my teacher did was she used the circle thing but only in parts so we had to memorize only parts of it at a time but not the whole thing so by the end we new it but we never used the whole picture if that helps explain it at all if anything we can skip this problem and I can ask my teacher in the morning I have some time in the morning to work on this but just not a lot and if anything we can always come back to it

Sorry for the confusion

[zenith maple]

https://youtu.be/mhd9FXYdf4s?si=1LP6LxTyrwM8Sb1D

I hope this helps you visualise @fleet frost

[fleet frost]

It most likely will I will take a look at it after we get a few more problems done if that’s ok and this is really been a big help

[zenith maple]

Ok once you understand the unit circle you wouldn't need to remember all those formulae nor all the values

[fleet frost]

Ok I think I get what you are saying that I need to just remember the numbers and not the formulas and I have that most of the way done I remember like the sin 30 is 1/2 like that stuff and the pie stuff too it’s just the 1’s are throwing me for a loop but I kind of what to just skip 16 for now if that’s ok so we can get a few more problems done if that’s ok with you instead of being stuck on one

[zenith maple]

I mean you need to remember only values till 90

[zenith maple]

And for 17 just reciprocate

and rationalize

[fleet frost]

Ok so i would flip it right then do you mean by rationalize like make it (2ratcal7/2)right or am i thinking of that wrong

[zenith maple]

I'm sorry but wdym 2ratcal7

[fleet frost]

This but with a 2 at the bottom

[zenith maple]

Why would you get 2 at the bottom?

[fleet frost]

Good question I don’t know

[zenith maple]

I mean to multiply by √7 to both numerator and denominator

[fleet frost]

I am remembering that now

That is can’t be on the bottom

[zenith maple]

So what do you get?

[fleet frost]

Answer A

And I am realizing now I called it the wrong thing to it’s a square root sign not a ratcal sign

[zenith maple]

Yup

[zenith maple]

Ohh so you meant a radical sign

Both work

[fleet frost]

Ya

[zenith maple]

And for 40 use the formula theta×r²/2

[fleet frost]

Ok ok So the next questions are 23,24 I have the answer and work to 23 because it gives the answer but I only have the first step to the work and I don’t know how to solve it

If you want I can start sending a better pic of the questions and new ones so you don’t have to scroll back as far

[zenith maple]

Yes please

[fleet frost]

Ok np

Is that ok or would you want the pic to be up and down

[zenith maple]

It's ok np

[fleet frost]

Ok

[zenith maple]

For 23 you use the same logic as 16

[fleet frost]

Ok ok that is what I was just realizing

Ok and for 24 I know you just put it in the calculator but I don’t remember how to do it for csc

[zenith maple]

Sorry but I don't work with calculators only by hand

[fleet frost]

Ok your all good I can just ask my teacher tomorrow

[zenith maple]

But shouldn't there be a button for csc

Or just put 1/sin

[fleet frost]

Man I was so close I new it was something like that I put in like 3 or 4 different things like that but just not that

I have done that a few times now

And here is 41 and again I can’t thank you enough for all this help

And counting this one I got 7 left

[zenith maple]

You did it correctly for 41

[fleet frost]

I get 67.5 for the answer but the correct answer is 424

How did I mess that up

[zenith maple]

Are you sure bc I'm pretty sure that the answer is A

[fleet frost]

The answer sheet on the back says it c for some reason

[zenith maple]

Maybe it is printed wrong or we are missing something

[fleet frost]

Maybe

[zenith maple]

But let's not waste time on that

[fleet frost]

Ya good pint

Point

Here are the next two

These word problems trip me up

[zenith maple]

Just use sec²x-tan²x=1

For43

[fleet frost]

Wait what do you mean I understand the function just not what to do with it

[radiant ice]

what do you know about cos in quadrant one

[fleet frost]

They pos

[radiant ice]

so can you remove 2 answers?

[zenith maple]

They have given you tanx value and you need to find the secx

Use that relation

[fleet frost]

So would you then treat it like a squared +b squared = c squared types thing so you just move the 3/4 over by dividing the 1 by it then square root it

[radiant ice]

ohh it wants u to use fundamental identities 😭

[zenith maple]

? U get sec²x - (3/4)² = 1

[fleet frost]

Ya that’s what I ment

[zenith maple]

Add both sides by 9/16

Then sqrt it

U will get the answer

[fleet frost]

Ok ok

[zenith maple]

For 44 remember if cos(ax) is given it's period is 2π/|a|

Same for sinx

And both their reciprocals

But for tan and cot replace 2π with π

This is explained by the graphs of them

@fleet frost do u get it

[fleet frost]

