help-26

[slow tundra]

Yea ik i wrote it

I dont understand how to tho

[odd jacinth]

you want to expand it, but in a smart way (so using Pascal's triangle or whatever you call the formula on the side)
because it avoids many calculations

[slow tundra]

Ok so

[odd jacinth]

the denominator of the fraction you must calculate will be 2^5, while the numerator will be (e^ix+e^-ix)^5 expanded

and when you expand it, you find

[slow tundra]

Yeaaa

[odd jacinth]

e^i5x + 5e^i3x + 10e^ix + ...

[slow tundra]

Ok this might be dumb but english isnt my first language

[odd jacinth]

finding that is a lot of calculations, but not if you use the formula on the side

[slow tundra]

Whats expansion…

[slow tundra]

Is it just this

[odd jacinth]

the opposite of factorisation

[slow tundra]

Ahhh

[odd jacinth]

(x+1)(x+2) = x²+3x+2, I expanded (x+1)(x+2)

[slow tundra]

Gotcha

So basically

Plugging the expanded equation into that formula

Will save time

Instead of having to

Calculate

That really big equation

[odd jacinth]

yeah you have the factored form (e^ix+e^-ix)^5
you could expand that by hand, but the formula is obviously more efficient

[slow tundra]

Okayyy

Thanks for your help i think i get it now

[odd jacinth]

and the coefficients (5 choose k) are in the Pascal triangle

they are in order, 1, 5, 10, 10, 5, 1

[slow tundra]

Isnt 5 the n not the k?

[odd jacinth]

it is the n yeah

[slow tundra]

Okok but what does the second part have to do with the first part?

[slow tundra]

Cause i see sin^5 x and cos^5 x

So it probably has something to do with it

[odd jacinth]

oh you meant 2nd question

[slow tundra]

Yea

[odd jacinth]

this is cos^5(x)

[slow tundra]

Oh

[odd jacinth]

cos(x) = (e^ix+e^-ix)/2

and sin(x) = (e^ix-e^-ix)/2i

[slow tundra]

So at the bottom for sin^5 x it would just be 2i instead of 2

[odd jacinth]

careful about the sign in the numerator

[slow tundra]

Okok

I think i get it now

.close

[topaz sinew]

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[rigid cloud]

.reopen

<@&286206848099549185>

[marble gate]

please read #❓how-to-get-help

[rigid cloud]

Done

[marble gate]

okay. so ask your question

[rigid cloud]

so in genuine i need help with fractions

i cant understand them

[marble gate]

a/b = a divided by b

[rigid cloud]

i mean as in

https://tenor.com/view/matematica-rio-professor-teacher-procopio-matematica-gif-14041339

those type

[marble gate]

do you have a question you need help with?

[rigid cloud]

so just normal questionsas in 2 - 3 4- 6 etc.

[marble gate]

sorry, I still don't know what you mean

if you have an example of a question then I/we can help explain it

[rigid cloud]

so, in like 3- 4 grade you learn fractiond normal fractions such as how 3 over 4 is the same as (whatever number) etc

[marble gate]

yep

[rigid cloud]

so, you can pick whatever i just need help understanding fractions in general

[marble gate]

so 3/4 can be seen as the ratio of 3 to 4

if you scale it up or down on one side, you need to do the same on the other side

if you have 2 lots of 3, and you have 2 lots of 4, then you still have the same relative amounts

[rigid cloud]

could u explain it how u woulde xplain it to a 5 year old im so confused

[marble gate]

2*3=6 and 2*4=8 so 6/8 is the same as 3/4

[rigid cloud]

how

2 3 is 5?

[marble gate]

2*3

[rigid cloud]

2 4 is 6

[marble gate]

2*4

[rigid cloud]

could u explain it in a kindergarden way im so confused

[marble gate]

no, I probably couldn't. sorry. if you have an actual question I can show you how to think about it, but "I don't know anything about fractions" isn't..something I can help with

[rigid cloud]

oh okay....thank you

how do i work out 3 over 5

<@&286206848099549185>

@topaz sinew

[blissful shard]

So like

3/5?

[rigid cloud]

ye

[blissful shard]

Oh

You can times both the top number and bottom by 20

no

2

[rigid cloud]

is it always 2

[blissful shard]

so you have

no no

What you want is a familiar bottom number

for example

[rigid cloud]

5

[blissful shard]

2/10

is 0,2

5/10

0,5

the same as 1/2

[rigid cloud]

huh

[blissful shard]

because when you times both "1 and 2" by 5, you get "5 and 10"

[rigid cloud]

huh wth where does the 5 come from

[blissful shard]

Lemme show

Ok i'm sorry if i don't know the correct words

english isnt my first language, i study in another

[rigid cloud]

is k

[blissful shard]

I'll try my best

[rigid cloud]

🙂

[blissful shard]

Im trying to visualise it for you

In the first circle

we have a whole pizza

it's simply 1

1 pizza

in the second circle we have 1 pizza divided into 2 pieces

so

1/2

let me show you 5/10

We have 1 whole pizza

divided by 10

We can call one of the slices 1/10

Basically 1 out of 10

I drew over 5 slices

So it is 5 out of 10 slices i have drawed over

5/10

and as we can see

it's half of the pizza

Like when 1/2

is half the pizza.

