help-49

[vague seal]

yes

[sullen bolt]

pls wait til i get the integral

(3 mins)

[last slate]

Factorize the terms.

(x-1)(x-2)(x-3)

[last slate]

The denominator is this.

And now use partial fractions.

[sullen bolt]

I know lmao

just figuring out the top part of the partial fractions

[last slate]

Bruh.

[last slate]

Yes.

[sullen bolt]

should I just matrix this lol

[red cloak]

4th one

[sullen bolt]

!occupied

[last slate]

Ahh, blackbook.

[midnight plank]

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

[red cloak]

Black book

[red cloak]

Turu

[orchid fossil]

u a jee guy?

[sullen bolt]

is the result perhaps
$\frac{\ln(x-1)}{2}-\ln(x-2)+\frac{ln(x-3)}{2}$

[last slate]

Yes.

[grand pond]

laestia

[sullen bolt]

unless I'm dumb

[last slate]

Lemme calculate.

[sullen bolt]

the pfd results is 1/2(x-1)-1/(x-2)+1/2(x-3)

[last slate]

Uhh, i gor both the 2's on the numerator.

OH WAIT.

No.

On the denominator.

[last slate]

Yeah.

[sullen bolt]

so ts is correct right

[last slate]

Yes.

[red cloak]

It's this one is basic qn of integration

Just some simplification

And it's done

[last slate]

Yes.

[red cloak]

Can you solve mine @sullen bolt

[sullen bolt]

!occupied

[midnight plank]

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

[sullen bolt]

if you want help, open a new help channel

[hexed barn]

#help-21｜아리스킨충1 is currently open by the way

[red cloak]

I sent it there

[sullen bolt]

.solved

[midnight plank]

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[last slate]

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[tame osprey]

is there any other way to do this without assuming ϴ = 90°

[last slate]

AM>=GM inequality.

[fathom onyx]

Note that cosec = 1/sin, solve the quadratic in sine-theta, and evaluate the desired value

[fathom onyx]

This would also work, assuming you know what that inequality is

[civic gazelle]

only applies when the terms are non-negative

[fathom onyx]

that too

[tame osprey]

oh that's smart lol

.solved

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[silk forge]

x + 1/x always greater than or equal to 2

[civic gazelle]

x = -1:

[silk forge]

yes for positive numbers

[civic gazelle]

is sin x always positive then

[silk forge]

for negative it is -2 to - inf

[last slate]

When sin is +ve, cosin is +ve.

So you can kinda use it.

[silk forge]

i mean

its obv

that u wont get 2

if sin takes a negative value

so i didnt bother

[last slate]

hi

[last slate]

u got doubt in this?

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[last slate]

oh nvm

[last slate]

...

[graceful drum]

you can close the channel with .close

[frigid gorge]

lol

[last slate]

.close

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[meager cypress]

Whatt

[meager cypress]

Close

[rich yew]

.close

[fathom onyx]

.close

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[shut canyon]

a guess is that maybe if the two curves intersect f(x)=x at the same two points something is …

[midnight plank]

@shut canyon Has your question been resolved?

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@shut canyon Has your question been resolved?

[shut canyon]

h(x) = α₁ x + 1/2.

h ∘ Qc = Fμ ∘ h

.close

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[leaden seal]

someone help me differentiate this i keep getting it wrong

[mystic condor]

what are the symbols?

is it theta and phi?

[leaden seal]

yes

and theres also alpha

[fathom onyx]

And you aim to differentiate with respect to...?

[fathom onyx]

What have you tried, if anything?

[leaden seal]

and thats about all

[sharp coral]

@vast citrus this channel is for helping zah with her geometry homework. please keep it on topic

[fathom onyx]

Wrong place, and I could swear this has been done multiple times.

[fathom onyx]

Can you spot this as a product?

Wait hold on, not even that

You're differentiating with respect to PHI, right?

[leaden seal]

yeah and so the other terms are constant

[fathom onyx]

Then the whole part before the cosine-reciprocal thing, relative to PHI, is just a constant, yh

[leaden seal]

right?

thanks

[leaden seal]

why is it just a constant?

[fathom onyx]

Because nothing there depends on phi

[merry shale]

cant u diff tan(a-0)

😒

[vast citrus]

Wrong server sorry

[fathom onyx]

And where here does the tangent function even appear?

[merry shale]

cant you combine sin and cos

[fathom onyx]

No?

[merry shale]

cus cos is to the power of negative one

so it becomes denominator

[fathom onyx]

That'd only apply if the arguments of both the sine and cosine matched

But they DON'T match

[merry shale]

oh damn i didnt read

mb

[fathom onyx]

But this, essentially

[fathom onyx]

I don't quite agree that this is the derivative of that section btw

[leaden seal]

oh

[fathom onyx]

The Chain Rule applies here too, so you have to multiply this result by the derivative of cos(phi - theta) [with respect to phi]

[leaden seal]

yeah

[fathom onyx]

So the outer function (the -1 power), that derivative was fine

The inner function is cos(phi -theta)

What's the deriv. of cosine?

[leaden seal]

-sine

[fathom onyx]

ye

So what's the d/dphi of cosine(phi - theta)?

