help-38

[astral ore]

1 2 3 4 5 6 7 8 9 10

that's 10 numbers not 11

(10 - 1) + 1 = 10

[wraith hinge]

However, you need to avoid to repeat things like 1111 (x=1, y=11, z=1) and 1111 (x=1, y=1, z=11)

[astral ore]

yeah that's the issue

idk how to do casework for double counting tbh

it's 50k but that includes double counting

exactly like the case u mentioned

[wraith hinge]

So, we need to get how many double counting we have

[astral ore]

there's probably 46320

idk tbh

i remember coding this before

but yeah

dk how to get it

[wraith hinge]

In this case, let's try to figure out with a simpler question first

(is a good technique)

So, let's just take x in [0, 10] and y [0,20]

[astral ore]

omg i have to go

😭

so how are we going to discuss this

ugh

can i dm you later?

bruh

[hollow lagoon]

yea

[wraith hinge]

Ok

[astral ore]

nah let's see this through

sure

let's consider x in the closed interval

and same with y

let's consider x = 1: y can be anything from 0 to 20 right?

[wraith hinge]

Yes

[astral ore]

but can't i argue like that for other x's as well

where am i excluding

[wraith hinge]

I meant [1,10] and [1,20]

[astral ore]

okay

okay i see

the three digit numbers is what we should be concerned about right?

cuz 2 digits are unique

consider 18 for instance

or 1 and 9 (19)

[wraith hinge]

I think so

[astral ore]

but 110 could work like 11 1 or 1 11

[wraith hinge]

4 digits as well

[astral ore]

yeah my point is

anything above 2 digits

there's no 4 digits in our example

lol

[wraith hinge]

1020

[astral ore]

oh there is

okay now

[wraith hinge]

1715, 1218...

[astral ore]

how do we exclude these

each 3 digit combination has 2 possilbities right?

[wraith hinge]

No

[astral ore]

yeah not in our domain ofc

cuz 110

[wraith hinge]

Because 132 has only one combination

[astral ore]

11 0 doesn't work

so we look at the prefix

if the prefix is more than our domain then it has only 1 possibility

so any 3 digit number with prefix [11] has only 1 combination for example

a|bc?

a divides bc?

[wraith hinge]

x=a and y=bc

[astral ore]

okay for x = 1

what do we have?

all 2 digit numbers only right?

everything else doesn't work such that x in [1,10] and y in [1,20]

well it works but as in 3 digit numbers would only be counted as 1 as well

110 is only one possible configuration -> 1 10 and not 11 1

[wraith hinge]

Yes

[astral ore]

so x = 1: 20 things

x = 2: would be hmm

wait

again 20 lol

what

[wraith hinge]

Yes

[astral ore]

yeah in your example i think it's all 20 * 10

is it not?

[wraith hinge]

I think so

[astral ore]

so what was the point of that

[wraith hinge]

Getting now x:[1, 10] and y:[1,50]

[astral ore]

this is just 20 * 10 yeah

[wraith hinge]

I'm trying to solve this with you as well

I don't know the answer

[astral ore]

the answer is

4620 or some shit

[astral ore]

oh lol

i think we can do the main problem now

x in [1,10] y in [1,20] and z in [1,10]

i reduced y's upper bound to 20 from 50

actually we can just do

x in [1,5] y in [1,10] and z in [1,5]

[wraith hinge]

Our problem is with the y and z

[astral ore]

this is hard lol

😭

[wraith hinge]

The x is not the problem at all

[astral ore]

yeah that's what we figured

form the small example

*from

[wraith hinge]

So, let's broke the ORIGINAL problem into parts

Part 1: y is under [1, 9] and z is under [1,9]

[astral ore]

yeah isn't that what we're doing

oh

[wraith hinge]

There's no double counting ( 19 can't be y=19 and z=nothing)

[astral ore]

this question isn't solvable in 2 minutes right?

i think i'll just skip it

all my questions are supposed to be solved in under 2 mins

this looks like it'd take a lot of time

more than 2 mins tbh

what's JEE?

