help-33

[nova sierra]

which is 0

so the lower bound is 1

and the upper bound is 0

[twin arch]

Ohhh

[nova sierra]

so for the first integral we have: $-\frac{3}{4}\int_{1}^{0} {\frac{1}{u}} d{u}$

[twin arch]

?

[nova sierra]

and the 3/4

[elfin berry]

Adam

[twin arch]

Why is it negative

Erm

[nova sierra]

u = 2 - x

du = -dx

dx = -du

and we can move the negative sign outside of the integral

since it is a constant (-1)

[twin arch]

Ah I see

Okay can I rewrite the question from beginning to make it easier

This will take 5 mins tops

[nova sierra]

of course

don't forget to do the same for the second integral

change the bounds of the integral using the substitution u = 2 + x in that case

[twin arch]

Almost finished

@nova sierra

Ok

@nova sierra

[nova sierra]

Alright

Good progress

now

we have made the substitution, but we have to change the limits

[twin arch]

@nova sierra

What do I do next

[nova sierra]

so another property of integrals

let me write it out for you

$$\int_{b}^{a}{f(x)} d{x} = -\int_{a}^{b}{f(x)} d{x}$$

[elfin berry]

Adam

[nova sierra]

when you switch the limits, you need to add a negative sign

if you want, you can prove this quickly

[twin arch]

How

[nova sierra]

so assume we have the LHS integral

when we evaluate, we get: F(a) - F(b)

where F(x) is just the antiderivative

and if we multiply that by -1

we get F(b) - F(a) ignore this

and the RHS integral, when evaluated, is F(b) - F(a)

and if we multiply this by -1

we get F(a) - F(b), which is equivalent to the LHS

therefore, multiplying a definite integral by -1 will swap the bounds

[twin arch]

Wait

Am i finished

[nova sierra]

yes

that is the singularity point

hence divergence

and that means the sum of those two integrals is also divergent

so the initial integral $\int_{1}^{2}{\frac{3}{4-x^{2}} d{x}}$ is divergent

[elfin berry]

Adam

[twin arch]

What should I write for the answer

Explaining why it’s divergent

[nova sierra]

ln(x) doesn't exist at that point

[twin arch]

Doesn’t exist at singularity point

[nova sierra]

and for an integral to be divergent, it either goes to +- infinity

or doesn't exist at that point

[nova sierra]

yes

[twin arch]

@nova sierra

I have another question for you

[twin arch]

Ty for helping me with this one

What do I do for this

The limit of 2

<@&286206848099549185>

Wait I’ll make a new ticket

Thanks again @nova sierra for helping me with that one

[marsh citrus]

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[upper sigil]

I'm trying to solve this differential equation:
P(x) is going to be$ \frac{-2(x^{2}+x)}{x+1}$, right?

[upper sigil]

which is just
$\frac{-2x}{x+1}$

[elfin berry]

Tomi

[upper sigil]

this is not correct

[marsh citrus]

@upper sigil Has your question been resolved?

[upper sigil]

I end up with this equation at the end, does it seem correct $I(x) being {e^{x^2-2x}}$
$\frac{1}{e^{x^2-2x}} [\int e^{x^2 - 2x} \frac{e^x{^2}}{x + 1} dx + c]$

[elfin berry]

Tomi

[upper sigil]

<@&286206848099549185>

[marsh citrus]

@upper sigil Has your question been resolved?

[marsh citrus]

@upper sigil Has your question been resolved?

[marsh citrus]

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[drifting wyvern]

Can someone explain to me what happened

I know its inverse

So I would type in sin-1(-2/5)

But It'd give me a smaller answer

and if I add 180 its not correct

(Number 17)

[proud basin]

your probably using radians

wait

[drifting wyvern]

ohh, Im supposed to minus 180

wait

but then wouldn't it be less than 180 or whatever

[proud basin]

erm wtf

[cobalt sedge]

203.6 seems right to me

But inverse is defined for sine in the domain -90 to 90

so arcsin gets you in 4th quadrant

you have to use suitable modifications on theta to get an equivalant angle in the 3rd quadrant that has the same value of sine

[drifting wyvern]

Wait, I realized theres a rule for this

[drifting wyvern]

Could you explain a bit to me

I know different quadrants mean different things

but I don't quite have the full rule of it

[cobalt sedge]

multiple values of angles can have same value in sine function

So you might haave heard of the identity sin(x) = cos(90-x)

there are multiple such identities, and one of them applies to your case. Find the right one

[marsh citrus]

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[iron meteor]

hello, how can i solve this equation?

[digital adder]

What's the b in the left

[eager tinsel]

that's t

[iron meteor]

its a t

i have to solve for t

[cloud field]

you probably have to solve for this with numerical methods

[iron meteor]

b is just the letter of the question

[digital adder]

Alr nvm

[iron meteor]

this is the whole question

how do i solve for t like i cant think of any other way

[digital adder]

Are you sure that the equation you provided earlier is correct?

If not then I'm gonna go through the actual question I guess

[cloud field]

have you learnt numerical methods before?

[iron meteor]

no we haven’t learnt it yet

how did you find t is 1?

[cloud field]

its not 1

[digital adder]

My bad

[cloud field]

then maybe graphical

[iron meteor]

yeah probably

i’ll try to solve it graphically

thanks for the help

.close

[marsh citrus]

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[marsh citrus]

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[still temple]

74627
-3193

71434

[still temple]

Method (s) = take 9 from 2 does not work. From 9 to 12 is 3. The dissolved hundredths place is not taken away from the minuend but is added to the subtrahend. This gives the same result from 200 to 600 is 400.

[proud basin]

what 😭

[lyric bay]

Do we have context?