[fathom onyx]

@leaden seal

[leaden seal]

so -sin(phi - theta)

[fathom onyx]

yee

Now multiply that with that ^-2 thing and the coefficient of that original cosine (that is, the w sin() thing)

[midnight plank]

@leaden seal Has your question been resolved?

[leaden seal]

so we use power rule for this part

.close

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[midnight plank]

@tidal turret Has your question been resolved?

[tidal turret]

i think i can simplify a bit more the second system

[midnight plank]

@tidal turret Has your question been resolved?

[midnight plank]

@tidal turret Has your question been resolved?

[midnight plank]

@tidal turret Has your question been resolved?

[signal ibex]

did you learn lagrangian?

i am not sure how much mileage ill get with the explanation but seeing the problem being up this long, can't hurt to try

start by finding all candidate optimizers (gradient = 0, which translates into 3x^2 - 2y = 0)

then check each boundary seperately

the harder one is $1/2 x^2 + 1/4 y^2 = 1$

[grand pond]

king of complex analysis

[signal ibex]

thats what i get for not using \frac

anyway

isolate for y^2

$y^2 = 4(1 - \frac{1}{2}x^2$

$y^2 = 4(1 - \frac{1}{2}x^2)$

[grand pond]

king of complex analysis

[signal ibex]

doing it this way, easier algebra

sub that in and find extrema along that boundary by taking the derivative with respect to x (you eliminated y)

then repeat with the other boundary

[tidal turret]

how

[tidal turret]

\textbf{2.} Find, if they exist, the absolute extrema of
[
f(x,y) = x^{3} - y^{2}
]
restricted to the region
[
A = {(x,y)\in \mathbb{R}^{2} : \tfrac{1}{2}x^{2} + \tfrac{1}{4}y^{2} \le 1,; y \ge \sqrt{2},x }.
]

[grand pond]

Renato

[worthy hill]

draw.

[frank wolf]

Hey Renato

[tidal turret]

hey

[frank wolf]

Still having trouble with the problem?

[tidal turret]

yes

[queen herald]

Okay, you got x^2=1. Hence x = 1 or -1, right? Then you can sub x into the first 2 equations and solve for lambda1 and lambda2

You learnt how to solve systems of equations, right?

[midnight plank]

@tidal turret Has your question been resolved?

[gilded viper]

@tidal turret é brasilieiro?

[signal ibex]

the original equation has y^2

like, the thing to find the extrema of

[signal ibex]

replace y^2 with the other half of this

easier to do it this way because there is no square root business, isolating x is harder

[midnight plank]

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[full grail]

hi can someone help idk how to integrate this

[radiant roost]

if you do integration by parts, can you get the sqrt into the denominator?

oh i thought that would make it look like the derivative of arcsin, but the sign is wrong

i think your simplification went wrong somewhere along the way

yeah i think the u-sub is not working

[native swift]

may put u = cos t
with Bioche rules

not -cos t

[wet pollen]

Eliminate the root with another substitution.

5u^2 = 9tan^2(v)

[radiant roost]

you didn't change the limits when you did the u-sub

[full grail]

yea I was fonna change everything back

Look

Bruh this is disguting

[radiant roost]

i think you can substitute u = t - pi

on the fifth line

then you get an odd function

[raven oak]

cant u rearrange cos double angle identity to replace cos^2

nvm i am blind

[wet pollen]

What's the original question?

[radiant roost]

,align S&=4\pi\int_0^{2\pi}\sin(t)\sqrt{9-5\cos^2(t)}dt\
&=4\pi\int_{-\pi}^{\pi}\sin(u+\pi)\sqrt{9-5\cos^2(u+\pi)}du\
&=-4\pi\int_{-\pi}^{\pi}\sin(u)\sqrt{9-5\cos^2(u)}du\
&= 0

[grand pond]

Axe

[midnight plank]

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[mint ravine]

please help

[mint ravine]

is there a smarter way than just doing casework for billy's position

@help

helper

[radiant roost]

I think you can make an argument for 1/3 based on symmetry

Like you can partition the space of orders based on the equivalence relation that permutes the triplets

[mint ravine]

can you give an example i don't really understand what you mean by that

[radiant roost]

Ex one part in the partition might be
138AB276C45
138AC276B45
138BA276C45
138BC276A45
138CA276B45
138CB276A45

The probability here is 2/6, and the same goes for every other part

[mint ravine]

oh ok that makes sense

thanks

.solvee

.close

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[stable halo]

yea cane someone do this for me i aint tryna do all that

[fallow scarab]

!noans

[midnight plank]

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.

[stable halo]

i know i know

i just think its stupid to set such a qwuestionb

.close

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[fallow scarab]

Also stupid to share just to complain

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[uncut maple]

$\int \frac{1}{1+\cos{x}}$

[grand pond]

felipe

[uncut maple]

I know u can substitute t = tan(x/2) or multiply the equation by 1-cosx/1-cosx

My question isn't solving, but why can you multiply it by 1-cosx/1-cosx? How can you guarantee that cosx isnt 1? In that case you would be multiplying by 0/0

prob a stupid question but i have to ask it lol

[polar mortar]

when you do this you generally (implicitly) assume you're excluding values where you divide by 0

[uncut maple]

oh, I thought about that but wondered if there was something else

Alr, appreaciate it zyb

Also @polar mortar could u help on a similar question?