[wraith hinge]

Are you indian?

[astral ore]

no

[wraith hinge]

Ok

Probably there's some technique I never ever have heard of

[astral ore]

probably

[wraith hinge]

So, I think you should ask another person to that

[astral ore]

as of now the question is pretty hard

[wraith hinge]

The problem is the details

You need to take care of many details

[astral ore]

it's the double counting nonsense i guess

if we can figure out how to get rid of those

then we're done

[wraith hinge]

Yes

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[graceful cairn]

In a triangle ABC, AC = 10, BC = a, and sin A = 3/5.

Decide the values for a in which angle B has two possible angles.

[graceful cairn]

It is unclear to me what to exactly do in order to achieve this. I do however know that if angle B is 90 degrees, then there is only one solution.

Here's is a basic figure that I've drawn for myself to illustrate the situation. Since sin A = 3/5, then we know that the height of the triangle is equal to 6, as sin(A) = 6/10 = 3/5

If I have understood this correclty, "a" needs to be equal to 6 in order for there to be only one possible angle for B, as it would then be 90 degrees.

This is where I reach a full stop at knowing how to find the other a values that make B have two possible angles...

[rigid tide]

111 to 499
Those are your duplicate cases
Cuz _ __ and __ _ can be identical
But __ and ____ are both unique. And obviously _____ is definitely going to be unique.

Now in between 111 to 499, you also have to skip the places where there is a 0 in the number because immediately allows you to identify the separation.

Also X is always identifiable so that can never contribute to a duplicate case.

Subtract that from all the possible combinations and you'll get your answer.

[graceful cairn]

If it helps: the books answer is:
one angle for a = 6 and a > 10, no angles when a < 6, two angles when 6 < a < 10

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[rigid tide]

Can't you simply do it bye sine rule?

a/sin A = b/sin B
a is side opposite to angle A, same goes for B.
So
5a/3 = 10/sin B
so a x sin B = 6
Now B has to be greater than 0
and B has to be lesser than 180.
Take a look at the sine curve from 0 to 180.
It goes up and then down, so each value from (0,1) in the y axis comes twice when going from 0 to 180 angle wise..
aside from a=6 when B=90, all values of a can give two angles for B.
Also a cannot be less than 6 cuz that's the perpendicular height
Although I think a can be greater than 10 and still have two angles for B.

oh wait right
it cannot
the angles repeat afterwards.

[graceful cairn]

I am unable to understand how you went from 5a/3 = 10/sin B to a x sin B = 6. Would you mind clarifying?

[graceful cairn]

Oh, I'm terribly sorry. Seems like I had accidentally replaced 5a/3 with 3a/5.

[rigid tide]

no worries
it just uses a bit of your reasoning after this point

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[flat wing]

for b if i use the discriminent and show that b^2-4ac is less than 0 when k^2 -2k + 6 = 0

[flat wing]

will i still get the marks?

[wraith hinge]

i guess you can say that, the coefficient of x² is 1 which is positive and D<0 so it will always be positive

well much easier to just say, k²-2k+6 = (k-1)² + 5 which is always positive

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[astral ore]

why will you not?

u just have to show det(B) != 0

det(B) is a quadratic so showing that it has complex roots answers your questions

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[grave slate]

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[wheat spruce]

just confused on notation here forget what this means

[wheat spruce]

probability of (A ??? B)

dont know what the line is

[rugged latch]

given

conditional prob.