Is it somewhat understandable so far?

I'm just gonna move on till the end till you come back then.

So by saying 1/2 is the same as 5/10 is true, and can be visualised.

The only difference, is the amount of slices and the amount of slices eaten

Now

Your question

3/5

Let's see what it looks like

We have a pizza divided into 5 slices.

Each slice is 1 out of 5 slices.

Basically: 1/5

If we eat 3 out of 5 (3/5) pizza slices

we now see this:

It covers this much of the 5 pizza slices.

Now let's try and see how many slices, and how many slices eaten it takes to cover as much as this pizza

Let's try and eat 3 slices

so it would be 3/10 slices eaten

Now we can see, it doesn't cover as much as 3/5

so lets eat more

Not quite yet

this is half the pizza

and we can see on the circle with 3/5 slices eaten, that it covers more than half

So lets just eat one more, and:

Both pictures compared

looks a little similar right?

Thats because they cover the same amount of the pizza

just different sized slices

and more slices in this case

But now, you want to know 3/5

in numbers

Basic Division

If you know division with 10 in this instance, you probably also know division by 5

5 is the half of 10

therefore, to make 3/5 reach a point, so we have division by 10

we can allow ourselves to times 3/5 by 2

on both numbers

It would look like this

This is as simple as i can explain it.

And by diving a number by 10, we simply move the comma to the left

so 6 will be 0,6

When we know that 6 can be 6,0000...

and so on and so with the zero's

I hope you understand it, and maybe have some questions.

I won't be here for that right now, as I will have to go now.

👋

[topaz sinew]

@rigid cloud Has your question been resolved?

[rigid cloud]

thank you:)

[topaz sinew]

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[steady bison]

hi please can i get some help on this? ( please ping when you are helping to solve it )

[topaz sinew]

Please don't occupy multiple help channels.

[steady bison]

so far i have gotten to this

[rigid ivy]

do not occupy multiple channels

[steady bison]

its 2 seperate questions

[rigid ivy]

Then stick to the one you want answered first, please

[steady bison]

@blissful shard

Please you think you can help?

<@&286206848099549185>

[hazy socket]

?

[hazy socket]

what do you need?

[steady bison]

To solve the q above

[steady bison]

I got up till here but I still don’t get it I got the q wrong 6x.

@hazy socket think you can help?

[topaz sinew]

@steady bison Has your question been resolved?

[steady bison]

<@&286206848099549185>

[main kiln]

hi

@steady bison what do you need help with?

[steady bison]

This q

This is my working

[main kiln]

wrong

hollup i’ll explain

[main kiln]

this part is wrong

[steady bison]

How comes

3 right angels

@main kiln

[main kiln]

that’s not what you do

[steady bison]

I’m not too sure. Can you show me your working?

The base for the bottom triangle is 1

@main kiln

Maybe this @main kiln

[topaz sinew]

@steady bison Has your question been resolved?

[topaz sinew]

@steady bison Has your question been resolved?

[main kiln]

yeah that looks about right

[main kiln]

can i show you tmrw

i’m studying for my ap calc test

[topaz sinew]

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[old zephyr]

It's just problem b what goes after this step

[old zephyr]

Ah I messed up

What goes after this I need finish the bottom part but ye

ah nvm im just confusing myself more atp

ah yet again another awkward situation

i should probably ping now

<@&286206848099549185>

ive done this too much otday

i feel guilty as hell

i need help with be

*b

just disregard every bit of work you see

[hollow path]

4x^4 /3

4x^3 times 2x^5 is 8x^8

4 times 2 is 8

And when multiplying exponents u just add them so that’s how I got the x^8

8/6 is 4/3

Then when diving exponents u just do subtraction so u have x^8/x^4

[old zephyr]

uhmmm

[hollow path]

So it’s just x^4

[old zephyr]

im so sorry you might need to go slower than that 😭

[hollow path]

What do u not get

[hollow path]

This was correct

U just needed to simplify it

[old zephyr]

Like Idk what I'm doing at all

[hollow path]

Do this

[old zephyr]

Uhh ok

[hollow path]

Do u have it written

[old zephyr]

Yeah

[hollow path]

Now go break it down simplify 8/6

[hollow path]

I mean ig u can do it like this

[old zephyr]

How did you do it?

[hollow path]

Wait nvm I looked closely what u did makes no sense

First start off with 8/6

[old zephyr]

Ah ok

[hollow path]

4/3

Simplify

[old zephyr]

Ok

[hollow path]

When dividing exponents u actually just subtract it

[old zephyr]

Yeah

But thst had a different base

[hollow path]

So u see the x to the power of 8 divided by x to the power of 4

[old zephyr]

Uhhh yeah

[hollow path]

They both x?

[old zephyr]

Oh I was getting confused

By the 3 and the 4

Ye sorry

[hollow path]

Nvm that

[old zephyr]

Oh

[hollow path]

But u get the exponent diving part

[old zephyr]

Yep

[hollow path]

So what’s ur answer

[old zephyr]

Uhh....

[hollow path]

But the 4/3 is needed for the answer

[old zephyr]

Ok

This?