[polar mortar]

sure

[uncut maple]

the best way for you to understand my question would be for you to solve it first

its a simple one, but it'll make more sense if u do it

[polar mortar]

isn't this a trig sub?

[uncut maple]

Yep, exactly that, but theres something bugging me

Well if just say x = sec(theta) and solve it, u'll get arcsec(x)

[polar mortar]

sounds about right, yeah.

[uncut maple]

I supposed 0 < theta < pi/2 U pi < theta < 3pi/2 so we have a inverse function, great

In this case it didnt say arcsec(|x|) because it already defined x > 1 implicitly

But why the restriction to the positive, why not simply arcsec(x) but rather arcsec(|x|)

Another example from a similar trig sub. In this case 0 < theta < pi/2 U pi < theta < 3pi/2

[polar mortar]

without having solved it, I'm going to guess they get a sqrt(x^2) in there which is |x|. But, as you said, you already need x>1, so the absolute value is redundant.

[uncut maple]

yeah the absolute value is redundant because it already considered x > 1

,w derivate arcsec(x)

[grand pond]

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[uncut maple]

Which would make sense, the |x|

But is there anyway I could figure that out without deriving my final answer?

[polar mortar]

you should be able to by looking at the domain and range of arcsec(x), that's where I'd start anyway.

[uncut maple]

well 0 < theta < pi/2 U pi < theta < 3pi/2

range x < -1 or x > 1 in this case

maybe because the function is odd?

[polar mortar]

maybe. I'm not sure. because now that I think about it, you don't need x>1, you need x>1 or x<-1 for the integral to be defined. Which is what secant does anyway. I'm not sure why they used |x| because it doesn't seem necessary.

[uncut maple]

Yeah, that's what im saying, but if you do int from -2 to -1 using arcsec(x) instead of arcsec(|x|) u get it wrong

,w arcsec(-1) - arcsec(-2)

[grand pond]

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[uncut maple]

,w arcsec(1) - arcsec(2)

[grand pond]

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[uncut maple]

-pi/3 would be the actual value of it

[polar mortar]

yeah, that is a weird little quirk.

[uncut maple]

im not crazy thankfully

[polar mortar]

this happens sometimes in stewarts book. It's likely a domain issue, or something like that.

[uncut maple]

Yeah but its not wrong

I just don't know why arcsec(|x|)

oh actually mb, you have to do the improper int to -1 bc it has a infinite discontinuity

but the idea still remains, like

-3 to -2

[polar mortar]

oh. It's being wildly pedantic.
when you sub in you get
$$\frac{\sec\theta \tan\theta}{\sec\theta \sqrt{\sec^2\theta - 1}} = \frac{\sec\theta \tan\theta}{\sec\theta |\tan\theta|}$$

[grand pond]

Zybikron

[polar mortar]

Which is +-1, the derivative of |theta|... maybe

[uncut maple]

yeah, since 0 < theta < pi/2 U pi < theta < 3pi/2, tan(theta) is always positive

one of the benefits of defining sec(x) for (0, pi/2) U (pi, 3pi/2)

I thought myb

idk honestly

if u use u = sqrt{x^2 - 1} you get in terms of arctan

,w \int \frac{1}{x\sqrt{x^2-1}}

[grand pond]

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[uncut maple]

then the problem wouldnt exist but

yeah

[midnight plank]

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[midnight plank]

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[twilit field]

Maybe I'm just too tired, but why isn't this 60

[twilit field]

it's a+<15> basically,right

[runic hamlet]

sure

what are the elements of 7+<15>?

[wispy mortar]

dont we have a direct formula for this

[twilit field]

{22, 37, 52,7}

oh right

These are basically translates

so 4 it is

[twilit field]

yup, number of cosets

[runic hamlet]

yes but no

correct brackets please

you arent first year

these mistakes are embarrassing

[runic hamlet]

what are the 4 elements of the group?

[twilit field]

{22, 37, 52,7}, {16,31,46,1},{15,30,45,0},{2,17,32,47}

I just realised there can be more

The order of <15> is 4, And o(Z_60)=60

so we have 15 cosets

[runic hamlet]

yes

[twilit field]

cool

thanks

.close

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[floral ruin]

guys I wanted to know if there is a method where I can solve this using calculus.

[floral ruin]

like somehow can i find out the points on both the functions where the slops are same?

[potent knoll]

Yes, you're able to

[woeful turret]

hmm if u equate the slopes ull only get one equation tho

[potent knoll]

Differentiate the curves and find the general equation

[woeful turret]

calculus is a bad approach for this question..just use standard form of tangent for both these curves

[floral ruin]

ohh

[floral ruin]

ohh i see. Was just curious if that would work better but thanks!

[slow thorn]

you could use line equation in parametric form if you know that

$yt = x + at^2$ for a tangent passing through $(at^2, 2at)$ for a parabola $y^2 = 4ax$

[floral ruin]

i solved it using y=mx+a/m and then applied distance formula from the centre of the circle

[slow thorn]

yeah that too

[floral ruin]

yeaa

What about this one?

I drew a rough sketch of the graph

[potent knoll]

What exactly are you confused about on this?

[floral ruin]

the approach

how should I tackle this one?

should I just assume parametric coordinates for P and Q and then continue with the algebra of centroid?

[potent knoll]

In general?