No not really I don’t really understand what you are talking about with this problem and the voltage thing is the thing the is confusing me the most

I have absolutely no idea what it’s asking me to find

Like I get it wants the period but how do I find that with this problem set up

[zenith maple]

Just forget the voltage they just mean to find the period of that func

[zenith maple]

Use the formula I gave you

[fleet frost]

I am starting to understand I bit

[zenith maple]

See if you do f(ax) for any function it gets compressed along the x axis by a factor of a

Same logic applies here

[fleet frost]

Give me one sec I think there might be a few examples in my notes that might help I will lyk if I find any

[zenith maple]

I'd love to stick around but it's getting late for me

I have to go

Bye

[fleet frost]

Your all good thanks for the help I should have enough time to get the rest done tomorrow

[zenith maple]

don't forget to watch that video on unit circle

[fleet frost]

Ok I won’t

[zenith maple]

K bye

[fleet frost]

Cya have a good day

Or night

[fleet frost]

.close

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[vast anvil]

Uh, I think I’m asking right

Hey, if someone would be so kind to help me on this problem? I think my bounds are right but I am unsure on which to go at first. If I go after ρ first that leaves me with the ρ^3* e^ ρ^2 , I am lost a little. This is me looking back on my exam, we have no key for it and what not

[zinc hornet]

You can just use a u substitution for rho^2 = u, it should be a straight forward calculation

[vast anvil]

If u = ρ^2 then du = 2ρ or 1/2du= ρ then do I use integration by parts to resolve?

[zinc hornet]

yes

[vast anvil]

Okay thank you, I wish fortune and happiness in your future

.close

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[full ridge]

b

How do I go about this

[worn delta]

some important features of a trig graph:

• the amplitude
• the period
• the phase shift
• the "translation up" y-intercept
• and with all of these features, you literally don't even need the equation anymore!!! but the equation is useful for checking your answer as a final step.

now, let's get into each one

• the amplitude is the vertical distance from the bottom to the top of the curve, or in this situation, from the bottom to the top of the motion right?

so we need to figure out the lowest and highest points.

oh that's genius ty

[worn delta]

this is related to the y-intercept too, except the y-intercept gives us an indication of where the motion starts (at t = 0) and then the amplitude is by how much we move up and down in the motion

okay personally I am not calling it the y-intercept from here because I'm thinking of a different y-value that's more relevant

I'm just going to call it the middle axis I think it was something like that

Right, so:

• the middle axis is the line that splits the curve in half. The curve oscillates up and down about this axis, with an amplitude

• to find the axis, we take the average of the lowest and highest values;

• to take the amplitude we take half the total change in vertical distance of the graph / distance from middle axis to the top, or bottom, of the graph

part 2: period and phase shift,

• period is the time taken for a full cycle from bottom -> bottom or top -> top

• phase shift is indicated a bit by the y-intercept; it's the first time at which the middle/highest value is seen for a sin/cos function respectively

yeah that's it! if i catch u online ill help u thru the specific question but i honestly have to go now i hope the detailed content dump helps

[full ridge]

This is all very helpful but I still have no idea how to determine the equation

[worn delta]

i did forget that bit that's mybad

[full ridge]

Like what do a, b, c, and d correspond to is what I’m confused about

[worn delta]

a = the amplitude

2π/b = the period

d is that middle value

[full ridge]

Middle value? You mean the peak of the wave?

[worn delta]

no, the average of the highest and lowest values

[worn delta]

it's this

[full ridge]

Oh ok

[worn delta]

c can be a lil complex. you can use the phase shift, but there are 2 different situations

• inside the cos function you have (bt-c)
• or inside the cos function you have b(t-c)
  personally I've always used the equation at this step because once you have 3 variables you can find the last. just choose a coordinate from your graph and substitute it

now

[worn delta]

is half the difference in highest & lowest

highest is 80, lowest is 4 (i should highlight it's not immediately that the highest is 80 because it's the highest in the table. ill explain why that's not always true after)

amplitude = (80-4)/2 = 38

did u find that?

[full ridge]

Why is it halved?

[worn delta]

that's just how we define amplitude, as the distance from middle to top, or middle to bottom

But not top to bottom

[full ridge]

Ok

[worn delta]

the axis is the average of the top & bottom values

= (4 + 80)/2

= 42

the axis has equation y = 42; and we also get d = 42

[full ridge]

Ok now how do I solve for c

[worn delta]

First we do b

[full ridge]

I got b

[worn delta]

the period is between t = 0 and t = 9, so period = 9 minutes

[worn delta]

as?