[proud basin]

i genuenly think hes asking for basic subtraction lmao

[still temple]

im trying to understand

basic subtraction yes

[proud basin]

wtf 😭

[proud basin]

err can you rewrite this

"dissolved" "minuend" "subtrahend"

[still temple]

they give method (t) =
take 9 from 2 does not work; one hundredth is dissolved into 10 tens places, and now: take 9 from 12 = 3. Next, instead of 6 hundreds there are only 5 hundreds left and take 100 from 500 gives you 400

this is the common way i think everyone is taught

im trying to understand the intuition of method (s)

minuend is the first number

[proud basin]

hm ok

[still temple]

subtrahend is the number subtracted from the minuend

7 - 3 = 4
7 is minuend
3 is subtrahend

dissolved is when you cross out the 6 in 600

and add 10 to 20

making it 12 - 9

74627
-3193

71434

ins this problem

[proud basin]

so lets focus on the 62 and 19 part
62-19 would be 60+2-10-9, 2-9 wouldnt be positive, so we split 10 from 60 to get
50+10+2-10-9
50+12-10-9
50+3-10
40+3
43

[lyric bay]

Is this some kind of abstract math or discrete math thing

[still temple]

its just elementary math

teaching how natural numbers work

[still temple]

this is method (t)

[proud basin]

wtf 😭

[still temple]

you are dissolving 1 tens place and adding it to the ones place

but in method (s) it says the dissolved tens place is not taken from the minuend

[proud basin]

i have no clue what your talking about man

[still temple]

but it is added to the subtrahend

the second number

[proud basin]

sorry, youll have to wait for someone else

[still temple]

wait sorry i understand now

i made a mistake

thank you for being patient with me sorry

.close

[marsh citrus]

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[still temple]

.reopen

[marsh citrus]

✅

[still temple]

have a different question

[digital adder]

What is it

[still temple]

can you explain what this is saying

.close

[marsh citrus]

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[marsh citrus]

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[formal blaze]

can someone help me with where to start

[calm harbor]

which question?

[formal blaze]

just the first one

or like what do i do to start any of them really

[digital adder]

I wish I didnt miss probability lectures in my last session because this question looks so interesting

[formal blaze]

its all good

[digital adder]

its not good for me lol

[formal blaze]

damn im still just tryna do this for highschool

[digital adder]

wait highschool?

do you have the answer key

[formal blaze]

ye no working out tho

[digital adder]

oh man

[marsh citrus]

@formal blaze Has your question been resolved?

[marsh citrus]

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[lapis basalt]

Im kinda stuck in this trigonometry problem. I think I somehow got to take advantage of sin2x denominator but i cant really work with the big fraction on the left side of the equation

[sharp sonnet]

you should be able to do something with the sin^4 +cos^4

you could aslo rewrite the cot(2x) with sin and cos

[marsh citrus]

@lapis basalt Has your question been resolved?

[digital adder]

nvm let them do it

[sharp sonnet]

yeah

[lapis basalt]

mb

[marsh citrus]

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[celest palm]

For d is the answer the limit from 0 to 4 ((sqrt y) + 4) - (y - 2) dy

[marsh citrus]

@celest palm Has your question been resolved?

[unreal geode]

I unforfunately have no idea what does with respect to x mean

[vast isle]

It just means the variable here is x

Like we vary the x values

[unreal geode]

Then create to integrals with the following limits:

-2 to 2
2 to 4

[celest palm]

to get the area wouldnt the limits need to be the range since were going right to left

with respect to x

-2 to 4 would be used in c

[vast isle]

Usually it's lower --> higher, so left to right for x, down to up for y

[vast isle]

No, the -2 to 4 is the x-values, so that would be for d

Identify which is the x-axis, which is the y-axis

[celest palm]

would u be able to set it up?

[celest palm]

https://www.youtube.com/watch?v=eYwfHEfreQ0

its not like this?

actually this is more for the disk method

[unreal geode]

$\int_{-2}^{2} \text{formula of the line} ,dx + \int_{2}^{4} \text{formula of the parabola} ,dx$

[elfin berry]

MetuMortis

[celest palm]

ok this makes sense thank you

.close

[marsh citrus]

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[true pulsar]

find the value of √x +1/√x if the vlaue of x = 9+4√5

[cyan peak]

Plug in and solve, what now?

[true pulsar]

how?

[marsh citrus]

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[cyan peak]

√(9+4√5)=√a+√b
<=> a+b+2√(ab)=9+4√5
√(ab)=2√5
ab=20
a+b=9
From guesswork a=4,b=5 works

Thus-
√x=2+√5
Thus thus-
1/√x=1/(2+√5)=(2-√5)/(-1)=√5-2
Thus thus thus-
√x+1/√x=2√5
qed
Might be slightly wrong

[cyan peak]

I was blurting that the question isn't clear, it's not trivial to simplify

[marsh citrus]

@cyan peak Has your question been resolved?

[cyan peak]

@true pulsar ifyw

[marsh citrus]

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[near burrow]

can someone explain this

[whole quiver]

the property?

[digital adder]

what do you need explained..