[hollow path]

Yeah

That’s the final answer

[old zephyr]

Ohh

Ty

.close

[topaz sinew]

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[limpid flame]

I got a remainder idk if I did something wrong idk what to do from here

[queen rampart]

you've made a mistake in the first step it seems

[limpid flame]

I just saw😭 😭 😭

Ty

.close

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[dry goblet]

Just got a question relating to forming a linear equation

A shopkeeper buys a crate of eggs at $1.50 per dozen. He buys another crate, containing
3 dozen more than the first crate, at $2.00 per dozen. He sells them all for $2.50 a dozen
and makes $15 profit. How many dozens were there in each of the crates?

[dry goblet]

I mainly just want to see someone else solve it and question their process

[woeful stone]

you solve it

[dry goblet]

I dont know where to start

apart from that you assign 'x' to being the number of dozens

[woeful stone]

total_cost = x*1.50 + (x+3)*2

total_revenue = (x+(x+3))*2.50

revenue - cost = 15

[dry goblet]

I've done it wrong but I'll show you what my little brain gears cooked up

Sorry about the handwriting im left handed on a right page

[woeful stone]

almost correct for cost you have 1.5x+2x + 6 = 3.5x + 6

[dry goblet]

so the cost = 3.5x + 6, right?

oh wait I see that whoops

I'm still lost on what to do next

so x should = 1, correct?

<@&286206848099549185>

[woeful stone]

yes x = 1

[dry goblet]

okay cool

what do I do with that from there tho

[woeful stone]

So what is x?

[dry goblet]

the number of dozens

Im literally head empty no thoughting rn

[woeful stone]

x is the number of dozen in a crate which is what the question is asking for.

[dry goblet]

so what do I do with that?

Im really having a moment

[woeful stone]

So that is answer. You can just write 1 or write a sentence saying how many dozens are in a crate?

[dry goblet]

It says the answer is 9 in the first crate and 12 in the second crate

[woeful stone]

The minus sign needs to be distributed -(3.5x+6) = -3.5x - 6.

[dry goblet]

I feel like I've been taught right and I finally understand it, then as I move into a new unit it's just counteracting what I did

I do get what you're doing

and Im awful at putting words to my thoughts

But beforehand I SWEAR they would've just done 5x + 7.5 - 3.5x + 6

[woeful stone]

Yeah that is not correct. What is being applied is called the distributive law i.e. a(b+c) = ab+ac.

[dry goblet]

yah

thank you

I think that should be all

alright thank you so much

have a good one

.close

[topaz sinew]

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[analog dome]

Ive been confused since yesterday

[rigid ivy]

What is the question?

[analog dome]

its the one on the image

l1 = 0.7m
m1 = 0.70kg
l2 = 1m
m2 = 0.50kg

the ones with the '2' subscripts is for the red one

i dont know how to find the height for the blue one

wait nevermind, i think i figured it out

[topaz sinew]

@analog dome Has your question been resolved?

[topaz sinew]

@analog dome Has your question been resolved?

[topaz sinew]

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[topaz sinew]

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[patent lynx]

I got x<5 but according to mathway x is also > -1

[patent lynx]

Can some explain why?

[opal vault]

because log(x+1) has to be defined

[patent lynx]

Oh that's right 😂

Thanks

I need to remember the basic rules

[topaz sinew]

@patent lynx Has your question been resolved?

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[topaz sinew]

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[neon iron]

I did the following;

,rccw

but I'm really unsure on how to translate this at all

[fluid belfry]

,rccw

[peak scarab]

well first of all

[fluid belfry]

the second eq is wrong ig

[peak scarab]

your friction is in the wrong direction

The thing is spinning

[fluid belfry]

this too ^

[peak scarab]

So the block is going up

[topaz sinew]

@neon iron Has your question been resolved?

[neon iron]

No

it shouldnt be

this is T_max we aer considering

meaning the longest time it takes for the thing to rotate around

meaning when the acceleration is the slowest

so it surely is going down

[neon iron]

what is a better representation of it?

[stable surge]

Since the block is also rotating with the cone you need the centrifugal force

[fluid belfry]

it would be at equilibrium as the question states "it has to maintain a constant height of h above the apex" and something like this

[peak scarab]

a = 0

[fluid belfry]

he asked the same question yesterday, i used the concepts of circular motion but i think he don't have to use them here

[stable surge]

Outward and perpendicular to the axis in rotating frame of reference

[peak scarab]

well actually z = h so v = 0 and a = 0

[stable surge]

But it affects the normal reaction and and friction you need to write them

[neon iron]

The book like gives a few hints to those questions

and to this it states this:

[peak scarab]

assuming the block stays in contact with the cone you can also use Coulomb friction law

[stable surge]

See I told you, you need the centrifugal force

[fluid belfry]

centripetal*

[stable surge]

I don't deal with centripetal, it's weird, I write all my equations in rotating frame of reference

[neon iron]

wait but isnt a centripetal force just like the net of the external forces of an object moving circularly? Like, it is not its own force is it?

[stable surge]

Yep

[fluid belfry]

true its not a real force, it a summation of all the external forces acting in the radial direction

[stable surge]

In your case centripetal force would be the resultant of all the horizontal forces

[neon iron]

also how can we conclude that the object moves in uniform circular motion?