[wheat spruce]

hmm okay

thanks

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[wraith hinge]

finding the bounds for this, found that 1≤r≤2cosø just fine but

the bounding for ø is where im struggling

given 1 = 2x
1 = 2rcosø

while I can see how ø would be -π/3≤ø≤π/3 why are they just ignoring the r

[tepid hamlet]

well it's a point on the unit circle x^2+y^2 = 1 so i guess r = 1 there

[wraith hinge]

ahhhhh

yeah that makes sense thennvm

coolio ty

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[sullen ore]

it is right

well

-x=90-9x

wait

did u do the multiplication right

0.1(-x*10)=-x

[violet pier]

i put that into wolfram alpha and it was x = 8

[sullen ore]

wait

-x + 90 - 9x = 0

-10x=-90

i get x=9

[hot verge]

x=9

That is correct

[sullen ore]

where did u get 8

theres a 10% chance they all fail

yes thats what we are doing

what

okay

theres a 10% chance that they all blow up

and its all or nothing

theres no some dying some not

you showed us the top case

wait

you want to do bottom case?

OH

oh

so we have -x+90-9x=10

-10x=-80

x=8

ohh

yeah 8 is right

i mean

its like

would you rather always win 10 bucks or have a 90% chance of winning 100 bucks

second case has more risk

but its almost always better

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[south badge]

this may seem silly

but i was doing this question

but

i dont get this LOL

i would have had the answer too

[vagrant prism]

$3^k + 2\cdot 3^k = 3^k \cdot(1 + 2) = 3^k \cdot 3 = 3^{k+1}$

[solid kiln]

artemetra

[vagrant prism]

does that make sense?

[south badge]

waiy

wait

oh..

[vagrant prism]

they factored out 3^k

[south badge]

ah i see

would you normally spot tha

instantly?

[vagrant prism]

with practice, yes

[south badge]

ok got it

[vagrant prism]

but in general it never hurts to try to factorize something like this

[south badge]

true especially in a proof question

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[vagrant prism]

make a rough sketch

[vale obsidian]

im not really sure

but both are exponential equations

[vagrant prism]

no

[vale obsidian]

no?

[vagrant prism]

4^x is exponential

x^4 is polynomial

[vale obsidian]

rihgt

so parabola

[vagrant prism]

x^4 yes

upward parabola

and 4^x

[vale obsidian]

right

[vagrant prism]

is just an increasing function

[vale obsidian]

is exponential

yeah

but

[vagrant prism]

so sketch them together

you don't need to know the roots to determine their count

[vale obsidian]

right

hold up

so 3 points

[vagrant prism]

why 3?

[vale obsidian]

because the parabola would intersect twice

and then would get catched up to

if that makes sense

[vagrant prism]

oh yeah good point

i forgot about that

lol

yes it's 3

[vale obsidian]

nice

thank you

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[crisp pond]

how

[delicate belfry]

Scientific notation.

300=3 * 10^2

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[cursive stream]

How do you solve this?

[cursive stream]

Translation:
Let $f: \mathbb{R}^{n+1}} \to \mathbb{R}$ be a differentiable function of $n+1$ variables and assume that there exists a differentiable function $g:\mathbb{R}^n \to \mathbb{R}$ such that:
$$f(x_1, x_2, ..., x_n, g(x_1, x_2, ..., x_n)) = 0$$ for all $x_1, x_2, ..., x_n$ (the thing in the paranthesis is irrelevant)

a) Show that: yeah, the thing in the picture

[solid kiln]

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

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[cursive stream]

<@&286206848099549185>

[warm depot]

what

[grand trout]

what do u want

[soft granite]

Lmao

[simple haven]

Aside from needing to juggle indexing, wouldn't this be an exercise in simply doing the chain rule?