[floral ruin]

yea or like any method

[potent knoll]

You can find the focus of the parabola and set up the centroid eq

[floral ruin]

yea

[potent knoll]

Then you should be able to relate them through sum of coordinates to the line eq

[potent knoll]

Yep

[floral ruin]

alright lemme try

done

thanks!

.close

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[onyx tide]

Can anyone check this before I proceed?

[bold peak]

Yup looks good

[onyx tide]

100% sure?

[bold peak]

As much as I can reasonably be

[onyx tide]

Fine then

[clever sedge]

Just put everything in one line, as it's clearly possible 🙃

As in x^4 + x^6 could easily be on the third line together with the rest of the terms

Makes reading much easier

[midnight plank]

@onyx tide Has your question been resolved?

[onyx tide]

.close

[midnight plank]

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[lyric charm]

holy shit the font sizes 😭😭

[woeful turret]

i was thinking since the function is strictly increasing, f(5x)/f(x) =1 when x-->infinity so ans = 0

but is it really meant to be that easy?

or am i missing something

[lyric charm]

well you have $1 \leq \frac{f(5x)}{f(x)} \leq \frac{f(7x)}{f(x)}$

[grand pond]

Ann

[woeful turret]

hm why will it be >=1

oh nvm

ohh ok got it

thanks

.close

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[tall crag]

In the game of Two-Finger Morra, 2 players show
1 or 2 fingers and simultaneously guess the number
of fingers their opponent will show. If only one of
the players guesses correctly, he wins an amount(in dollars) equal to the sum of the fingers shown
by him and his opponent. If both players guess
correctly or if neither guesses correctly, then no
money is exchanged. Consider a specified player,
and denote by X the amount of money he wins in
a single game of Two-Finger Morra.
(a) If each player acts independently of the other,
and if each player makes his choice of the
number of fingers he will hold up and the
number he will guess that his opponent will
hold up in such a way that each of the 4 pos-
sibilities is equally likely, what are the possi-
ble values of X and what are their associated
probabilities

[tall crag]

i didnt understand how the game works

[obsidian anvil]

essentially each player has two tasks

pick a number, either 1 or 2

and then guess the number your opponent picks, either 1 or two

does it make sense how there are 4 outcomes?

[tall crag]

ye

[obsidian anvil]

so what would each outcome be

[tall crag]

{1,2}x{1,2}

[obsidian anvil]

yes

[tall crag]

mhm

[obsidian anvil]

now we know that each possibility is equally likely right

because they choose independently of one another

[tall crag]

yes

[obsidian anvil]

i think u get it u just needed the game itself to be further elaborated

[tall crag]

yeah thanks bro

[obsidian anvil]

i’m assuming part b asks for expected value

[tall crag]

no

[obsidian anvil]

oh

well just remember that X is the amount of money you can win

[tall crag]

Suppose that each player acts independently
of the other. If each player decides to hold up
the same number of fingers that he guesses his
opponent will hold up, and if each player is
equally likely to hold up 1 or 2 fingers, what
are the possible values of X and their associ-
ated probabilities?

[obsidian anvil]

so what are the possible values for X

[tall crag]

X = 0 if both are either correct or incorrect and X = i + j for i fingers up from the first person and j by the sec

[tall crag]

only if the player guess correct tho

[obsidian anvil]

yes but it might help to think about what values X can take on

[tall crag]

that'd be smth like 16?

[obsidian anvil]

because ultimately you’re finding P(X=0), P(X = 7) etc

[obsidian anvil]

wdym

think about the outcomes where X ≠ 0

[tall crag]

X = 2, 4, 3

[obsidian anvil]

i don’t know if X can = 2

[tall crag]

why not?

[obsidian anvil]

oh wait can u hold up no fingers

no u can’t

[tall crag]

nope

[obsidian anvil]

I lied x can be 2

if you hold up 1 and ur opponent holds up 1

[tall crag]

yep

[obsidian anvil]

yes so x can be 0, 2, 3, 4

so now we have to find out the prob. of each outcome

the way i might do it is figure out how many diff ways you can get each value or X, sum them, and then do naive probability

easiest to start top down

how many ways can X = 4

[tall crag]

one

for all

its symmetric

[obsidian anvil]

yup

X = 3?

[tall crag]

for X = 2, 4, 3

all 1

wait

[obsidian anvil]

arent there multiple ways to win 3 dollars

[tall crag]

X=3 is 2

i think

[obsidian anvil]

i think so too

and by similar logic, x=2 is also 2 outcomes

[tall crag]

hm ye

no

its 1

[obsidian anvil]

yes

my bad

[tall crag]

lol

[obsidian anvil]

bc outcomes are 11, 12, 21, 22

[tall crag]

yes

2 wait

so for

[obsidian anvil]

2, 3, 3, 4

[tall crag]

X=0

we have how many ways?

[obsidian anvil]

let me think

it’s smth like all the ways where your guess ≠ your opponents hand or where they are equal

bc it’s 0 if either both guess incorrectly or both guess correctly right

[tall crag]

P(X=0) = 1 - sum(P(X=i)) , i = 2,3,4 right?

[obsidian anvil]

yup

[tall crag]

so what is the whole outcomes?