[neon iron]

but yeah i see

[stable surge]

They have given the time period of revolution of them cone so I assume it's moving with constant angular velocity

And then the block is stuck at the place

So it's also moving in uniform circular motion

[neon iron]

yeah okay

alright so i guess to work on the problem i need to figure out how to set up my FBD correctly

hmm

wait

so what was exactly wrong with my first FBD?

that i drew

[stable surge]

You should have a resultant horizontal acceleration

[neon iron]

the radial acceleration i take it?

[stable surge]

Yes

Mw²R

W=2π/T

[neon iron]

okay so

lets recap everything conceptually before getting into the math

we have a box rotating with the cone circularly with no change in height vertically

as T is constant, this implies the velocity is constant, and the box is in uniform circular motion

[stable surge]

Yes

[neon iron]

at T_max, the radial acceleration is slowest and there will be a strong friction force parallel to the wall of the cone

this is to prevent the box from sliding down the cone

[stable surge]

Hm

I think when the radial acceleration is max the friction available will also be max

[neon iron]

at T_min, the radial acceleration is highest and there will be a strong friction force again parallel to the wall but going down

[peak scarab]

This is what I found using frenet’s coordinates

The second line is the projection on T

[neon iron]

oh why is that

,rccw

[neon iron]

wait for which case is this?

T_max?

[peak scarab]

Tmax

Wait

on second line might be d(v_t)/dt

[stable surge]

If you increase radial acceleration, normal reaction would increase, and hence more friction

[neon iron]

hmmm really?

[stable surge]

Also

You should make a function of T with respect to friction

Since static friction varies from 0 to μN

Then we talk about maximum and minimum

[neon iron]

by the way ill send them anyways for the sake of verification but this is the answer key

[neon iron]

oh okay this is smart yeah

start with a general form

[stable surge]

It's not necessary that the friction being applied is the maximum available

[neon iron]

but how are u meant to represent static friction in its non-maximum form?

because what im used to is like [
f_s \le \mu_s N
]

[stable surge]

You can only find its value in terms of other forces by putting acceleration in that direction to be zero

[neon iron]

ah i see so basically solve for f?

[stable surge]

Yea

[neon iron]

okay cool lets do this

thanks for the help by the way much appreciated

how should i start with constructing this function

[stable surge]

Make free body diagram

And make equations of forces

[neon iron]

yeah

but the friction is variable here

so how should i draw that 😅

[stable surge]

Basically this, but make components of forces in horizontal and vertical direction s and sum horizontal forces would be the centrepetal force

[neon iron]

yeah but having f be in that direction would be implying T_max wouldnt it

or...something

because f should be in the other direction during T_min

okay actually

it doesnt matter let me make it 😅

[stable surge]

You can just change f to -f for that

[neon iron]

yeah its like

a vector function then

just swings back and forth

okay gotcha

one sec lemme write

[fluid belfry]

okay wow, the question says that it should be in terms of only h and B

[stable surge]

If you are having problems I would suggest going into rotating frame of reference and making centrifugal force radially outward and then just balance the forces since block would be at rest in rotating frame

[fluid belfry]

and that's exactly i told you yesterday

[fluid belfry]

this was the method i was talking about yesterday btw

[neon iron]

,rccw

[neon iron]

hopefully i didnt fuck up my trig but is this correct

[stable surge]

What is theta

[neon iron]

Beta

[stable surge]

N sin beta in first equation and N cos beta in second one

[neon iron]

oh huh

but like

the angle those two make is like exactly the same as the black triangle as a whole

no?

[stable surge]

i was talking about N

[neon iron]

okay trig fact check time lets see

what

oh im an idiot i see

[neon iron]

okay yes i agree

so we have

,align
N\m\sin\theta - f_s\m\cos\theta &= m\f{v^2}R \
N\m\cos\theta - f_s\m\sin\theta -mg&= 0

[stable surge]

yes

[neon iron]

woooo progres1

[stable surge]

now you dont want N

[neon iron]

yeah

[stable surge]

and T is 2piR/V

[neon iron]

so solve for N in both equations and equate them

[stable surge]

yes

[neon iron]

okay algebra time

,align
N &= \f{\ff{mv^2}R + f_s \m\cos\theta}{\m\sin\theta} \
N &= \f{f_s\m\sin\theta + mg}{\m\cos\theta} \
&\Implies \f{4m\pi^2 R}{T\m\sin\theta} + f_s \m\cot\theta = f_s\m\tan\theta + \f{mg}{\m\cos\theta}

[stable surge]

oh wait

[neon iron]

oh?

[stable surge]

you would have to put f as +muN beta for max time and -mu N minimum time

so we havent eliminated the N yet

[neon iron]

Is finding a general function not possible then?

[stable surge]

no

you have to use general sense to see at high angular velocities the block can slip in upward direction, to prevent that we do max friction in opposite direction

and slow angular velocty the block would slip down so you need friction in the direction you have taken

[neon iron]

yeah

so like

we need to split it up

alright no worries

lets do that

[neon iron]

this would still be correct for the T_max though right?