[full ridge]

4π/9

[worn delta]

^^

[worn delta]

It's 2π/9

[full ridge]

Shit

[worn delta]

^^

did you get the period as 9?

[full ridge]

I thought the period was 4.5

Idk how I got that

[worn delta]

Full period is from one point to the same point again

[full ridge]

Ok

[worn delta]

@lavish dagger i think you want this?

[lavish dagger]

correct i think

um

hold leme elaborate

ok see how the amplitude says 3 KEEP THAT IN MIND

[worn delta]

ill be a lil busy getting ready for school U two so might respond slower

[lavish dagger]

(the top graph) where tf does it say 3 anywhere so where did they get 3 from

[lavish dagger]

oki

[worn delta]

so you're clear that the amplitude is 3?

[lavish dagger]

yes

that’s what mrs mazzy said

[worn delta]

it's 3 bc you can count 3 units/boxes from middle axis to top

They don't give us any values you're right

[lavish dagger]

😵

[full ridge]

Ok so 4=38cos((2π/9)(9-c))+42 is the equation I got how do I isolate c

[worn delta]

We take each square on a grid as 1 thou

[worn delta]

Okay so

Subtract d, 42, from both sides

-38 = 38cos((2π/9)(9-c))

Then divide by the amplitude in front, 38

-1 = cos((2π/9)(9-c))

Clear?

[full ridge]

Yeah

[worn delta]

Dyk the cos^-1 function?

If not just think "when does cos(x) = -1?" and you'll get x = π

so π = (2π/9)(9-c)

divide both sides by 2π/9

9/2 = 9-c

and finally c = 9/2.

I'll tell you now that there are actually an infinite number of possibilities for c

We uuuuusually want the smallest (which we happened to get!)

but to check if your c is as small as possible, just make sure it's less than the period ( = 9)

if it isn't, just keep subtracting 9 from your answer till it's small enough

e.g. if we got 27/2 = 13.5 i'd subtract 9 to get 4.5

sound good?

I gotta go goodluck 🤘🏾

[full ridge]

Wouldn’t c = 27/2 then?

[stoic torrent]

??

no?

its going to be 9/2 only

[stoic torrent]

.

[stoic torrent]

.

[full ridge]

Wtf happens to the 9??

[stoic torrent]

subtract 9/2 both sides

add c both sides

see what you get

[full ridge]

I don’t get it how can 9/2 = 9-c and c??

[stoic torrent]

did u get this eqn?

[full ridge]

Yeah

[stoic torrent]

okay, so you did get pi = (2pi/9) (9-c)?

[full ridge]

Yes I got that

[stoic torrent]

and dividing by 2 pi /9 both sides?

no?

[full ridge]

Yeah I did that

[stoic torrent]

so then you managed to get 9/2 = (9-c)?

[full ridge]

Yes

[stoic torrent]

nice

now lets add c to both sides

we get,

9/2 + c = 9 - c + c
9/2 +c = 9+0

[full ridge]

Oh I fucked up on like the last couple steps

I’m a fucking moron

Sorry for wasting your time

[stoic torrent]

dw happens to best of us gng😭

[stoic torrent]

All good!

[full ridge]

Thought I had to add 9 to both sides instead of c 😭

Brainfart

So you’re telling me all this time c was just half of the period?

Or is that just a coincidence

[stoic torrent]

.

[safe radish]

@full ridge Has your question been resolved?

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[safe radish]

Please don't occupy multiple help channels.

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[minor meadow]

yo, quick question about infinite limits, with this rule, on the case I have

(2-3x)/(1+x²), I have to look for the term that has X and the biggest exponent, right?

[outer pollen]

correct. essentially you want to determine the degrees of the numerator and denominator polynomials.

[minor meadow]

If I follow this, by the 2nd rule it's zero, and if I solve it by dividing everything /x² it also gives me zero

[minor meadow]

nice ty, just wanted to confirm

chatgpt was very sus

i appreciate it!

.close

[safe radish]

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[safe radish]

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[soft lava]

I don’t undedtahd this problem at all. I don’t understand how to fit the wind into this one triangle. We never learned any equations for speed so idk how we know the speed . And I thought we already knew the speed 325mph. And I don’t undedtahd what it means to find a direction. We already know its direction

[outer pollen]

hint: consider that wind (or water currents, or any moving medium in which an object is itself moving through) affects the velocity of the object as seen from a stationary point of view.