[obsidian anvil]

that’s why we’re trying to find |x=0|

[tall crag]

look

we should take one guess and bring either 1 or 2 fingers

right?

each player can do it tho

so its just |{guess,finger}|^2

[obsidian anvil]

wouldn’t it just be the probability that your guess is not equal to your opponents hand

[obsidian anvil]

yes i think

[tall crag]

so its like

we bring 2 fingers up either 1 or 2 and take 2 guessues

so the cartesian would be 4^2

= 16 ways

now we have probs for X = 2,3,4

so sum of them is 3 * 1/16

1 - 3/16 = 13/16

@obsidian anvil i think we got part a

oh wait

its 4/16

i forgot X = 3 has. outcomes

[obsidian anvil]

that’s for x=1

and then the same for all the other ones, just find the prob

[midnight plank]

@tall crag Has your question been resolved?

[tall crag]

X=1????

[obsidian anvil]

x=0 has p 12/16

[tall crag]

Hm

[obsidian anvil]

the w asks for each value of x and its probability

[obsidian anvil]

isn’t that what u just solved lol

[tall crag]

Didnt we say it for X=0

[obsidian anvil]

oh right i mistyped

x cant = 1

[tall crag]

Yep

[tall crag]

b part?

[tall crag]

What would it be

[midnight plank]

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[pale osprey]

Very simple and silly question I know but

I did 3/2 x 160 and got 240 ml for this but the answer is 540ml idk what I missed

[lusty python]

you dare steal my name

[lusty python]

anyways

[outer saffron]

Not only the height but the base also increases in size so its isnt just a 3/2 multiple

[gaunt jetty]

the volume doesnt scale by a linear factor

it scales by a factor of k^3

[pale osprey]

How would I do that again? I’m helping my sibling out and literally forgot

💀

[lethal path]

volume ratio = (length ratio)^3

[lusty python]

goodness

[gaunt jetty]

just cube the scaling factor you found

[pale osprey]

Same with this I got 1080g but the answer is 750g

[lusty python]

let the volume be $l$

[grand pond]

1 divided by 0 equals Infinity

[pale osprey]

Ah ok

[lusty python]

so we got an equation $\frac{14}{160} = \frac{21}{l}$

[grand pond]

1 divided by 0 equals Infinity

[lusty python]

simplest way to understand

[outer saffron]

The box with 900 grams is the bigger box they are asking you to find the original size

[wooden badger]

or you could do longer by actually making equations of volumes

[lusty python]

like dis

[pale osprey]

Omg

🤦‍♀️

That was mb

[outer saffron]

Nw

[wooden badger]

wht r u even doing

[lusty python]

isn't that forming the equation?

[wooden badger]

tht would give 240...

she did the same i believe

[pale osprey]

I got 360ml by doing this which is wrong

The answer is 540

[outer saffron]

No like you got 3/2, cube that and multiply it to 160

[gaunt jetty]

,calc 160*3.375

[grand pond]

Result:

540

[pale osprey]

Bruh I got 720g when it’s 750

[outer saffron]

Try again i got 750

[wooden badger]

calculation mistake maybe

[pale osprey]

Oh nvm

[outer saffron]

Yeah just recalculate both the answeres and you should get it

[wooden badger]

this would be the longer method

ignore my camera quality i prob need a new phone

[lofty knoll]

1.20 * standard box = 900 so special is 750

[outer saffron]

Lets form the equation properly ,we have 120 percent of a variable Xwhich gives 900,meaning (120/100)×X=900

[pale osprey]

120%=900

100%= (20% of 900) - 900

720

[outer saffron]

Nope that isnt how it works

Percentagws dont work that way

[pale osprey]

Other way around srry

[lusty python]

okay

let $x$ be such that $120% \cdot x = 900$

[grand pond]

1 divided by 0 equals Infinity

[lusty python]

so if you want to find $100%$, then go find $x$ yourself

[grand pond]

1 divided by 0 equals Infinity

[lusty python]

@pale osprey

that's the trick for you to remember how

[pale osprey]

Thank u very much!

[midnight plank]

@pale osprey Has your question been resolved?

[midnight plank]

Channel closed

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[wicked mauve]

i have the curl

[wicked mauve]

i dont know what do to since its a cube

[unreal finch]

what did you get as the curl

[wicked mauve]

<1,0,-x>

there

[unreal finch]

shoot, ive just had my multivariable exam, stokes theorm can indeed be tedious

you do know how to solve a surface integral tho right?

[wicked mauve]

😂

i have even more troubles with it lol

ever since we started going over line and surface integrals ive lost an understanding in what we're doing in class

[unreal finch]

ah well in our case youd have to verify the line integral and the corresponding greens theorm equivalent as proof, it messed me up

the thing about stokes is that you just need to "choose" a convenient surface

[wicked mauve]

good golly

😭

i am cooked

idk what to choose for the surface here

and idk how to get ru and rv

[unreal finch]

well theyve specified the faces use unit outward normal

[wicked mauve]

i really have no clue what that means

my prof doesnt use the normal vector

well explicitly

[unreal finch]

me neither, i rly dont wanna confuse you with this further because of how difficult for me it is too 😭

but good hting im done with that sem

[wicked mauve]

my final is on saturday

and there is gonna be an frq on stokes surely

but the thing is the homework is so much more difficult

so i have no clue how to do this

the textbook doesnt help and my prof only does problems from the textbook lol

[thorn crown]

Since it's telling you to use Stokes's theorem should you be trying to calculate a certain line integral?