[whole quiver]

wait

[still temple]

$Log_5 5^8$

[elfin berry]

Sukiyaki

[still temple]

LOL ya

[whole quiver]

source: mathisfun

[still temple]

$log_a(a^x) = x log_a(a)$

[elfin berry]

Sukiyaki

[still temple]

the Log_a(a) will cancel

If you learn natural logs, $Ln(e) = 1$

[elfin berry]

Sukiyaki

[near burrow]

how does the exponet of a log equal the function

[whole quiver]

see if you can connect

[whole quiver]

or with this

[sinful thistle]

that's how logarithms are defined

logs invert exponentiation

like division inverts multiplication

so generally, a logarithm for base a (as in the base of the exponent) of another number, b, is written as log_a b

and this number is basically what power "a" must be raised to (exponentiated) to get "b" as a result

basically, let's say that a^x = b

and we want to find x

well, x is defined as log_a b

@near burrow hope this helps

[near burrow]

.close

[marsh citrus]

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[sinful thistle]

uh
no response?

okay

[marsh citrus]

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[vital sequoia]

i need help to find x

[marsh citrus]

Please don't occupy multiple help channels.

[vital sequoia]

.close

[marsh citrus]

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[marsh citrus]

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[inner grove]

srsly?

[jagged oracle]

Just multiply

,w 8*17

[elfin berry]

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Requested by oppailol.

[jagged oracle]

Still simple

8*17

[digital adder]

wait guys

[inner grove]

8 * (10 + 7) = 80 + 8*7 = 80+56

[jagged oracle]

8*(10+7)

Yes

[digital adder]

were you not in his previous tickets?

[inner grove]

you just wanted to occupy channel I've been writing a question, didn't you?

[jagged oracle]

No

I came here just now

[inner grove]

yes.

[jagged oracle]

Is he a troller

Or a and seeker

[inner grove]

probably

[digital adder]

he has been asking questions like HUGE NUMBER + HUGE NUMBER and HUGE NUMBER - HUGE NUMBER

[marsh citrus]

Channel closed

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[eager tinsel]

@jagged oracle OP is suspected to be a troller

[jagged oracle]

Fr

I k this trick

People ask fricking calculations

[digital adder]

its his 3rd ticket under my watch

could be more

[jagged oracle]

To the point where people give answers

This helps people without calculators

[digital adder]

he said youtube is IP grabber

[jagged oracle]

And wolfram knowledge

Lol

[digital adder]

and you were never tought how to carry?

in addition and subtraction

[jagged oracle]

Everything is a IP grabber

Even DC has your ip

[digital adder]

respectfully, thats wild

[jagged oracle]

Fr

[digital adder]

🫡

[jagged oracle]

[digital adder]

im out.

[jagged oracle]

Me too

[marsh citrus]

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[lyric bay]

Hello

[lyric bay]

Why isn't the sign for f_1 + f_4 being flipped?

[cloud field]

why should they

f1 + f4 = 40 is the same as 40 = f1 + f4

[lyric bay]

But they are inputs that are flowing in

[cloud field]

?

[lyric bay]

Okay I understand

Thanks

.close

[marsh citrus]

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[storm swift]

hey can someone help with how to go about b

[limber hearth]

So the difference between the maximum and the minimum

[storm swift]

yeah

so find the y vales fir maximum and minimum

ok, i dont have the x values for them tho

[limber hearth]

Its in terms of what ?

In terms of c ?

[storm swift]

yeah,

[limber hearth]

Then keep it in terms of c

Can you show what you have done in a) ?

[storm swift]

ok so, how do i get the y values then?

yeah sure

its a bit messy

i tried to find stationary point by derivative first but it became a mess so did it a different way

thats just the summary of it, there is pages of working

workes out ther was three roots

and a cubic has max of 2 stationary points

the gradient at the y intercetp is negative always

therefore a max and min

[limber hearth]

Ima try it by derivative

[storm swift]

we can also look at what happens when x goes to negative and positive infinity which confirms it goes in opposite directions

[limber hearth]

E'(x) = -(4-3x^2) -6x(c - 1/c -x)

[storm swift]

yeah and i tried to put that equal to zero

i put c=z becuse it looked better

but i just thought after doing this must be an easier way idk cos i wasnt sure what i was doing was right

[limber hearth]

It is the way to go, it gives x to stationnary

[storm swift]

ok so is what i have done so far ok then?

[limber hearth]

Use product rule maybe instead of developping

[storm swift]

is 3/2x=c correct?

or should i have got something else?

[limber hearth]

Try it, let x = 2/3c

[storm swift]

okie

[limber hearth]

And see what you get for b)

[storm swift]

cool

[limber hearth]

Wait shouldnt there be two sol ?

Since there is two stationnary points

[storm swift]

hm yeah

good point

[limber hearth]

,w derivative of (c -1/c -x)(4-3x^2)

[elfin berry]

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Requested by yakubros

[storm swift]

im just not sure how i get x from that

[limber hearth]

,w derivative of (c -1/c -x)(4-3x^2) = 0

[elfin berry]

Results provided by WolframAlpha

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Requested by yakubros

[storm swift]

ahhhh

i close to that lol

[limber hearth]

That leads to yours answer too

But there is the second one

[storm swift]

yeah

[limber hearth]

So all good ?

[storm swift]

yeah

i just would never of seen that factorisation

[limber hearth]

That is a hard one tho

[marsh citrus]

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

[storm swift]

.close

[marsh citrus]

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[proper widget]

hello, could anyone help me with this, recently i study 'factoring higher degree polynomial' but encountering a question

x⁶-1

(x²)³-(1)³
a³-b³ = (a-b)(a²+ab+b²)

(x²-1)( (x²)² + x²×1 + 1² )

(x²-1)(x⁴+x²+1)

x²-1 is factor of (x-1)(x+1) because:
(x)²-(1)² is a²-b²
a²-b² = (a-b)(a+b)

and (x⁴+x²+1) is factor of:

(x²)²+2x²-x²+(1)²

using comutative:

(x²)²+2x²+(1)²-x²

using a²+2ab+b² = (a+b)²

( (x²)²+2×x²×1+1² ) - (x)² = (x²+1)² - (x)²

(x²+1)²-(x)²

using a²-b² = (a-b)(a+b)