[soft lava]

I dont know where to put it on the picture tho

I just don’t even know where to start

[outer pollen]

consider drawing the movements of the airplane and wind separately first. but get their angles right.

[outer pollen]

,rccw

[outer pollen]

this is the movement for what?

[soft lava]

Wind

[outer pollen]

so I presume the slanted arrow on the right is true north.

[soft lava]

I don’t really know what I’m doing I tried looking at the answer key but kn it even mroe confused

[outer pollen]

not a very convenient direction to point north in, but sure.

[soft lava]

It said it was pointing west to north

[outer pollen]

I know.

but normally we draw north directly up. you made your north arrow slanted.

but I digress.

now then, draw another north arrow, and draw in the velocity of the airplane.

also, don't forget to label the actual velocities of the wind and plane.

(with the numbers.)

[soft lava]

I was just trying to make it look like forty degrees

[outer pollen]

I never said that was wrong.

my whole point is about the direction the north arrow is pointing in.

I never said a single thing about the size of the angle or the direction of the velocity of the wind because both of them are okay.

[soft lava]

ok

What about the competent form of the velocity it never gave us the velocity so idk what it is referring to

[outer pollen]

the component form of the velocity of an object is the velocity, broken down into the vertical and horizontal components.

[soft lava]

Yea but how do we know which side to use for velocity which side to use for speed , etc

[outer pollen]

example: if the black arrow is the velocity of some object, then its component form is the blue and red arrows.

[outer pollen]

velocity is speed with direction.

[soft lava]

I thought that was the speed tho

[outer pollen]

there's no side to use for speed.

speed is a scalar. you don't usually draw vector diagrams with speed.

[soft lava]

I thought speed was like this I put v but I would not know which letter it is

[outer pollen]

well yeah, speed is the magnitude of the velocity vector.

[soft lava]

I’m sorry I’m completely lost 😭

I just can’t keep tract of which variables belong to which part of the triangle

[outer pollen]

then I'll step back, since I saw another helper potentially trying to help.

[soft lava]

So is velocity always the diagonal side or smth

But some of the problems have multiple of those so idk

I don’t understand the answer key for this. Because they said the current is between s and west. But in this photo it is between north and west

[vague phoenix]

Well what is the ship's velocity vector?

[soft lava]

I don’t know how to know that

I thought the velocity vector was also the speed vector but on the next problem they switch so idk what to think now

[vague phoenix]

Ship is heading North for 12mph.

[soft lava]

And I thought the velocity vector is the magnitude vector

Does it have to be north to be a vector?

[vague phoenix]

Well, what do you define a vector?

[soft lava]

Magnitude and direction but I’m confused because I thought the speed was determined by the vector but on other problems it changes so not sure how to know when it changes

and also for 5 we have no magnitude or direction

I thought

[vague phoenix]

A vector is a line with direction.

[soft lava]

so isn’t that every line

I’m so confused

[vague phoenix]

Ehh that's why they invented the arrow :))

[outer pollen]

this line has no direction. (sorry for stepping back in just to make this comment.)

[vague phoenix]

-# No problem there -.-

Well, from Chiaki's illustration, it have no direction, since no arrow is indicating the direction it is going to.

[soft lava]

I’m just confused because the vector line keeps changing on every problem it feels like. Like I know they are going to be different but sometimes the vector represents speeds and sometimes it doesn’t

[soft lava]

But then there are multiple ?

[vague phoenix]

Multiple?

That line there, suppose it have 2 endpoints A and B, we do not know if we are going from A to B, or B to A.

[soft lava]

How would we go from

Sorry I know this is off topic it’s the same problem I’m just trying to figure it out because I think I’m confusing them

[vague phoenix]

What are you writing there? I can't read...

[soft lava]

Is that better?

[vague phoenix]

That is a vector transforming into it's magnitude, or norm.

[soft lava]

ok but then for this one I lost track of what I was orginally asking. I don’t understand where I went wrong with my drawing

And for the next one I have the same issue. I don’t know if I start the next line on the side with the arrow or the side witho it the arrow

[vague phoenix]

Suppose you have a compass... Then from the centre you would have 4 different arrows pointing toward West, South, North, East.

We are given several informations for the current flows and ship's velocity, since they are having direction and magnitude, it is sufficient to transform them into vectors.

Start with the ship, it is heading North, then it is moving up on the y-axis (a.k.a. positive side), then it would have positive magnitude, etc.

[shut willow]

a "positive magnitude" doesn't really make sense here since the magnitude of any vector is always non-negative, you may be referring to the sign of the y-component of the vector

[vague phoenix]

That's what I am mentioning, you can re-read the conversation to understand which scenario I'm I referring to.