[unreal finch]

honestly stokes becomes pretty easy if youd just know how to solve a surface integral

[unreal finch]

yeah i feel you

[wicked mauve]

im trying to do a surface but idk how 😭

[thorn crown]

What do you mean you're trying to do a surface?

[wicked mauve]

isnt what what stokes is

a line integral into a surface

[thorn crown]

It can also be viewed as a surface integral into a line integral

[wicked mauve]

oh yeah

the reverse

so what would i do

[thorn crown]

Stokes's theorem doesn't directly hold here but you can show that Stokes's equation will be true nonetheless for this problem by considering each face of the cube separately and you will see certain line integrals (remembering the orientation rules of the theorem) cancel such that you can calculate it as the line integral around the "base" on the xy-plane

I'm not too sure how familiar you are with Stokes's theorem so I tried to include extra details

[wicked mauve]

so it is a line integral of a square

[thorn crown]

Yes

My quick calculations had three of the four line integals being zeroes

[wicked mauve]

what would i do to do them line integrals

oh

nvm

[thorn crown]

You would calculate them like you normally would

[wicked mauve]

okay let me try

[wicked mauve]

what would i parameterize it as?

[thorn crown]

What is "it"? I assume you are referring the boundary of the square on the xy-plane described by 0<= x <= 1, 0 <= y <= 1

[wicked mauve]

well tbh im not really sure myself

[thorn crown]

Ok.

[wicked mauve]

but in my notes i have

hang on

[thorn crown]

Ok

I will try and use paint to draw a graph so you can see precisely what I'm trying to describe

[wicked mauve]

this

or is this not the same

that is equal to the line integral of F dot dr

[thorn crown]

It is equal to that if the curve is "nice" yes

[wicked mauve]

i am still stuck

[thorn crown]

My diagram has gone a bit crazy

[wicked mauve]

its okay

[thorn crown]

It's not got the back faces stuff but this is the idea I was trying to say with words and this is why you end up with only the line integral over the base as all the other cancel

[wicked mauve]

okay i understand that part

what a nice cube

[thorn crown]

So we want to calculate the line integral over this which I have split into four

[wicked mauve]

can i do any line integral?

[thorn crown]

Yes we will need to do all four over these four curves

We desire

[wicked mauve]

or it has to be

[thorn crown]

$\int_{c_1} xy \dd{x} - z \dd{y} + ... + \int_{c_4} xy \dd{x} - z \dd{y}$

[grand pond]

stabulo

[wicked mauve]

oh okay

so

[thorn crown]

We will need to come up with four paramaterisations for each c_i

[wicked mauve]

greens theorem

no

yes?

[wicked mauve]

how

[thorn crown]

You could actually, I never thought of that option

For c_1:

x needs to go from 0 to 1 in a linear way

The obvious choice would be x = t

We can see from the graph that y = 0 over the entire segment c_1

c_1: x = t, y = 0, 0 <= t <= 1

This is the easiest one to do but you should try to extrapolate this logic to c_2, ..., c_4

[wicked mauve]

so for c2 x=0 and y=t?

[thorn crown]

Close

[wicked mauve]

oh

[thorn crown]

But x = 1 over the entirety of c_2

[wicked mauve]

x=1

[thorn crown]

Yes

I'll give you this trick ahead of time

But say if instead we needed to go from x = 0 to x = 2 it's best to use x = 2t so we can keep is having 0 <= t <= 1

This is because the eventual integral we end up with is almost always easier if the bounds are 0 to 1

[wicked mauve]

ah okay

that makes sense

[thorn crown]

So far we have
c_1: x = t, y = 0, 0 <= t <= 1
c_2: x =1, y = t, 0<= t <= 1

Just two more

[wicked mauve]

and then c3: x=-t y=1

and c4: x=0, y=-t ?

[thorn crown]

Are you using the same 0 < = t <= 1?

[wicked mauve]

yes

[thorn crown]

If so c_3 looks wrong

and c_4

[wicked mauve]

oh

😩

dang

[thorn crown]

This is because you will, for c_3, start at x = 0 and end at x = -1

You want to start at 1 and end at 0

x = ?

[wicked mauve]

oh so x=t ?

y=1

[thorn crown]

No

x = 1 - t

This starts at x = 1 and ends at x = 0

[wicked mauve]

ohhhhh

okay that makes sense

😭

damn

[thorn crown]

This one is more of a thing where you need to see it at least once before to be fair

c_4 will have a similar paramaterisation trick

[wicked mauve]

okay then c4: x=0, y=1-t

[thorn crown]

Yes

So we have
c_1: x = t, y = 0, 0 <= t <= 1
c_2: x =1, y = t, 0<= t <= 1
c_3: x = 1 - t, y = 1, 0 <= t <= 1
c_4: x = 0, y = 1 - t, 0 <= t <= 1

[thorn crown]

To will use this formula for each here

[wicked mauve]

is the z irrelevant here?