(x²+1-x)(x²+1+x)

my question is, at the time this was found:

a³-b³ = (a-b)(a²+ab+b²)
a²+2ab+b² = (a+b)²
a²-b² = (a-b)(a+b)

is it found by bruteforcing(multiply), for example:

(a+b)×(a²+ab+b²) = a³+2a²b+2ab²+b³
(a-b)×(a²-ab-b²) = a³-2a²b+b³
(a²+b)×(a+ab+b²) = a³+a³b+a²b²+ab+ab²+b³
(a²+b²)×(a+ab+b) = a³+a³b+a²b+ab²+ab³+b³

so... on

and saving up the important one, like:

(a+b)×(a+b) or (a+b)² = a²+2ab+b²
(a-b)×(a+b) = a²-b²
(a-b)² = a²-2ab+b²

(a+b)×(a²-ab+b²) = a³+b³
(a-b)×(a²+ab+b²) = a³-b³

or, there is method (step by step method) for converting:

a³+b³ into -> (a+b)(a²-ab+b²)

I hope someone help me, Thank you so much for your help.

[severe owl]

(a³+b³) = (a+b)³-3ab(a+b) = [(a+b){(a+b)²-3ab)}] = (a+b)(a²-ab+b²)

[marsh citrus]

@proper widget Has your question been resolved?

[proper widget]

hello @severe owl, thank you so much for the help but could you please breakdown the answer a little bit. and how about a³-b³. my apologies🙏

[marsh citrus]

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[calm harbor]

!status

[marsh citrus]

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

[gilded ember]

1

idk where to start

am i supposed to just sub the x values in and that gives the area

[cobalt sedge]

y values

but yes

this is just the basic definition from the riemann integrals' limit

[lyric bay]

Is this using just Riemann's sum

[cobalt sedge]

technically riemann's sum is using the function values at the center of the interval

this is using the maximum values of the function within the intervals

which is the upper limit of riemann sum/integral

[gilded ember]

so would i have to integrate first

[cobalt sedge]

no no

only multiply calculate the areas of rectangles as they asked you to do

with standard formula for area = width * height

[marsh citrus]

@gilded ember Has your question been resolved?

[marsh citrus]

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[proper widget]

hello @severe owl , i try to solve a³-b³, here my result. please correct me if need correction

a³-b³

so the founder need greater value than a³-b³ so he choose

(a-b)³

since a³-3a²b+3ab²-b³ > a³-b³

the founder need to get rid of
-3a²b+3ab²

so the result would be
a³-b³

(a-b)³ + 3ab(a-b)

(a-b) out

(a-b) ( (a-b)²+3ab )

(a-b) ( a²-2ab+b²+3ab )

(a-b) (a²+ab+b²)

this is my result, any correction is greatly appreciated. thanks once again @severe owl for helping me 🙂

[knotty trellis]

(a-b)^3 is not greater than a^3 - b^3 in general

(2-1)^3 = 1
2^3 - 1^3 = 7
1 is not greater than 7

[gilded ember]

so does the y=2x^2 even matter

[knotty trellis]

You will need that to calculate height of the rectangles

[gilded ember]

how does that give the height tho

[knotty trellis]

Can you open another channel btw? I'll explain it there

[gilded ember]

okay

[proper widget]

here i close it, thnks btw

.close

[marsh citrus]

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[gilded ember]

ive never seen a question with dx in the middle if that makes sense

[gilded ember]

i have no idea how to approach this question

[rough smelt]

it is the same

[tough ibex]

1/f(x) dx = dx/f(x)

[rough smelt]

dx/sqrt(2+x) is the same as 1/sqrt(2+x) dx

[gilded ember]

oh okay

how do i antiderive that tho

should i put (2-x) in the numerator and make it ^-1/2

[tacit trail]

You mean 2 + x?

[heavy grail]

Just use u-substitution.

[gilded ember]

i never learnt that method

theres another way of doing this is ther enot

[tacit trail]

You can see (2 + x) just as a shift

$$\int_{a}^{b} f(x) dx= \int_{a - s}^{b - s} f(x + s) dx$$

[elfin berry]

Pluton

[marsh citrus]

@gilded ember Has your question been resolved?

[marsh citrus]

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[lucid turret]

can it be said that:

The rate of the rate of change of the displacement of a particle moving in a circular path uniformly, changes as cos(x) in a graph changes?

[lucid turret]

rate of rate of change of displacment = rate of change in cos(x)

reference is taken as the initial point P.

the rate of rate of change in displacement is negatie

negative

Okay os

so

the slope in negative is increasing

So I assune cos(x) changes as sin(x) changes but in the negative direction.

so -sin(x)

Is my deduction correct?

ding dong

<@&286206848099549185>

KINDly write your answer here> I will come here after 10 minute exercize.

that is also related to the unit circle where hypotenuse = 1 (let)
and ddrawing a perpendicular gives us a referenec. for measurement.

therefore, to trig ratios.

helpers?

I am back helpers.

<@&286206848099549185>

Can I ping at a period of 15 minutes?

[marsh citrus]

@lucid turret Has your question been resolved?

[velvet cedar]

I don't understand your question tbh. Does displacement refer to a particular component of the particle's displacement?

[proud ice]

!noans

[marsh citrus]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[lucid turret]

I use words vaguely

[lucid turret]

displacment of this sort from C where the other points are the position of the object at different intervels.

[proud ice]

What is rate of change in respect to?

[lucid turret]

time

[proud ice]

So how do you define uniform movement in a circular path? The direction is always changing

[lucid turret]

uNiform motiton = uniform speed in a circular path

[proud ice]

And finally, what do you mean by "changes as cosx changes in a graph"

[lucid turret]

As the slope of cos(x) changes.