@cursive stream ^

[cursive stream]

Not quite sure

[simple haven]

Hint, what is $\dv{f}{x_i}$ (the full derivative)

[solid kiln]

OmnipotentEntity

[cursive stream]

help with a), if that wasn't clear (should've probably said it explicitly)

[simple haven]

(In this case, specifically when g is the final argument, so the value of f' is obviously 0, because f is 0 when fed g irrespective)

[cursive stream]

Ehmm

yeah I don't know, this task just boggled my mind honestly. I thought I understood the chain-rule, but now I'm a little unsure

[simple haven]

So

$\dv{f}{x_i} = \pdv{f}{x_i} + \sum_{i \ne j} \pdv{f}{x_j} \pdv{x_j}{x_i}$

[solid kiln]

OmnipotentEntity

[simple haven]

Does this look familiar? @cursive stream

[cursive stream]

Yeah I guess? The book generalised it in this way:

[simple haven]

Ok

So apply that formula here

You should get $0 = \pdv{f}{x_i} + \pdv{f}{g} \pdv{g}{x_i}$ which should look familiar.

[solid kiln]

OmnipotentEntity

[cursive stream]

alright, but I'm also having trouble with what to choose as the function corresponding to F here

[simple haven]

(Considering that g is x_{n+1})

In this case it's f(x1, x2, ..., xn, g(x1, ... xn))

[cursive stream]

But how do you get to this?

Could you just write it out?

[simple haven]

From here #help-38 message

[cursive stream]

hmm

Ok let me think

[simple haven]

All of the x_i are independent, so it i ≠ j then pdv x_i/x_j is 0

So the only remaining thing left is g

(which is x_{n+1})

[cursive stream]

What do you mean with independent?

[simple haven]

Independent variables

They have no dependency on each other

[cursive stream]

Oh yeah, like that. Yes true

but why does that mean that pdv x_i/x_j is 0?

[simple haven]

Because if there is no dependency between them then there's no dependency on the derivative either

[cursive stream]

and to clarify, i and j are just numbers from 1 to n, right?

[cursive stream]

Oh okay, that makes sense

[cursive stream]

(I'm not so used to working with generalised notations)

[simple haven]

If you have independent variables, then if you do have some kind of equation governing the variable, such as x_i = p(t), and x_j = q(s), then taking the derivative of x_i with respect to x_j is 0, because there is no x_j in the expression

[simple haven]

Yeah

[cursive stream]

Alright yes, very true

[cursive stream]

Ok but one thing now. I don't see how this formula corresponds to this #help-38 message

soo yeah... still not sure how to apply the formula you wrote out

[simple haven]

I'm not sure either, are you certain it's the same formula? Looks different

[cursive stream]

apologies for all the questions, it's a little late

[cursive stream]

No clue 😅

[simple haven]

I need to go afk

[cursive stream]

ah okay

Well thanks for the help

I'll let it cook in my head a little

I'll come back to it tomorrow.

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[rotund fable]

im having trouble figuring out what to after this or if im even doing it correctly

[left oriole]

what is the goal?

[rotund fable]

to find the gradient function

gradient/slope

[wraith arch]

so basically the derivative?

[rotund fable]

yes

[wraith arch]

what is the original function

is the one in brackets

or the one under it

[rotund fable]

its the one under

i can send it sepertately

[wraith arch]

alright

so we can just differentiate term by term here

so what’s the derivative of 2x?

[rotund fable]

2

[wraith arch]

yep

and now for the second term

we have to rewrite it in a way that is x raised to some power

which it looks like you’ve done correctly in the work above

well the third line is correct but the the fourth

so $\frac{-3}{x^{\frac{3}{2}}}$ is what we have

[solid kiln]

y0shi

[wraith arch]

how do we move the x up to the numerator?

[rotund fable]

make its exponent negative?

[wraith arch]

yep

and now we can do the power rule

[rotund fable]

would the 3 stay in the numerator?

[wraith arch]

yep

[rotund fable]

ohh i see

[wraith arch]

treat it as a constant multiplier

it’s actually -3

don’t forget the negative here

[rotund fable]

so would it be like this?

[wraith arch]

close

[rotund fable]

or wait i think i did the exponent wrong

[wraith arch]

yeah

[rotund fable]

-5/2

?