[stable surge]

yes but put f = mu* N

[neon iron]

hmm lets expand on why T_max implies f_s.max again

it means f_S.max means N is greatest

why is N bigger here

ohh wait

its because like

when T_min for example

you have N_y - w - f_s.y

so you have a component of the friction pulling down stronger in the y direction

[stable surge]

hmm yes

[neon iron]

wait hmm

[stable surge]

for Tmax you would fsy upward balancing the weight of the body

[neon iron]

yeah

so N_y is not reduced

i guess i see that yeah

so lets continue

[
\f{4m\pi^2 R}{T\m\sin\theta} + f_s \m\cot\theta = f_s\m\tan\theta + \f{mg}{\m\cos\theta} \
\f{4m\pi^2 R}{T\m\sin\theta} = \mu_s N\p{\m\tan\theta - \m\cot\theta} + mg\m\sec\theta \
T = \f{4m\pi^2 R}{ \mu_s N\p{\m\tan\theta - \m\cot\theta} + mg\m\sec\theta}
]

we still need to get rid of this N and mu s

[stable surge]

its more real life experience than maths, like if you spin the cone really fast you have friction so that body doesnt fly out and if its rotating slowly you need friction so it doesnt slip in ( i think we were supposed to think this before making equations)

[neon iron]

yeah

i get that

[neon iron]

okay so we have this bad boy now

we technically got our answer

but we need all the variables to be in terms of theta and h

[stable surge]

you should put f = muN in these equations i think

would be easier

to eliminate N

[neon iron]

oh okay

,align
N\m\sin\theta - N\mu_s\m\cos\theta &= m\f{v^2}R \
N\m\cos\theta - N\mu_s\m\sin\theta -mg&= 0

[neon iron]

,align
N\p{\m\sin\theta - \mu_s\m\cos\theta} &= m\f{v^2}R \implies N = \f{4m\pi^2 R}{\m\sin\theta - \mu_s\m\cos\theta}\
N\p{\m\cos\theta -\mu_s\m\sin\theta} &= mg \implies N = \f{mg}{\m\cos\theta -\mu_s\m\sin\theta}

[stable surge]

wait

mg on the other side

[neon iron]

yes my bad

[neon iron]

hopefully no bullshit happening

[
\f{4m\pi^2 R}{T^2\p{\m\sin\theta - \mu_s\m\cos\theta}} = \f{mg}{\m\cos\theta -\mu_s\m\sin\theta}
]

[stable surge]

t missing

[neon iron]

oh right

[neon iron]

how do we get rid of this mu s tho

[stable surge]

you cant

it was given to you in the ques

it should be there in the answer too no way to get rid of it

you sent the answer here right?

[neon iron]

yeah

right

we have to get rid of R with h though

[stable surge]

they have it in the answer

[stable surge]

use the triangle

[neon iron]

yeah

let me solve for

T

first

[
\f{4m\pi^2 R}{T^2\p{\m\sin\theta - \mu_s\m\cos\theta}} = \f{mg}{\m\cos\theta -\mu_s\m\sin\theta} \
T = \s{\f{4m\pi^2 R\p{\m\cos\theta - \mu_s\m\sin\theta}}{mg\p{\m\sin\theta - \mu_s\m\cos\theta}}}
]

[neon iron]

so uh

triangle

am i meant to do this

[stable surge]

yes

[neon iron]

that angle at the bottom is 90-beta

so like

the one at the box is beta

so

tan(beta) = h/R

R = h/tan(beta)

[
\f{4m\pi^2 R}{T^2\p{\m\sin\theta - \mu_s\m\cos\theta}} = \f{mg}{\m\cos\theta -\mu_s\m\sin\theta} \
T = 2\pi\s{\f{h\p{\cos\theta - \mu_s\m\sin\theta}}{g\m\tan{\theta}\p{\m\sin\theta - \mu_s\m\cos\theta}}}
]

[stable surge]

yep

cancel out the m and there's your answer

[neon iron]

omg

[stable surge]

yep

[neon iron]

wait hol up

[stable surge]

there was a +

[neon iron]

they have a plus

yeah

wait what went wrong

[neon iron]

oh 💀

woopsie

okay perfect

we dont have to do T min

but lets just understand what goes on in there

so what is happening with f_s in that case

[stable surge]

the cone is rotating really slow, so centripetal force is low, so the N is low, now fs must balance the weight of the object

or it will slip in to the cone

so fs would become max

oh wait shitt

tmax has fs inside, you calculated tmin

[neon iron]

no wait why

t max the thing isn't going outside

it surely will be going down

so the friction is up

[stable surge]

oh wait my max

bad*

im got confused

sorry

[neon iron]

centripetal force is max no? t min means like it takes the lowest amount of time I think

ah yeah no worries

[stable surge]

so we are dicussing tmin now?

right?

[neon iron]

yeah we finished t max

[stable surge]

ok ok

so the cone is rotated really fast

[neon iron]

yeah

[stable surge]

so the box would slip out

if theres no friction

[neon iron]

yeah

[stable surge]

it would slide up the thing

so in this case friction would be mu*N in the opposite direction

[neon iron]

and then same story i guess

[stable surge]

yes

[neon iron]

okay i gotcha

btw on like an unrelated note

with physics question are u meant to like not be rigourous with stuff if that makes sense?