I mean

Yeah

[proud ice]

[still temple]

what you wanna say is; velocity is a function of cosx

[lucid turret]

wdym?
that the v-t graph would be cos(x)?

[proud ice]

Here's your problem

[still temple]

depends upon the frame of reference too

[proud ice]

You're describing motion in two dimensions

[lucid turret]

the reference point is point C

in the figure

see

[proud ice]

But are only describing a single output variable

So your question is not making sense

[lucid turret]

add time

[proud ice]

output variable I said

[still temple]

@lucid turret its a 2-d motion
not like an SHM where you can see sin curve in it

[lucid turret]

yes?

[proud ice]

How are you describing "displacement"

[lucid turret]

...

it's not angular displacment that is linear

I mean the graph

[proud ice]

Do you mean distance?

[lucid turret]

No

[proud ice]

Or do you mean two dimensional displacement?

[lucid turret]

Distance is the lenght of the path covered by an object

That would be the circumference

[lucid turret]

yes, displacement, i.e., the straight line distance from point C at given position on the circumference at different times.

[proud ice]

I meant your length like CD, CB, CE, etc

[lucid turret]

yes

[proud ice]

That's what I meant by "distance". Not circumference

Okay

[still temple]

i see, i get it what you are tryin to say

[proud ice]

I understand your question now

[lucid turret]

👍

[still temple]

so if we try to plot a graph of x-t taking C as origin ( assuming the particle rotates in anti-clockwise direction) the graph would rise until the point which is diametrically opposite to C; then the graph would go down to 0, again same proccess
I dont see a cos,sin fucntion here; as it never goes below the time axis
its is an oscillatory motion

@lucid turret

[proud ice]

I can give you a thorough explanation when I'm at my PC in a few minutes

[lucid turret]

It goes.

HERE:

When you start to move, you face the east direction.
So, after half the circle, you have a negative displacement.

Not at the center yo uface.

The direction is tangent.

Okay.

[proud ice]

How do you get a negative distance?

[lucid turret]

Idk

I didn't mention distnace.

[proud ice]

Here's you mentioning distance

[still temple]

bruh

[lucid turret]

I just defined displacment as the straight line distance from C to other points That distance changes as a period. But the actual distance, which is the path, never decreases.

[proud ice]

What is "the path"?

What is the "actual distance"

[lucid turret]

that was a little unclear sorry.

PAth = actual distance
Path or actual distance I am talking about is the lenght of the path that the object covers.

as it's a uniform circular motion scenario

Here it can clear:

The actual distance after one cycle = 2 * pi * r

but the displacment (wrt C) is 0.

after n cycles, the actual distance is n(2pi*r)

but displacment (if final position is C) is 0.

[still temple]

well, true displacement would be zero; and for that thing to happen thr curve must go below the time axis
it might be a sinusoidal wave, but we are unclear right now if it really is a cos fucntion. It can have straight lines too goin up and down

[lucid turret]

it is a sinusoidal wave; that's confirmed.
I am unsure about the order.

Straight lines? No

[still temple]

how are you sure about it ?

[lucid turret]

um

sorry for lateness

https://www.quora.com/How-do-I-plot-a-velocity-time-graph-of-uniform-circular-motion#:~:text=In uniform circular motion%2C the,of body moving in circle.

I can myself with 100% confirmity say that it would NOT be linear.

[still temple]

lemme google now

[lucid turret]

wa

https://www.quora.com/How-is-the-displacement-time-graph-and-velocity-time-graph-of-a-uniform-circular-motion

[proud ice]

@lucid turret do you want d(t) like I've done it here?
https://www.geogebra.org/calculator/satqtkbq

[lucid turret]

Here

what is d(t)?

displacement * time?

[proud ice]

The length of the line segment AC, with respect to time

[still temple]

thats what i am thinking too

[lucid turret]

YES. That's the displacment I am talking about.

Woah, you that was the animation I wanted.

[proud ice]

okay you need to stop using the word "displacement"

because that's a vector term

[lucid turret]

ok

lenght AC

[proud ice]

You want the "distance from C" as your particle is moving in a circular path at uniform speed

[lucid turret]

yes

[proud ice]

okay, what do you know so far about uniform circular motion, or vectors in general?

[lucid turret]

I don't know about vectors. I know what are vector quantities.

ABOUT UNIFORM MOTION:

1. Uniform speed.
2. Magnitude of veclotiy stays the same.
3. The direction changes of course!
4. The change in displacment is not linear.

NOTHING MORE

4. when plotted in a graph

vectors are in 11th

class

[proud ice]

Okay. There will be quite a bit to cover

[still temple]

fr

[lucid turret]

fast fast as I need to go 🚤

I will respond fast

[proud ice]

Sadly, you need vectors to solve this problem. At the very minimum, you need a decent understanding of coordinate geometry

short version is that the position of $A$ with respect to time can be expressed as $\vec{A}(t)=(r\cos(\omega t-\theta_0), r\sin(\omega t-\theta_0))$

[elfin berry]

SWR

[still temple]

oh yeah i forgot this

[lucid turret]

there are many vairables

variables*

[still temple]

no

maybe you arent familiar with those

[lucid turret]

uh uh

[still temple]

angular velocity ?