[shut willow]

Better to use the correct terminology then to avoid confusion

[safe radish]

@soft lava Has your question been resolved?

[shut willow]

The magnitude of any 2D vector is the square root of the sum of the x component squared and y component squared

$$
|\vec{v}| = \sqrt{ v_{x}^{2}+v_{y}^{2} }
$$

[flat frigate]

proxy_warfare

[shut willow]

assuming you're familiar with component form of vectors

$$\vec{v} = \begin{pmatrix}
v_{x} \
v_{y}
\end{pmatrix}$$

[flat frigate]

proxy_warfare

[vague phoenix]

You can takeover the channel then, thank you.

[shut willow]

No problem

@soft lava if you can tell me what you are confused with specifically I can help you better

[safe radish]

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[proven smelt]

im so confused rn

oh

.close

[safe radish]

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[azure harbor]

Bro this is the second self solve already

[proven smelt]

embarassing atp

[azure harbor]

Fr

[safe radish]

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[formal wolf]

im so worried this should be easy

[formal wolf]

i forgot how to do this though

because i ended up doing IBP twice which is obviously overcomplicating

theres no way in hell they would ask me to do IBP twice like that

[gilded star]

isnt it the integral of y from 0 to 1?

[crystal wigeon]

idk what you mean by "IBP" but just integrate it

[crystal wigeon]

seems like that to me

[golden nebula]

I think IBP is required. But you can use that one DI method that makes it easy

[golden nebula]

IBP is integration by parts

[safe radish]

@formal wolf Has your question been resolved?

[mild plaza]

Put u = 8x^2 and v = e^-3x

As e^-3x never becomes zero

[formal wolf]

so yes

but it's the integral of x^2(e^-3x)

which is annoying because it's if the only way to do it is IBP it's long and easy to make mistakes on

and if i make a mistake well i lose my place at my university because in the uk i have to perform flawlessly to meet the conditions of my offer

so yeah

[formal wolf]

you mean dv/dx = e^-3x

[golden nebula]

https://youtu.be/2I-_SV8cwsw?si=Z5hmLkroNlTLeeRc

[golden nebula]

Take a look at this

[formal wolf]

oh wait actually yeah i remember a table for this

[golden nebula]

It makes integration by parts very easy

[formal wolf]

thanks ill have a look at it

because that will save me a lot of time

[golden nebula]

Its a very useful trick. I think a couple of books have it too. Not sure if they teach it in school

[mild plaza]

I've never heard of it

[golden nebula]

Its very useful

[mild plaza]

Its amazing

It is so easy

Wish we could use it

[golden nebula]

Its a pretty common shortcut

[safe radish]

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[opaque hamlet]

Are there any errors in my solution?

i feel like the integral in part a might be a bit too hard

[agile scaffold]

your solution for a should be correct even if it looks tedious however in part b, there is an error in your setup for this part cuz you forgot to square the function when you substituted $r = \theta$ into the integral

[flat frigate]

reze ♡

[opaque hamlet]

ohhh

i see

[agile scaffold]

yep shouldve been theta squared

[opaque hamlet]

thanks a lot, appreciate it.

[agile scaffold]

anytime!

-# (oops you didnt need this reminder sorry LOL)

[opaque hamlet]

.close

[safe radish]

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[opaque hamlet]

nah its fine no worries lol

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[strong nexus]

Consider the potential $V(x) = ax^{-7} - bx^{-3}$, where $a$ and $b$ are positive constants. A particle of mass $m$ is initially located at the minimum $x_M$ of the potential. No other forces are acting on the particle.
\[10pt]
Sketch the potential. Where is its absolute minimum $x_M$ located?
\[10pt]
The system is evidently conservative; that is, the total mechanical energy $E_{kin} + V$ is conserved. Graphically determine the allowed values ​​of $x$ as a function of the kinetic energy $E_{kin}(x_M)$ that the particle possesses at the potential minimum $x_M$. Hint: You must distinguish between the cases where the sum $V(x_M) + E_{kin}(x_M)$ is positive and negative.

[strong nexus]

Regarding the last part of this:

I found out what x_M is

Now how do we approach that? What x are allowed?

[glad phoenix]

wym what x are allowed? @strong nexus

[strong nexus]

"allowed values of x ..."

I need to graphically determine those

[glad phoenix]

u got a graph?

@strong nexus

[strong nexus]

Im not given a, b

I can just take a = b = 1 as an example and plot it with GeoGebra

[glad phoenix]

so ur just gonna simulate things?