[stable surge]

you dont have to do the calculation again just replace f by -f everywhere so mu by -mu

[neon iron]

like i feel like we are making plenty of assumptions that its kinda eh

[stable surge]

i dont think we made any assumption

can you give example?

[neon iron]

"without friction the box will slip out of the crate if the thing is going really fast, so friction must be going down"

i get that intuitively

but its like

how do you show that that will happen mathematically

like im thinking about it

if this was a more complex system and i wasnt so sure where the thing we are considering would be going without friction, what should i do?

i cant just "see" it like we did here

[stable surge]

go to diagram where you made forces and assume mv^2/r is a lot bigger than the weight

you will realise fs must go down to keepp the block at rest

and that's what friction does right? prevents things from moving

[neon iron]

hm

i mean we can consider our original equations without friction

lets see

[
N\m\sin\theta = m\f{v^2}R \
N\m\cos\theta = mg
]

[neon iron]

wait huh

wait i guess this doesnt even work

because it wouldnt be uniform circle motion

anymore

[stable surge]

no it does work

[neon iron]

oh

[stable surge]

N is adjustable since normal reaction by a body can be anything agreed?

[neon iron]

yes

agreed

[stable surge]

so now v is high

hence N is high

[neon iron]

agreed

[stable surge]

but mass of object is fixed

[neon iron]

right

oh this system becomes non inertial

[stable surge]

yes

[neon iron]

i mean i know that like

in elevator problems for example where you have N > w the thing is accelerating upwards

oh right

because acceleration needs to be in the direction of the net force

N being the sole force (so the net force) and going upwards implies that whatever you will be having will be in the direction of N's components

i guess?

[stable surge]

i dont get what you mean

[neon iron]

ok can u explain why N is high implies acceleration upwards then?

[stable surge]

note that the object would wanna move outwards like any rotating object would wanna outwards cuz of the centrifugal force

[stable surge]

yea it would upwards since Ncos beta is upwards, but it will also be radially outward

because centrifugal

why we didnt draw it in our diagram?
because it doesnt exist in the frame of reference from which we made the diagram

you basically need an external force to keep the object rotating, if the force is not enough the object "appears" to move radially outward from rotating frame of reference

in reality, the objects tangential velocity is just to big for the force to handle, so the radius of curvature slowly increases

(because the tangential velocity was big, the object also move away from the center of rotation, since the force wasnt able to rotate the object's velocity completely so it moved away a little bit and slowly it goes to the correct radius of curvature)

now i know this is all very confusing and i wasnt able to explain this properly, but thats the whole story

[neon iron]

what is a rotating frame of reference exactly

[stable surge]

a rotating frame of reference is basically what a person rotating with the same angular velocity and angular acceleration would see the object to be

llike if a box is rotating around you, but you are also rotating with it

you wouldnt see it rotating

[neon iron]

oh i see i guess thats true

hmmmm

alright

i guess i get the main takeaways here

i guess i will get around to trying to understand this better when i involve myself in angular motion. Because technically we have not covered that at all just yet chronologically in my textbook

but anyways thank you so much for ur patience i totally get the idea now i believe

[stable surge]

i mean what i told you, you dont need to know all that, it just my way of understanding centrifugal force, nobody told me all this, you probably wouldnt find it in a book, because it doesnt really matter

i just like to know how things work physically too

[neon iron]

yeah i get that

im trying to achieve that too to the best of my abilities

[stable surge]

yea, its a good thing, it helps you in new situations where its just you and your understanding of the subject

[neon iron]

yeah i get that

honestly no joke i have been trying to rationalise this problem for a while now

been 2 days i believe

finally got the idea

oh yeah by the way

wanted to ask

what if we had picked our coordinate system to be with x parallel to the incline?

i guess that wouldnt work?

[stable surge]

ofc that would work

the equation would just look different

[neon iron]

i feel like that would have been a much simpler approach

since we would only need to represent w in its components

but how do the equations work then? we wont have ma_c

since its not radially horizontal

[stable surge]

but you owuld have net accelerationg both along the and perpendidular to the incline

like this

now you would add forces along the incline = macos beta
sum of forces perpendicular to incline = ma sin beta

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[north canyon]

I don't understand this solution, can anyone elaborate on what is done after q^(n-1) | a => a ≥ q^(n-1)?

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[north canyon]

Ahh this is really confusing random stuff is popping out of nowhere

<@&286206848099549185>

[candid fable]

yes

idk how to do this sorry ill see if i can get other helper

<@&286206848099549185>

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[gray basin]

how?

[junior pumice]

first step would be drawing a rough sketch of the situation described.

[gray basin]

ok

so

something like that i guess

so it’s like a truncated pyramid, flipped upside down

[gray basin]

so now what should I do?

nvm got it

.close

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[acoustic pecan]

welp

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[smoky trellis]

If I'm told that the ratio of BD to DC is 3:5, can I do (3x)^2 + (5x)^2 = (3sqrt(5))^2 and solve for BD & DC?

[grave vigil]

Yes

[smoky trellis]

thanks

.close

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[floral skiff]

Yo guys I need help with homework, do you guys know the difference between nonfactorable with real numbers and non factorable with integers/prime?
From what I’ve figured out it’s in the grouping method where prime is when there aren’t any integers to multiply to ac or add to b. And all real numbers is when a there’s no true numbers that work for both.