[lucid turret]

oh

I know it

[proud ice]

$\omega$ is the constant angular speed, $r$ is the radius, and $\theta_0$ is the initial angle of $A$ at time $t=0$

[elfin berry]

SWR

[proud ice]

Oh, it should be $\vec{A}(t)=(r\cos(\omega t+\theta_0), r\sin(\omega t+\theta_0))$

[elfin berry]

SWR

[proud ice]

Plus, not minus

[still temple]

its just phase, wont matter much
anyways

[proud ice]

The position of $C$ is unchanging: $\vec{C}(t)=(0, -r)$

[elfin berry]

SWR

[proud ice]

So the distance is just $$d(t)=\norm{\vec{A}(t)-\vec{C}(t)}=\sqrt{(r\cos(\omega t+\theta_0))^2+(r\sin(\omega t-\theta_0)+r)^2}$$

[elfin berry]

SWR

[lucid turret]

☑️

[still temple]

according to me its quite difficult to figure out the actual curve

[proud ice]

"curve"?

[still temple]

x-t curve

[proud ice]

x?

[lucid turret]

x = displacement

[still temple]

displacement

[lucid turret]

s = distance

[proud ice]

,w plot y=sqrt(cos(x)^2+(sin(x)+1)^2)

[elfin berry]

Results provided by WolframAlpha

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

Requested by spelwerdzrite

[still temple]

we need to check behavior of dy/dx

oh well

not really a cos function, i thought that at the start but was not sure

wait

mhm

[lucid turret]

I am really sorry

I have to go now

my parents are calling me

Sorry

I may come tomorrow again

[proud ice]

[lucid turret]

THanks

.close

[marsh citrus]

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[still temple]

this thing solves it

thanks to SWR

omega is constant too

i think there would be some transformation depending upon the radius of circle and velocity ?

@proud ice

[proud ice]

what?

[still temple]

w =v/r

nvm mybad

bye

[marsh citrus]

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[reef topaz]

How do I do this, I thought it would be 40/60

[cobalt sedge]

How many ppl are ther in the group who are older than 10 yrs

in all three groups A, B and C combined

[reef topaz]

180

[cobalt sedge]

not 180

180 are all of them

[reef topaz]

Oh 60

[cobalt sedge]

No not 60

think again

[reef topaz]

40/60 in group A?

[cobalt sedge]

no no no

Among all 3 groups, you have how many 10+ yo ppl

[reef topaz]

oh

120

[cobalt sedge]

we check for group A after we select them

yes 120

and of those 120 how many are in group A?

[reef topaz]

40

[cobalt sedge]

yes

so that makes it...

[reef topaz]

oh 1/3

[cobalt sedge]

Yep

[reef topaz]

I thought it said in group A itself

how many were 10+

[cobalt sedge]

This is conditional

so you have to consider the base condition first

and then the group A

[reef topaz]

yeah I see, it means across all groups and then compare it to group A

alr ty

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[drifting leaf]

hi, need some help understanding how to get the bounds for the integrations

[sharp sonnet]

you know that its spherical coordinates which is good since you know what you should look for

actually i don't think im good enough to help with this lmao kinda crammed my exam for that

sorry

[drifting leaf]

rip all good

[sharp sonnet]

i checked my notes and i found this

thats for the radius

but you have 2 equations

so i'm not sure how to do it really

[azure temple]

Wikipedia has generalizations to convert from Cartesian to spherical but I'm not sure how much that helps

[sharp sonnet]

i got these too they can help if you put them in the equation since you can find phi and theta

disclaimer! for me phi is the spin angle and theta is the angle from the z axis downwards

these are not the correct names for them but i think that makes sense

[grave badge]

The bounds of integration are going to come from the reparameterization of the balls. Instead of x, y, z, you'll have some intervals for theta, phi, and r.

[marsh citrus]

@drifting leaf Has your question been resolved?

[drifting leaf]

yeah im not entirely sure how to get those intervals tho

these three specifically

[sharp sonnet]

maybe try to plug in the values of x,y,z from the spherical coordinates

[drifting leaf]

hm i dont have expllicit values of x yz tho

just the two spheres

[hot sequoia]

8 cos phi comes directly from spherical coordinates of the first sphere

[drifting leaf]

orz

[hot sequoia]

Like if you rewrite it you’ll get rho = 8cos(phi)

Bottom bound is radius of the other circle since like

You want to be outside it

[drifting leaf]

oh really

hm lemme

see

[hot sequoia]

Yes

[drifting leaf]

ok i see

that is nice

[hot sequoia]

Arccos comes from

Um

Drawing a picture/imagine in ur head

Basically phi is the angle starting from +z and moving clockwise (at least I think that’s how it’s defined)

So in this problem you go from 0( the x axis) down to the intersection of the spheres

[drifting leaf]

mhm

[hot sequoia]

If you can imagine the angle, the adjacent side is 32/5 and the hypotenuse is 16/5

Hypotenuse is radius of small sphere, 32/5 is the z value where the two spheres intersect

[drifting leaf]

sorry how do we find where the two spheres intersect

[hot sequoia]

Expand the top one out and substitute 256/25 for x^2 + y^2 + z^2

Do you understand why the equation of the big sphere in spherical is also equal to the bound for rho?

[drifting leaf]

i get 256/25=8z so 32/25?

[hot sequoia]

Yep

[drifting leaf]

yeah i think so

[hot sequoia]

So the intersection is a circle on the z=32/25 plane

[drifting leaf]

yeah

ok i think i got it

thank you

orz

.close

[marsh citrus]

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[hot sequoia]

Np

[marsh citrus]

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[steady mural]

is there any way to estimate an answer of a cumulative frequency curve without having to draw it

[steady mural]

I mean thing is im not so good at drawing

So i want to somehow get the correct answers i need then draw the curve by marking it

Basically i want to cheat it in a way yes

I have software that can draw the curve and find points feom it

That would mean theres a specific way to find the answer through calculation right?

Does anyone know?