[strong nexus]

I mean, it says "graphically"

So I think thats what we need to do?

[glad phoenix]

graphically means they must give u something to be based on i guess

[strong nexus]

Nothing is given, no

[glad phoenix]

hmmm

well if u did successfully plot it

[strong nexus]

But that was just for a = b = 1

is this going to be invariant under a, b or what

[glad phoenix]

no

it is variant

well if u can solve Ex^7 +bx^4 -a=0

sure

u can find the turning points hahaha

[strong nexus]

I did, to get x_M

[glad phoenix]

but for total energy E = 0? u already got a turning point x = (a/b)^(1/4)

[glad phoenix]

no u only found the minimum

[strong nexus]

Yeah, I did not plug back

[glad phoenix]

yes so V(x_M) is only the minimum value for V(x)

so since the system is conservative

E = E_kinectic + V(x) = E_kin(x_m) + V(x_m)

you need to study 0 = E- V(x)

@strong nexus they gave u E_kin(x_m)?

[strong nexus]

Right

[strong nexus]

No

[glad phoenix]

then try simulation use python

[strong nexus]

Can I use GeoGebra?

Should work the same way I guess

[glad phoenix]

use whatever u want

[strong nexus]

So, we want E_kin >= bx^(-3) - ax^(-7)

[safe radish]

@strong nexus Has your question been resolved?

[strong nexus]

Ok, got it. Now assume that the kinetic energy is relatively small, such that for all $x$ permitted, $|x - x_M| = \delta \ll x_M$. Expand the potential around the minimum up to the first non-vanishing order in $\delta$.

So that means we need to Taylor expand, right?

[glad phoenix]

i think so because the minimum is a stationary point, the first derivative vanishes, so you Taylor expand in the small displacement but the second one doesnt

[strong nexus]

Is there a neat way to do this?

Do we actually need to compute the Taylor expansions of 1/x^7 and 1/x^3?

[glad phoenix]

yes

[strong nexus]

So i got V(x) ~ a * (1/x_M^7 - 7/x_M^8(x - x_M)) - b * (1/x_M^3 - 3/x_M^4(x - x_M))

Now >Set up the equation of motion for small \delta, using the result from the expansion. What do you notice?

How do we set up the equation of motion?

[glad phoenix]

try using (1+x)^n for small x

[safe radish]

@strong nexus Has your question been resolved?

[strong nexus]

How does that help?

We need an equation in terms of x, dont we?

[glacial ibex]

I think so

[safe radish]

@strong nexus Has your question been resolved?

[plucky elk]

better off asking in the physics server. see #network

[safe radish]

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[vale chasm]

Can someone explain the Functions defined by branches and module functions, like the picture?

[visual linden]

Get a new channel please

[vale chasm]

how?

[visual linden]

Send this in #help-20 (or any available help channel)

[vale chasm]

ok

[safe radish]

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[upper forge]

how is g, i and k incorrect?

[tardy mango]

Me when {1,2,5} is a subset

[upper forge]

it is a subset?

[tardy mango]

It’s a collection of elements of A (1,2,5)

So it’s a subset

But because {1,2,5} isn’t among the elements of A, g is false

[swift fjord]

epsilon means contained in {1,2,5} isnt an element of A

nor is the empty set

[upper forge]

oh my god

[tardy mango]

The elements of A are 1, 2, {3,4}, 5

[upper forge]

.close

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[upper forge]

purge this chat

[swift fjord]

lol

[upper forge]

too emb

ngl

[swift fjord]

itll be documented for the ages

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[shy osprey]

Can anyone verify if I did this right? It’s been a while so I want to mke sure I didn’t mess up anywhere. Thank you!

[plucky elk]

method looks right

[shy osprey]

okay thank you!

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[plucky elk]

just missing the 1/2 on the semicircle in the beginning

[shy osprey]

.reopen

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✅ Original question: #help-23 message

[plucky elk]

you added that later so it's fine

[shy osprey]

ohhh okay

ty!

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[shy osprey]

Okay this is what I remember being taught but this definitely can’t be the answer, right? Please help, ty!

[noble mango]

Yeah - you can probably see that <BDC looks larger than <BAC just by inspection.

The trick to this question is to use inscribed angle thereom of a circle. Does this sound familiar?

[shy osprey]

Yeah, it sounds familiar but I don’t remember how to do that for some reason

[swift fjord]

bac is 29

[noble mango]

Basically, inscribed angle theorem says that the inscribed angle (the angle at the circle) is double the angle at the center.