[floral skiff]

If elaboration is needed I can provide

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.close

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[elfin sparrow]

consider the case where k+1 divides n

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[elfin sparrow]

oh, a bit more complicated than I first thought, I misread it

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[soft escarp]

hello

i just have a simple vocabulary question

[soft escarp]

in this image, does period mean fundemental period?

is 'fundemental' just skipped and implied in it?

[sonic dawn]

for sinusoidal fns this is generally the implication yes

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[soft escarp]

thank you!

@sonic dawn

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[rotund hawk]

how does one solve $$3^3x-2=4^5x$$

[thorny flame]

adarw

[rotund hawk]

thats not how it works

the 3x and 5x are exponents

:(

[sonic dawn]

$$3^{3x}-2=4^{5x}$$

[thorny flame]

IV

[rotund hawk]

$$3^(3x) -2 = 4^(5x)$$

[sonic dawn]

is the question what i typed?

[rotund hawk]

yeah

I dont know how to solve it

@sonic dawn any idea on how to do it? lol

[sonic dawn]

oh i solved it but im figuring out how to get you there
i guess start by figuring out a minimum value for x for solutions to even have a chance to exist

[rotund hawk]

yeah I dont know how

how can a smaller number with a smallelr co efficent equal

[sonic dawn]

correct, it cannot

which is why I’m trying to get you to start there because properly showing a solution dne requires a tiny bit more effort

[rotund hawk]

do we do log of each side?

and then uh

I dont know

wait let me think

3x log 3 -2 = 5x log 4?

@sonic dawn is that a good start?

[sonic dawn]

the existence of -2 makes what you’re doing to lhs invalid

[rotund hawk]

yeah, i was thinking about that

can you give me a hint?

[sonic dawn]

ignore the equality for a moment and just consider the range of each side only

[rotund hawk]

range?

[sonic dawn]

if you recall the range of exponential this should take a few seconds at most

ye, range of 3^(3x)-2 and similarly for rhs

[rotund hawk]

do we simply devide the sides?

[sonic dawn]

i mean you can try but it’s probably not simple for all x

[rotund hawk]

hmm

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[severe vector]

need help

[severe vector]

critical at x=-1 and x=4

[raven sparrow]

It should be right. Maybe it doesn't like the U for union?

[severe vector]

there is another symbol for union but it doesn’t mark it right either :/

[raven sparrow]

That is rather odd.

[severe vector]

.close

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[calm jackal]

How do I use the Math Induction to prove 2 + 6 + 18 + ... + 2(3)^n - 1= 3^n - 1

[calm jackal]

Wait

[opal vault]

oh the parentheses were lacking

3^(n-1)

[calm jackal]

$2 + 6 + 18 + ... + 2(3)^n - 1= 3^n - 1$

Like this

[opal vault]

no that's not right

[calm jackal]

Up bruh

[opal vault]

maybe like this : $2 + 6 + 18 + ... + 2(3)^{n - 1}= 3^n - 1$

[thorny flame]

rafilou2003

[calm jackal]

Yes

Thats the one

got me trapped tp the part I tried proving it since I did not have enough knowledge to prove em equal

I got to the point my answer was

[opal vault]

Well per definition of reasoning by induction, there are 2 steps :

• Base case: prove the result is true for the first n possible, here n=1
• Induction : suppose the result is true for some n>=1, prove it then for n+1

[calm jackal]

$3 + ksqrtk = 3 + ksqrt2$

[thorny flame]

ICANTCOMEUPWITHANAME

[calm jackal]

Now how do I say squareroot

[opal vault]

\sqrt{}

[calm jackal]

Oh ok

$3 + k/sqrt{k}= 3 + k/sqrt{2}$

[thorny flame]

ICANTCOMEUPWITHANAME

[opal vault]

but how is it related to this?

[calm jackal]

I did the 5 steps to prove it

[calm jackal]

But it ended like this instead

[opal vault]

how did a square root ever come up?

please show your work

[calm jackal]

$3 + k\sqrt{k}= 3 + k\sqrt{2}$

[thorny flame]

ICANTCOMEUPWITHANAME

[calm jackal]

I hit my head just to fail

[calm jackal]

Give me a second

[opal vault]

+k+1??

I don't think you understood very clearly what was asked of you*

[calm jackal]

I did what I did

Enlighten me sir

[opal vault]

If you let $A(n) = 2 + 6 + 18 + ... + 2(3)^{n - 1}$, please tell me what is $A(n+1)$

[thorny flame]

rafilou2003

[calm jackal]

No its not a function

Cause you need this for step 4

If I change that $2(3)^k-1$ then it wouldnt be equal

[opal vault]

That is not what I was saying

Look at what you wrote beginning of step 3

[calm jackal]

That k + 1?

n = k + 1?

[opal vault]

are you SURE we ask you to prove in induction that 2 + 6 + ... + 2(3)^(k-1) +k+1 = ...?