[thorn crown]

You should see many are zeroes quickly especially since z = 0 everywhere

[wicked mauve]

okay

[thorn crown]

z = 0 over the entire curve so really you can just get rid of it

[wicked mauve]

gotcha

thats basicallyyyy what i meant lol

[thorn crown]

$\int_{c1} xy \dd{x} + ... + \int_{c_4} xy \dd{x}$

[wicked mauve]

okay let me try

[grand pond]

stabulo

[wicked mauve]

would it be wrong to make them all 1 integral?

since its addition

now im confused

[thorn crown]

x will have different values in each of them

You could eventually I suppose when everything is in terms of t

But you will see quickly the line integrals over three of the four are zeroes fast

Either because x, y or dx = 0

[wicked mauve]

can you help me with c1

[thorn crown]

Yes

[wicked mauve]

okay thank you

[thorn crown]

c_1: x = t, y = 0, 0 <= t <= 1

[wicked mauve]

i just plug them in?

[thorn crown]

Essentially, yes.

[wicked mauve]

ohhh okay

[thorn crown]

x = t, y = 0 and dx = dt

[wicked mauve]

so 0

[thorn crown]

Yes

[wicked mauve]

c4 is 0 also

[thorn crown]

I agree with that

The only non-zero one I find is the line integral over c_2

[wicked mauve]

is the answer 1?

[thorn crown]

I don't think so but why do you?

[wicked mauve]

for why isnt c3 nonzero

[wicked mauve]

because i got c3

[thorn crown]

c_3 is zero I think

[wicked mauve]

the integral would be t- 1/2 t^(2)

no?

[thorn crown]

Yes you're right

I mixed up c_2 and c_3

But I don't get the value to be 1 for this integral

I get -1/2

[wicked mauve]

let me show what i did

hang on

[thorn crown]

Ok 🙂

[wicked mauve]

thats for c3

[thorn crown]

dx = - dt is the difference since x = 1 - t

So the actual line integral will equal - 1/2

[wicked mauve]

how did you get dt though

[thorn crown]

If x = 1 - t then dx = (dx/dt) dt but dx/dt = -1

[wicked mauve]

ohh

okay i see

yes

so

then i get 0?

overall

[thorn crown]

No

What do you get for c_4?

[wicked mauve]

0

for c2 i gto 1/2

[thorn crown]

So why would the sum of 0, 0, - 1/2 and 0 be 0?

[wicked mauve]

i got 0 + 1/2 -1/2 + 0

[thorn crown]

dx = 0 dt for c_2

Since x = 1

[wicked mauve]

oh

okay

i see now

so overall -1/2 is the answer then

[thorn crown]

Yes I think so

[wicked mauve]

you're a genius

okay can you help me with another one?

[thorn crown]

I've been studying this stuff more carefully lately is all

[thorn crown]

I don't know

[wicked mauve]

ill show you and mayhaps you can

[thorn crown]

First try applying what you've learnt here

[wicked mauve]

i will try

i am just confused by the wording maybe

ok hang on

[wicked mauve]

okay so far i've only done part B

woudl they be the same ?

[thorn crown]

Probably not the easiest problem if new

[wicked mauve]

answer

[thorn crown]

What did you get?

[wicked mauve]

36pi

[thorn crown]

I get 63 pi

[wicked mauve]

they would be the same no?

[thorn crown]

Yes both answers should be the same as you are verifying the theorem for a particular problem

[wicked mauve]

makes sense

which one did you do

A?

[thorn crown]

I did the surface integral

[wicked mauve]

well i dont want the problem to reset so ill show my work

hang on

[thorn crown]

Ok

I get 63 pi for the line integral path too so not sure what's making the difference

The surface integral path is much more tricky

[wicked mauve]

hopefully its legible

[thorn crown]

I get a different curl

I get <2, 4, 7>

[wicked mauve]

hm

okay hang on

fuck

[thorn crown]

That error propagates but is a quick fix

[wicked mauve]

i wrote 4x instead of 7x

😭

[thorn crown]

Happens 🙂

[wicked mauve]

what is even society

okay well

i will try the line integral now

[thorn crown]

Once that is fixed you should get 63 pi

[wicked mauve]

i am having troubles with the line integral

does it have to be in t?

oh wait

[thorn crown]

Doesn't have to be t

Any paramaterisation variable works

[wicked mauve]

can i do x = rcos theta?

but then thats 2

things

[thorn crown]

This is the obvious choice, yes.

Can even say x = r cos(t)

[wicked mauve]

and then that will be r(t)

and then i differentiate them

?

[thorn crown]

r(t) = < 3 cos(t), 3 sin(t), 0>

[wicked mauve]

okay

okay i got it

wow

[thorn crown]

Nice 🙂

[wicked mauve]

i feel my neurons

thank you for all your help

i really appreciate it

[thorn crown]

Glad to help

[wicked mauve]

too kind i wish you nothing but the best

thank you

.close

[midnight plank]

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[midnight plank]

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[dreamy prairie]

Can u help me too

[midnight plank]

Please don't occupy multiple help channels.

[astral canyon]

.close there's already an open channel

[midnight plank]

Channel closed

Closed by @astral canyon

Use .reopen if this was a mistake.

[dreamy prairie]

wow

thanks for your help bro

appreciate you

[midnight plank]

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[tall crag]

Suppose that two teams play a series of games that
ends when one of them has won i games. Suppose
that each game played is, independently, won by
team A with probability p. Find the expected num-
ber of games that are played when (a) i= 2 and (b)
i= 3. Also, show in both cases that this number is
maximized when p= 1/2

[tall crag]

for i = 2 i said $P(2) = p^2 + (1-p)^2$ and $P(3) = 1-P(2) = 2p(1-p)$ is that right?