Refer to the top two theorems :)

[shy osprey]

Oh! It’s been briefly mentioned but I’ve never seen this before. I guess it’s somewhat an introduction to something new I haven’t seen yet.

Now I remember, it taught be about the central angle but not the inscribed angle.

Tysm! I’ll keep this in mind now! It’s definitely super useful.

[noble mango]

No problemo 👍

[shy osprey]

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[tropic hatch]

How to show that for compact spaces the pointed gromov Hausdorff convergence implies the ordinary gromov Hausdorff convergence?

[safe radish]

@tropic hatch Has your question been resolved?

[ebon hatch]

!status

[safe radish]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

@tropic hatch Has your question been resolved?

[tropic hatch]

1

[ebon hatch]

https://arxiv.org/pdf/1703.09595

Begin here. Read this and see if it helps.

If not then maybe this will: https://math.stackexchange.com/questions/4103931/definition-of-pointed-gromov-hausdorff-convergence-for-metric-spaces

And if neither of those help then I'm sorry, I can't help you. I don't know this.

[glacial ibex]

Are you still stuck op

[ripe raven]

u cant tell me these words arent made up

[azure harbor]

It isnt

Gromov and Hausdorff are real mathematicians

[safe radish]

@tropic hatch Has your question been resolved?

[tropic hatch]

Thank u very much 🙏🏻

[tropic hatch]

Its okay

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[fringe drift]

how do I express it in Cartesian form

[fringe drift]

this is what I've done

[spice furnace]

I suspect it could be referring to (f_x(t), f_y(t), f_z(t))

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[fringe drift]

can someone help me with ii

im not sure how we figure out where it intersects

[safe radish]

@fringe drift Has your question been resolved?

[lavish magnet]

Is this A level mathematics?

[glad phoenix]

but u already got the right answers

[fringe drift]

hy but I don't understand why

[glad phoenix]

so u just copied it?

[fringe drift]

no this was provided

[glad phoenix]

ah okay

so ur struggling with (ii)?

u can try plotting f(x) = exp(-x) and g(x) = -y , x = 0, x = 2

,w plot exp(-x) , -x , x=0, x=2

[flat frigate]

Results provided by WolframAlpha

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Requested by kira086868

[fringe drift]

but I won't have a graphic calc

[glad phoenix]

ofc u dont u wont need it, because they asked for a slight sketch not smth rigorous

exp(-x) & a linear function are easily plotted generally

so after u do that, the area is basically the integral of the difference between those two functions

[fringe drift]

how would I know which function is which though

s in which one to minus the other one from

[glad phoenix]

you dont know properties of exp function?

[fringe drift]

I do

[glad phoenix]

tell what do u know

[fringe drift]

as in which curve is on top

of the other

[glad phoenix]

yea

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[abstract inlet]

Ok so (if we presume our Function is always continious and integralable)
such as if "∫ f(x) = A" is true, "∫ f(2x) = A/2" is also true
Ok great But why i cannot do this if there is x in multiplication? Such as
if "∫ xf(x) dx = A" is true, "∫ f(2x) dx = A/2" is false
And is there a way to solve this withOUT doing substution?
(Such as)
if "∫ f(x) dx = A" is true "∫ x^n f(x) dx = uA" is true too (n and u are reel numbers I guess)

[stoic dune]

We can try a simple f(x) to see why it isn't true.

For example, let f(x) = 1:
∫ xf(x) = ∫ x = 1/2 x²
∫ f(2x) = ∫ 1 = x

[abstract inlet]

Well, is it possibke to solve that

[stoic dune]

There's no A that works

[abstract inlet]

So, I must do substution?

[stoic dune]

Some integrals need ipb or substitution. Are you struggling with u-sub?

[abstract inlet]

Well, not really
But lets say despite it works, I try to avoid it as much as I can do

It feels like I might miss somethig (even a simple co efficient) while doing it

Like for example, a basic one
∫ (2√x + 3)⁴/√x
I do x = u², dx = 2u du
than i find the answer 2/5(2u+3)⁵ du, but the answer is supposed to be 1/5(2u+3)⁵ du

Hello?

Ok uhhh i wont get an answer I guess

.clode

.close

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[calm cloak]

uhm do notice that the coefficient of u after substitution is 2 and not 1, I guess thats where you made the mistake

[abstract inlet]

.reopen

[safe radish]

✅ Original question: #help-23 message

[abstract inlet]

Ok so uh

Does that mean I always have to subside the final resulr with coefficient

[calm cloak]

technically cause after the substitution you again have to make another sub say z=2u so du=dz/2

I hope you get it

[abstract inlet]

Hopefully

Alright thanks

.close

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[calm cloak]

[safe radish]

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[stone shale]

11th Problem. I could only make this much progress that, b is odd definitely.