[opal vault]

so please, think about this again

[calm jackal]

No

[opal vault]

and YES it is a function, over the positive integers

[calm jackal]

But they taught us this

When they started teaching us this I lost all reasoning once I saw step 3 being taught

[opal vault]

We will go through what went wrong

Suppose that you have a statement defined for all natural integers, named P(n)

[calm jackal]

like n = k on step 2 but n = k + 1 on step 3

[opal vault]

we're getting to that, please listen

[calm jackal]

I am

[opal vault]

Here, you will agree that the statement we want to prove, $P(n)$, is "$2 + 6 + 18 + ... + 2(3)^{n - 1} = 3^n-1$", right?

[thorny flame]

rafilou2003

[calm jackal]

Yes

YEAH

[opal vault]

Ok

So

We suppose P(k) is true

(in induction step)

What is P(k+1), the statement we want to prove?

[calm jackal]

We use step 2 and step 3 to substitute them for step 4 to prove it is equal

[opal vault]

Ok, but then tell me, can you write exactly what P(k+1) is?

[calm jackal]

Wdym?

[opal vault]

Write out exactly what "n=k+1" means for our statement

[opal vault]

Rewrite this, with n = k+1

[calm jackal]

Im not sure if I got the answer for that

[opal vault]

then that's where your problem is

[opal vault]

hint, it's exactly what you said, you have to copy this, and swap every instance of "n" with "k+1"

[calm jackal]

Yeah thats the point but it shows the step to not do it 😭

Then we wouldn't be able to substitute step 2 to step 3 for step 4

[opal vault]

No but my question is

can you write P(k+1)

the statement we WANT to have in the end

just write it

[calm jackal]

Ok

While I write it ill send you how the book did it

[opal vault]

yes so you want to show here that $S_{k+1} = 1 + 4 + ... + \mathbf{(k+1)^2} = \frac{\mathbf{(k+1)}(\mathbf{(k+1)}+1)(2\mathbf{(k+1)}{6}$

every instance of k is swapped with k+1

[thorny flame]

rafilou2003
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[calm jackal]

Yeah but example 3 here didnt wrap with k+1 but added another term

He wrapped it but cloned it instead

[opal vault]

he didn't "add" another terms

S_k is the sum of squares between 1 and k

so S_(k+1) is the sum of squares between 1 and k+1

so compared to S_k, it's like we added another term in the sum which corresponds to (k+1)^2

[calm jackal]

Why did I not realize this 🥲

I thought $S_{k+1}$ was 1 function

[thorny flame]

ICANTCOMEUPWITHANAME

[opal vault]

no, S is the function

k+1 is like the input

[calm jackal]

Oh ok

[opal vault]

So think of S_k as S(k)

[calm jackal]

So where did this come from

[opal vault]

again, this is (in this context) $S_k$ being the sum from 1 to k of all 2m+1

[thorny flame]

rafilou2003

[opal vault]

so $S_{k+1}$ is the sum from 1 to k+1 of those terms

[thorny flame]

rafilou2003

[opal vault]

so there is an additional term in k+1

And 2m+1, when m=k+1, becomes $2(k+1) - 1$

[thorny flame]

rafilou2003

[opal vault]

I was trying to figure out what your book was saying but here S_k is more like a statement for "k"

so S_k is not "the sum" but makes the sum intervene

but the intention is the same

[calm jackal]

So that one I underlined came from that $(2k - 1)$ but substituted $k + 1$ on it again to add more value than than $(2k - 1)$

[thorny flame]

ICANTCOMEUPWITHANAME

[calm jackal]

So that it wouldnt intervene for substitution for $S_{k + 1}$

[thorny flame]

ICANTCOMEUPWITHANAME

[opal vault]

in all the examples you showed for the left hand side, you could always write the sum in "k+1" as the sum in "k" + some additional term

[opal vault]

in this, the additional factor is $(k+1)^2$

[thorny flame]

rafilou2003

[opal vault]

In this, the additional factor is $2(k+1)-1$

[thorny flame]

rafilou2003

[opal vault]

because it corresponded to the last term of the sum up to k+1

[calm jackal]

yes

Its different examples

[opal vault]

and it's precisely what goes wrong with your induction

[opal vault]

because "k+1" is not the last term of the sum in this case

[calm jackal]

So its $2(3)^{k}$

[opal vault]

Everytime, the "additional factor" corresponds to "what happens if the sum goes 1 step further, what do we add?"

[thorny flame]

ICANTCOMEUPWITHANAME

[opal vault]

Yes!!!!!

[calm jackal]

I overlooked something dumb

[calm jackal]

So how do I handle this

[opal vault]

You wrote brackets for the 3

why did you forget about them on the line under?

(hint, I'm trying to say that $2(3)^k$ is VERY different from $(2\cdot 3)^k$)

[thorny flame]

rafilou2003

[calm jackal]

Its supposed to be like that

[opal vault]

what is?

[calm jackal]

This

[opal vault]

yes

[opal vault]

I'm just saying that line 1 is true

line 2 is not

[calm jackal]

Which line

[opal vault]

because

[opal vault]

there are literally 2 lines

line 1 is true

line 2 is not

[calm jackal]

Ok bro I just got confused

OH YEAH 2(3)^k cant be done to 6^k

Unless its (2^k)(3^k)

[calm jackal]

Or that

[opal vault]

yes