[marsh citrus]

@steady mural Has your question been resolved?

[steady mural]

<@&286206848099549185>

[vital oracle]

post the data

also, "the answer of a cumulative frequency curve" is vague, it just means the frequency curve is the answer youre looking for
you should say something more specific like "I need to find out the number 60% of the data lies below" or "I need to find out how much of the data lies between here and there"

[steady mural]

Ye that's what i meant sorry

[vital oracle]

I said two things there, do you mean both of them?

[steady mural]

Ye

If i can learn how to find one point i can also find the other no?

[vital oracle]

if youre doing things on paper, youre better off not doing that

you also havent posted the data

[steady mural]

For eg

[vital oracle]

no no

you need to tell me in advance whether the data is discrete or continuous

[steady mural]

This one for eg

[vital oracle]

are you going to ask about doing this for continuous distributions?

or only for discrete data only?

[steady mural]

Ye but im asking is there any alternative through pure calculation so that i can confirm my answer rather then drawing

[steady mural]

That woukd help as well but lets do this one first

I believe if i can do this one i can figure out the other one myself

Thanks for asking though 🙏🏼

[vital oracle]

,rccw

[vital oracle]

@steady mural what did you get for (b)? I havent done a question like this before

[steady mural]

least is 0 and greatest is 28

[vital oracle]

thats correct

[vital oracle]

now when I said these two things up here, I only did mean specifically those two things, they each have a simple pen-and-paper method of calculating if youre given a table (or histogram) of the data

[steady mural]

how can you teach me

[vital oracle]

Im still typing, what do you mean by that

[steady mural]

the pen and paper method

howd you do it

[vital oracle]

hold on there

Im not done typing yet

the "example question" you gave though is different, because it gives a cumulative table instead, and it also asks a different kind of question for (b)

if you wanted to do it through pen and paper, it would require a unique method just for (b)

[steady mural]

actually lets go with this

its simpler

how would you do c. i.

using pen and paper

[vital oracle]

from the info youve written down so far,

you know that 16 books have a width less than 19cm

and then after that are 25 books that are from 20cm to 29cm

now you need to find the amount of books that are < 27cm

[vital oracle]

now, the boundary here is better read as "20cm to 29.9999cm"

so really its "20cm to 30cm" or more accurately "width has a 2 in the tens place"

can you see how the boundary has to be read that way?

[marsh citrus]

@steady mural Has your question been resolved?

[steady mural]

1 sec

[steady mural]

ye

then ?

[vital oracle]

so there are 25 books that are 20cm-30cm

from that, we're looking for the # of books that are 20cm-27cm

(once we get that number, we can add it to the number of books that are < 20cm which is 16
to get the total number of books that are < 27cm)

[steady mural]

yes

[vital oracle]

so this is now the problem we have to figure out
do you understand that

[steady mural]

ye

that is where the issue lies

[vital oracle]

now usually this question isnt possible

[steady mural]

cause in drawing the curve shapes in or out

by calculation its tough

[vital oracle]

oh thats not where we're taking it

if you purposefully drew a curve instead of a histogram or straight lines, youre doing something that only a graph can do

thats not necessary though here

usually this question isnt possible, because you dont know the shape at all, you guessed it

its not because you drew the curve

the information they tell you is only enough to make a histogram

it cannot make a smooth curve for you to take areas of

do you understand that

[steady mural]

i guess

[vital oracle]

@steady mural so the issue here can only be answered if we estimate on how the data behaves

[steady mural]

yes

[vital oracle]

now on pen-and-paper, we make the following assumption

the 25 books are uniformly distributed from 20cm to 30cm

so that should mean 70% of 25 books should be from 20cm to 27cm

after rounding, we essentially take a guess that 18 books are from 20cm to 27cm

you see how this guess works?

[steady mural]

thing is

i tried this method

it doesnt work cause you said

we cant estimate how the data behaves

[vital oracle]

waiting for you to realize why we got to do this

[steady mural]

do you know of any method to find how the data behaves without the curve

[steady mural]

yes i know

i did this

[vital oracle]

because?

[steady mural]

wdym because

you think the data is equally distributed like in a histogram

but its not

its a cumulative curve so ye

we assume it

continue

[vital oracle]

wrong reason

before we get to that, do you think graphing the curve would let you figure this problem out easier?

or does the problem that we dont know the distribution of the data block out any way of guessing the 20cm-27cm book number, graphing or otherwise?

[steady mural]

This ig

So your sayng graphing is also an assumption?

[vital oracle]

its a bigger assumption

you guessed how the data was distributed

its "more" of an assumption than assuming the data was uniformly distributed

either way we had to guess

do you agree?

[steady mural]

Yes

[vital oracle]

now the real reason we made an assumption is because the question explicitly told us to do so:

that is the only reason

it has nothing to do with anything else

what we estimate the distribution as is another question

here, we have some leeway, which is where I can show that graphing and not graphing can give the same estimation

lets say you graph the data like this

the data here is made to match the table, but then to draw straight lines from one point to the other

[steady mural]

Yes

[vital oracle]

when assuming the books are uniformly distributed inside each interval, this is the graph that results from doing that

so the question of "is it fair to assume this" = the question of "is it fair to estimate the graph by drawing straight lines"

either way, you will get 17.5 books that are 20cm to 27cm

which rounds to 18 books

[steady mural]

Ok

[vital oracle]

now with this, we have a way of finding out the cdf from one value to another

this method will answer "how much of the data is between this and that"

now, using the same data, I can show you a second method that instead answers "what data point is at the 60th percentile" or "what number is 60% of the data below"

are you with me on that

[steady mural]

Ye sure

But one thing

[vital oracle]

yea

[steady mural]

Can you summarize your point for what you said previously

Like we assume it to draw lines then our answer should be close enough?