[grand pond]

Viͥђaͣnͫ

[visual tiger]

what is "P(2)" and "P(3)"

probability that x games are played?

it looks good

[tall crag]

P(X=2), P(X=3) yes

[tall crag]

i computed $\mathbb{E}[X]$ and got $2+2p(1-p)$ then i took the derivative and got $\frac{d}{dp} \mathbb{E}[X] = 2p(1-p) = 0 $ hence $p* = 1/2$

[grand pond]

Viͥђaͣnͫ

[tall crag]

but idk what to do for when i = 3

[visual tiger]

uh wrong derivative

[tall crag]

why

[visual tiger]

how do you take the derivative of 2p(1-p)

[tall crag]

wait yes sorry

2(1-2p)

i wrote it wrong

[tall crag]

this

[visual tiger]

how would you have that

[tall crag]

sorry i meant P(3)

not EX

[visual tiger]

yes P(X=3) is that

we can keep going a bit

what are the other possible values of X

[tall crag]

X=4,5 right?

bc if

[visual tiger]

yes

[tall crag]

we hit 5

it either be draw

sorry

either lose or win

so like

[visual tiger]

maybe X = 5 is easier to understand

[tall crag]

X = 4 idk

[visual tiger]

what happened if X = 5

what happened to the first 4 games

[tall crag]

two games split and in 5th its either lose or win

[visual tiger]

ok

so 5th game can either be lose or win, we don't care

1st-4th games are 2 wins, 2 losses

[tall crag]

C(4,2) p^2 (1-p)^2 right?

binom

so

for X=4?

can we sum up

and 1 - sum?

that would be hard to compute

[visual tiger]

it's 3 terms, you don't need to expand

all we care about is solving dE/dp = 0 in the end

[tall crag]

hmmm

am calculating

4 - P(3) + P(4)

$\mathbb{E}[X] = 4 - (p^3 + (1-p)^3) + 6 p^2 (1-p)^2$

[grand pond]

Viͥђaͣnͫ

[tall crag]

$\frac{d}{dp} P(3) = 3p^2 + 3(1-p)^2(-1) = 3p^2 - 3(1-p)^2$ and $\frac{d}{dp} P(5) = 12 p(1-p)^2 - 12 p^2 (1-p)$

[grand pond]

Viͥђaͣnͫ

[tall crag]

$\frac{d}{dp} \mathbb{E}[N] = - (3p^2 - 3(1-p)^2) + \big(12 p(1-p)^2 - 12 p^2 (1-p)\big)$

[grand pond]

Viͥђaͣnͫ

[tall crag]

bruv

[visual tiger]

wait a bit

[visual tiger]

yeah ok this is good

[tall crag]

calculating this

i simplified it a bit

[visual tiger]

you can factor the second big term by p(1-p)

[tall crag]

$\frac{d}{dp} \mathbb{E}[N] = (1-2p)\big(3 + 12p(1-p)\big)$

[grand pond]

Viͥђaͣnͫ

[tall crag]

so just 1-2p = 0 so p* = 1/2

[visual tiger]

almost

[tall crag]

warum

[visual tiger]

you need to show why you can't have 3 + 12p(1-p) = 0

well it's easy

but small bit to show you saw that

[tall crag]

it just wanted us to show max is 1/2 so sure it doesnt

[visual tiger]

but if there's another critical point

what tells you that 1/2 is the max

and not that other critical point

[tall crag]

hm

ok ig i found it out

p(1-p) = -3/12 which is impossible

im so dumb

smh

[tall crag]

@visual tiger thx bruv 🫂

.close

[midnight plank]

Channel closed

Closed by @tall crag

Use .reopen if this was a mistake.

[midnight plank]

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[brittle cedar]

can someone help with the ticked numbers? start with 4 pls

[brittle cedar]

how do I start 4

are these 2 triangles the same?

[wooden badger]

No not necessarily

@brittle cedar

Try to equate alternate interior angles

[brittle cedar]

ohhh

is the 100 angle the same with T?

am i correct @wooden badger

[wooden badger]

Yes

[brittle cedar]

ok

im guessing r and s = 40?

@wooden badger

mb for pinging

[wooden badger]

Yep

[brittle cedar]

wait but how

they're not marked congruent though

[wooden badger]

S and b do u think they'd be equal?

[brittle cedar]

wdym

[wooden badger]

Angle s and B

Ive marked them

[brittle cedar]

s looks like

4

ok anyways

[wooden badger]

No thhts A lmao

[brittle cedar]

OH

ok

so and B?

[wooden badger]

Yes

[brittle cedar]

is b referring to the outside or insied

[wooden badger]

The angle thts marked in the diagram

[brittle cedar]

yeah

[wooden badger]

So they're equal ryt?

[brittle cedar]

yea

because of corresponding?

[wooden badger]

Co exterior angles

[brittle cedar]

what's that

i dont think we were taught that

[wooden badger]

Wait no what r they called

[brittle cedar]

consecutive????

[wooden badger]

Yeah corresponding you're right

R is equal to a

@brittle cedar can u see how

[brittle cedar]

i still dont see how thats A lol

i mean

the A doesn't look like an A

that's why i got confused

anyways thanks