[wraith arch]

you kinda added 1 instead of subtracting it

yep

[rotund fable]

so then after that how do i get the x back down to the numerator?

[wraith arch]

just negate the exponent again

basically to move it up and down

we just negate the exponent

[rotund fable]

oh i got it

thank you!!

[wraith arch]

yw!

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[azure wagon]

did i draw this correctly?

helloo???

<@&286206848099549185>

thanks for the help....

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[upbeat ridge]

Can someone help me understand 59

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[iron vector]

How would you do this Q as I tried to set D as the origin then each of the other vertices 1 away from this point and I can only get$\frac{1}{\sqrt{2}}$

[solid kiln]

TEDDY

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[iron vector]

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[kind forum]

Hi

What vector methods do you know?

Are you familiar with the dot product?

[iron vector]

yes dot product

only

[kind forum]

First we need to find a vector perpendicular to the face ABC

(1,1,1) is a good vector to choose

We just need to give the angle between this vector and the ground

[iron vector]

I dont understand where this vector is in relation to the other points

[kind forum]

Okay let's call the midpoint between B and C "E"

What are the coordinates of point E

If the length is 1, like youve set

[iron vector]

if D=1,1,1) then E(1,1.5,1)?

@kind forum

[kind forum]

Why don't we set D to be the origin

[iron vector]

Okay then E is 0,0.5,0?

if we assume C is (0,1,0)

[kind forum]

Not exactly

Let c = (0,1,0) and b=(1,0,0)

[iron vector]

whoops its (0.5,0.5,0)

?

[kind forum]

Yes

Now find the vector EA

[iron vector]

Just did it, thanks for your help

@kind forum

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[plucky lily]

for part d,

the answer is supposedly

why doesn't E[W] = 1(100) + 2(0) + 3(-100) = -200 work

hmm

oh wait X is the # of boxes you open

X = 4 would be 0 (2 no prize, 2 prize)

X = 5 would be 100 once again (2 no prize, 3 prize)

so E[W] = 1(100) + 2(0) + 3(-100) + 4(0) + 5(100) = 300 which is still wrong hmm

oh seems it's because i'm doing E[W] = kW(X = k) instead of E[W] = kP(W = k)

for all values actually not just 1

E[W] = $100 * P(W = $100) + $0 * P(W = $0) + (-$100) * P(W = -$100)

P(W = $100) = P(X = 1), P(W = $0) = P(X = 2), P(W = -$100) = P(X = 3)?

E[W] = $100 * 3/5 + $0 * 3/10 + (-$100) * 1/10 = $50

<@&286206848099549185>

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[weak yacht]

hey, how would I find CDE

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[vague phoenix]

how would i show a function 2xsin(1/(y^4 + x^2)) - 2*(x^2 + y^2)xcos(1/(y^4 + x^2))/(y^4 + x^2)^2 is or isn't continuous (x,y) = (0,0)?

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[vague phoenix]

<@&286206848099549185>

[vague phoenix]

<@&286206848099549185>

[vague phoenix]

<@&286206848099549185>

[sacred cove]

You have to show that it is discontinuous at x,y = 0,0?

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[vague phoenix]

how do I do that?

[vague phoenix]

<@&286206848099549185>

[vague phoenix]

<@&286206848099549185>

[hot coral]

i am coming patrict

just finishing something with a other guy

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[pale swift]

could i please get help with b please

[wispy totem]

let's say, 2theta = x, then cos(2theta) = cos(x) = 7/25

cos-1(7/25) = x

since x = 2theta, cos-1(7/25) = 2theta

theta = sin-1(13/5) as it says

then 2theta = 2sin-1(13/5)

[pale swift]

i dont get it sorry

😭

[wispy totem]

if cos(x) = 1

then what is cos-1(1)

the -1 means inverse function

[pale swift]

i know

but like

i didnt really get the description of the other message

thank you for explaination

[wispy totem]

it says cos(2theta) = 7/25

[pale swift]

yep

[wispy totem]

then what is cos-1(7/25) equal to?