So just think of it as equal distribution?

[vital oracle]

the reasoning behind drawing lines doesnt have a reason beyond "its the simplest guess"

the data we're given only says this

so graphing or not, stuff will be made up to estimate

so the simplest guess is to drawing lines = assuming uniform distribution

[steady mural]

Again you mean this right?

[vital oracle]

oh right, yes

[steady mural]

Aight understood sensei

[vital oracle]

lol

[steady mural]

We can continue

[vital oracle]

lets say that we need to figure out the length where < 70% of the books fall into

a length that 49 books fall into

when you graph, this consists of drawing a line at y = 49 books

from here, it seems that 33cm or 34cm should work according to the graph

does this line up with the way you do it

(we're just showing the graphing way for context)

[past atlas]

unrelated and sorry for interrupting but what is the name of that graphing tool?

[vital oracle]

https://desmos.com/calculator

[past atlas]

thank you

[vital oracle]

@steady mural so yes or no to this?

[marsh citrus]

@steady mural Has your question been resolved?

[marsh citrus]

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[still vector]

Just checking because I have a quiz tomorrow, piecewise functions are just graphing like normal then getting rid of the side of each line that isn’t in the domain right

[late geode]

you can view it like that

[marsh citrus]

@still vector Has your question been resolved?

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[small stream]

Let the function ( f(x) ) have a continuous second derivative. If the curve ( y = f(x) ) passes through the point ((0,0)) and is tangent to the curve ( y = \ln x ) at the point ((e,1)), then evaluate ( \int_0^e x f''(x) , dx = ?)

[elfin berry]

riyobi

[latent coral]

if y = f(x) is tangent to lnx at (e,1), then what's its slope?

that's f'

[small stream]

1/e

[latent coral]

right

so f'(e) = 1/e

set up IBP for the integral

differentiate x and integrate f''

[small stream]

$$\int_0^e x f''(x) dx = \left. x f'(x) \right|_0^e - \int_0^e f'(x) dx$$
$$\int_0^e x f''(x) dx = e f'(e) - f(e) + f(0)$$

[latent coral]

try dollar signs

right

[elfin berry]

riyobi

[latent coral]

right

and you know f'(e), f(0) was given, and it shares the same value at x = e as lnx

[small stream]

1-1+0

Tyyy

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[marsh citrus]

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[wet crystal]

I don't understand where the number "48" is coming from or how to get it, that is all I need

[late geode]

if he pays every 3 months,
how many times does he pay over one year?

[wet crystal]

Would that not be 36

[late geode]

how are you getting 36

[wet crystal]

Sorry nevermind I figured it out, I'm very tired thank you

.close

[marsh citrus]

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[marsh citrus]

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[marsh citrus]

Please don't occupy multiple help channels.

[thin cove]

is there any technique on how to solve

decimal exponents

[granite knot]

no you need a calculator under all practical/realistic circumstances

[proud basin]

not really

[thin cove]

so its js estimate?

in context of no calculator exams nd all multiple choice

[marsh citrus]

@thin cove Has your question been resolved?

[latent coral]

as long as x is small, (1 + x)^n is approximately 1 + nx

for example, doing that approximation here gives 1 + 12 * 0.03 = 1.36 rather than 1.425

x has to be really small for it to be accurate but this is the best you can do without manually multiplying by hand or with a calculator

doing a binomial expansion and taking only the first three terms gives roughly 1.42 as well

[marsh citrus]

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[burnt cosmos]

This is what I have done

please ignore the first step

[marsh citrus]

@burnt cosmos Has your question been resolved?

[marsh citrus]

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[idle vessel]

https://artofproblemsolving.com/wiki/index.php/2024_AIME_II_Problems/Problem_11

[idle vessel]

What does a/b/c=100 mean? Is this legit?

[cobalt sedge]

its just or sign

[sand fable]

it means a or b or c is 100 and yes it is legit

[cobalt sedge]

from the secondline you see one of the factors has to be 0

so its either a or b or c thats 0

[idle vessel]

Isn't it ambiguous?

[sand fable]

which part

[crystal lintel]

i've never seen / mean or myself

[idle vessel]

If that's used like a/b=1 for example

[sand fable]

i mean no not really it's pretty obvious what it means here

[sand fable]

yep sometimes aops solutions writers post really weird things

[sand fable]

in general, yes using informal notation in this way should be avoided, but aside from the initial shock, i don't think writing "a/b/c = 100" is going to confuse anyone

[crystal lintel]

counterexample: M_

[sand fable]

ah you got me

[idle vessel]

.close

[marsh citrus]

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[novel juniper]

For question 28, is this answer right? ${ \frac{3n}{4}| n \in \Z}$

[elfin berry]

(why am i here )= idk

[knotty trellis]

Looks right

[novel juniper]

Thanks

.close

[marsh citrus]

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[muted narwhal]

how do you figure out the 4 attempt probability?

[muted narwhal]

I understand you need a situation where he passes 2 and fails 1 to pass it in the 2nd attmept but how do you consider all the possibilities where he could fail in 1st, 2nd, or third?

[marsh citrus]

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[marsh citrus]

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[hardy prairie]

just a simple question i can't understand the second step

[whole thorn]

do you know about double angle identities

[hardy prairie]

no

[whole thorn]

cos(2t) = cos^2(t) - sin^2(t)

use that

[hardy prairie]

okay

[acoustic bramble]

sin²(t)+cos²(t)=1 is also useful

[marsh citrus]

@hardy prairie Has your question been resolved?

[marsh citrus]

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[still temple]

what is a conjection