[timber karma]

hint; the digit on the right must be divisible by 5, what can multiples of 5 end in?

[stone shale]

Oooh

0,5

only

So b=5

Damn

[swift fjord]

whats the bar over ab mean

[stone shale]

Its a number

a,b are its digits

[swift fjord]

why is there a bar over it does it not mean anything else

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[barren fjord]

How do I raise something to the power of e

[azure harbor]

$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$

[flat frigate]

∫ (larp larp) dx

[barren fjord]

This is my working srry

Im just a bit stuck

[azure harbor]

jokes aside asking that question feels irrelevant to the problem at hand, maybe try asking "how to make my answer fit that form" or something like that

[barren fjord]

Sorry I thought I sent my working but I didn’t press the send button

😭

[nimble wyvern]

hm

[nimble wyvern]

hi so

[barren fjord]

Can someone help

[nimble wyvern]

do you remember that we can write A as $e^{\ln A}$

[flat frigate]

Itsuki

[nimble wyvern]

as $A=e^{\log_e A}$

[flat frigate]

Itsuki

[nimble wyvern]

||@barren fjord||

[primal bone]

-# did you just spoiler a ping?

[nimble wyvern]

||must've been the wind||

[primal bone]

-# puts bucket on your head and sneaks past

[nimble wyvern]

oop, i gtg urgent work can you take over please?

[primal bone]

soz might be too drained atm

You've suggested what I'd suggest as well tho so so far anyways

[barren fjord]

Yeah

<@&286206848099549185>

[dense lark]

wzup

lemme look into it

[simple galleon]

which part are you on?

[dense lark]

how do u do tht text

[simple galleon]

-#

followed by the text

[dense lark]

-# bla bla bla

wow

[barren fjord]

I showed my working

[dense lark]

-# this is so cool lol

[barren fjord]

-# cool

[dense lark]

-# lmao

-# 😭

😭

WOA

😭 😭 😭 😭 😭 😭

we're doin everything except the problem 😭

-# we're doin everything except solving the problem 😭

[barren fjord]

That’s all u

Ok can u focus now pls

[dense lark]

-# tsk tsk

[dense lark]

yea

[simple galleon]

why 9B?

[dense lark]

whts the problem

[barren fjord]

Oh yh I made that mistake but my values for ABC r right

I just wrote down the wrong thing

[simple galleon]

[problem](#help-23 message) [working](#help-23 message)

[dense lark]

woaaa

how'd u do tht

[barren fjord]

Omg focus

[dense lark]

sry

[dense lark]

q9 right?

[barren fjord]

Yes

[dense lark]

first subpart

yea?

[barren fjord]

No ii

[dense lark]

alr

[dense lark]

bro...

there is no iii

[barren fjord]

I showed my working

[dense lark]

was tht sarcasm

[barren fjord]

No

[dense lark]

typo

alr

[dense lark]

A must be all the constants

[barren fjord]

Yh

[dense lark]

gosh i remember studyin this havent done calculus since a long time

@patent citrus @stone shale

try this

[barren fjord]

😭😭😭😭

Bro

[dense lark]

alr ykw

i think i can do this

[stone shale]

I'm in 10th hrade

[simple galleon]

right click message
click copy message link
type out 'problem'
paste the link in the parentheses

[barren fjord]

.close

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Channel closed

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[patent citrus]

im in 10th grade

[dense lark]

😭 gng

[dense lark]

wowwww

ur smart

[dense lark]

is ur query resolved?

[barren fjord]

Whats the point in having the helper role

If ur js gonna waste time

Like actually lock in or js go

[dense lark]

mb mb

[dense lark]

srry

[barren fjord]

.reopen

[safe radish]

✅ Original question: #help-23 message

[dense lark]

but i serously pinged them cuz they know calculus

[barren fjord]

Clearly not

[dense lark]

anyways

lemme solve

[barren fjord]

Ok

Finally

[dense lark]

they're faking it, they both know calculus....but anyways lets focus on the problem

[stone shale]

I SWEAR IM IN 10TH GRADE

[barren fjord]

Just shush pls

[dense lark]

you couldnt solve it?

[barren fjord]

…

[dense lark]

uhm

[barren fjord]

No

People don’t typically ask for help when they understand it

[dense lark]

some ppl yk can solve their solutions, they have a doubt in method or wanna know a simpler technique

....