[pale swift]

uhh 1/2 cos-1(7/25) ?

[wispy totem]

no

Shouldn't it be 2theta?

[pale swift]

oh

[wispy totem]

because you have to ask yourself "What should i write to the argument for the cos function to get the value 7/25"

as it showed us, you should write 2theta

because cos(2theta) = 7/25

do you understand why the answer should be 2theta?

[wispy totem]

for this

[pale swift]

no sorry

thank you for explaining though

i appreciate it

[wispy totem]

no need to sorry it's okay

if cos(x) = y

then arccos(y) = x

did you understand this well?

[pale swift]

yes! thank youu

🤗

[wispy totem]

if you don't understand tell me

there is no problem

[pale swift]

thank you so much

[wispy totem]

we can go step by step

you can do this

[wispy totem]

you are welcome but you sure you understand it, right?

or no

[pale swift]

yess

🫡 😆

thank you

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[maiden glacier]

I'm not sure how to approach this without trial and error

[dull temple]

well you clearly can't try infinitely many numbers

how do you know when you can stop looking?

[fathom scroll]

Maybe proof by contradiction? Idk

[worldly wing]

Try making bounds for both sequences

Or just set the nth term of one equal to the mth term of the other and then use some sort of number theory

[fathom scroll]

Contradiction works

[maiden glacier]

What does that mean

[tepid hamlet]

you can also notice that theyre both monotonic and first one is increasing and 2nd one is decreasing, they will only have terms within [-1, 40] in the first couple of terms then quickly move out of that range in opposite directions, so you can just examine the first couple of terms

[fathom scroll]

So basically you let a and b are the 2 numbers that are in these sequences and a is not equal to b

And then you work things out

There will be something that contradicts

Like a = b

Or smth

[tepid hamlet]

just list the first 5 terms of each one 😄

or 6 dunno if youre starting at n = 0 or 1

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[frosty heron]

Can anyone help me how we got these values for the first question

[rain citrus]

do you mean question (a)?

As you can see, the first thing we can put in is the middle one, which is 4 in this case. (4 ppl like all 3 sports)

And since there are 7 people who like both R and F, the ones that are in R and F but not in C are 3.

isw, the ones that are in C and F but not in B are 2.

since n(F)=12, the ones in ONLY F are 3.

[frosty heron]

Yeah

[frosty heron]

Oh gotcha

[rain citrus]

btw the answer to (b) is ||x=7||,
(c) is ||19 and 10||

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[trail zinc]

find the minimum of A

0 < x < 4

[rugged latch]

@trail zinc whatve u tried

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[rough goblet]

@trail zinc just differentiate and set to 0, then check for maxima/minima

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[jagged sage]

Im a bit stuck… can you check my work I don’t think I am even on the right track

[signal ether]

breakdown the limit into where v(t) > 0 and <0
And integrate separately

[worldly wing]

You can’t integrate absolute value like that

[signal ether]

e^(t-1) > 1
t-1 > 0
t> 1

So anything less than 1 is negative

[jagged sage]

I’m confused so how do you want me to write it out

[worldly wing]

Make two integrals, one from t=0 to t=1 and one from t=1 to t=3

Between 0 and 1 the function is negative

So you want to integrate the negative of the function

Since negative of a negative is positive

[jagged sage]

How do you know this

[jagged sage]

What do you mean by this

[worldly wing]

That’s what @signal ether showed above

[jagged sage]

I didn’t quite understand what Deshak did there

[signal ether]

@gaunt belfry

[gaunt belfry]

yes?

[signal ether]

You want to break down the integral because your velocity is negative for a while (Negative velocity will reduce the distance covered)

So I'm basically checking at what point does the function go from negative to positive and then using those limits to integrate the function

[jagged sage]

Ah Okay I understand that part! But how do you know to break it down from 0 to 1 and 1 to 3

[signal ether]

You want to see where v(t) is positive
So you check v(t) > 0
e^(t-1) - 1 > 0
e^(t-1) > 1
e^(t-1) > e^0
Thus,
t-1 > 0
Hence
t>1

[jagged sage]

Ahh okay now I see it. So then at 1, that’s where it changes from - to +?

[cyan oxide]

i prefer checking when v(t) is equal zero, then can make a sign chart for when the sign changes cuz it might change several times

[signal ether]

Yes

[signal ether]

Fair

[jagged sage]

Now I’m thinking is the answer D? Because if it’s asking for total distance so bc going from 0 to 1 is negative, the negative will make it positive and then u add to get total distance?

[signal ether]

D is correct, but I don't follow the second half of the reasoning (The negative will make it positive)

[jagged sage]

Hmm can u explain why D is correct bc maybe I explained it wrong

The part why there is a negative in front

[signal ether]

So yes, the distance is negative for the negative velocity (You're running in the opposite direction)
And positive for the positive velocity

I agree with your explanation

[jagged sage]

Okay thanks

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[maiden glacier]

what exactly is part 3ii trying to get me to do?

[lean spear]

Maybe it wants you to solve a system of equations?

Question kinda broad

[maiden glacier]

yeah

both equations have the root x=-3 if that means anything

idk

[maiden glacier]

what does this mean

[lean spear]

I mean it says place an appropriate straight line

So ig make one up and intersect it with the parabola

And find point of intersection

Idfk tho

[maiden glacier]

ohhh

i think it just means y=0

then i can solve for x

[lean spear]

So it just wants the roots?

[maiden glacier]

i guess

weird way to ask it

[lean spear]

Yea

[maiden glacier]

thank you anyways

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[fossil cobalt]

how does it make sense to use intregals hee

here

i do not understand why we used intregals

[wraith arch]

that’s the graph of f’

we need to find f

and their relationship is via the antiderivative

or the integral

[fossil cobalt]

ok but how does it make sense to use intregals

intrregals give area

not the point

[wraith arch]

yes

but the integral is also the antiderivative of the function

remember the ftc where it states that

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

where F(x) is the antiderivative of f(x)

[solid kiln]

y0shi

[fossil cobalt]

ok i know that

but at the end it gives the area

not a point

thats my question

[wraith arch]

well it does give a point

F(b) is a point

F(a) is a point

and F is the antiderivative of f

so it gives the antiderivative of f at the point b and a

[fossil cobalt]

then why do we say intregal gives the area

[wraith arch]

it does both

the area and the antiderivative

they are the same

but it’s just that we use both definitions

[fossil cobalt]

ok thanks

[wraith arch]

yw!

and you can prob figure out why they use those bounds for the integral via the FTC

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[clever chasm]

Without using L’hopital, any recommendations on strategies?

Tried conjugate

[wooden mural]

Sure.

[limpid heart]

the conjugate should work!

[clever chasm]

But with conjugate you get 1-x..?

[wooden mural]

Conjugate works.

[clever chasm]

Oh god

Yeah my brain did not think about cancelling

Well thank you

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[wooden mural]

no worries.

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[late vector]

so i have these parameters or what you call them

[late vector]

they look something like this

and i have to find the volume under

the function is

i know if i find a function for the area as a function of x, i think its like the disc method or something

then i can integrate from 3/2 to 5/2

but i seem to get that part wrong

[limpid heart]

don't need a disc method for this, this is just a 2D integral! Disc and stuff is for volumes (or surface areas)

what integral do you get with respect to y?

[late vector]

ln(x^2+y)

then i put in 1/2x^2 for y

and i get ln(5/2*x^2)

[limpid heart]

right so your integral is $\int_{1.5}^{2.5} \int_{0}^{x^2/2} \frac{1}{x^2+y} dy , dx$

[solid kiln]

OssihLikesBlue

[limpid heart]

and you're saying that when you did it you get

[late vector]

yes

[limpid heart]

$\int_{1.5}^{2.5} \ln (5/2 x^2) - \ln (x^2) dx$

[solid kiln]

OssihLikesBlue

[limpid heart]

now you can use ln a - ln b = ln(a/b) to help here

[late vector]

what

[limpid heart]

this becomes $\int_{1.5}^{2.5} \ln(5/2) dx$

[limpid heart]

you need to evaluate the inner integral at both y = 1/2x^2 and y = 0 right?

[late vector]

this is the right answer btw

[solid kiln]

OssihLikesBlue

[limpid heart]

I guess the question is how?

[late vector]

damn i dont know

[limpid heart]

this should have been ln(3/2) and thats what I get here too

when you evaluate the inner integral yoy get

ln(x^2 + 1/2x^2) - ln (x^2)

right?

which is

[late vector]

wait wtf yes

[limpid heart]

ln (3/2x^2) - ln (x^2)

= ln(3/2)

[late vector]

basic math fail

[limpid heart]

happens lol

[late vector]

i added x^2 and 1/2x^2 in my head

and got 5/2x^2

its 3/2x^2

yes

[limpid heart]

great, then ln(3/2x^2) - ln(x^2) is just ln(3/2) by properties of logarithms

and so you get $\int_{3/2}^{5/2} ln(3/2) dx$

[solid kiln]

OssihLikesBlue

[limpid heart]

$= \ln (3/2) * (5/2-3/2) = \ln (3/2) * (1) = \ln 3 - \ln 2$

and you're done

[solid kiln]

OssihLikesBlue

[late vector]

wow i didnt know this

[limpid heart]

ln(a) - ln(b) = ln(a/b)

that's all it is

same as ln(ab) = ln(a) + ln(b)

[late vector]

where did you get -ln(x^2) from

[limpid heart]

evaluating ln(x^2+y) at y = 0

[late vector]

yes

ok

usually you dont have to think about this in 2d right?

when you integrate from 0

but here you have to

[limpid heart]

you always think about this

whether 1 or 2D

think about integrating x from 1 to 5

you take the value of x^2/2 at 5 and then subtract the value of x^2/2 at 1 from it

no difference here

[late vector]

usually when you put in 0 for x it becomes 0

like the area or volume

cause i guess you add an x to all variables when you integrate

[limpid heart]

that usually is doing some heavy lifting

[late vector]

but not when its $1/(x^2+y)

like the x variable doesnt get a y

[limpid heart]

that's heavily dependent on the type of function I mean even integrating just x+1 from x = 0 to 2

this doesn't have anything to do with y

if you integrate x+1 from 0 to 2, so (x+1)^2/2, at x = 0 the value is still 1/2

[limpid heart]

this is only sometimes the case and is not a usual at all

like you need to be really sure before you say its 0 and you always write it out to check

[late vector]

hm ok thanks i didnt know

i guess in my head i assumed that something cant have an area if it has no width

and no volume if it has no thickness

well i gues if you did integrate from 0 to 0 you would get 0

[limpid heart]

bounds are arbitrary! x = 0 does not correspond to no volume. But you're right that if you integrate anything from x = a to x = a you would get 0!

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[dim kite]

,rotate

[dim kite]

can someone tell me what's wrong?

i used the disk/washer method

and why is it negative?

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[dim kite]

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[solar parcel]

hi

[solar parcel]

I'm curious about my proof

so I'm trying to prove
$(A^C \cap B^C) = (A \cup B)^C$

[solid kiln]

Abdullah Zero 🇵🇸

[solar parcel]

I'm still not even sure about my first half of the proof so here it is

let x ∈ (A^C ^ B^C)
then x ∈ A^C and x ∈ B^C
then x ∈/ A and x ∈/ B
that means x is in neither A or B
hence x ∈/ A U B => x ∈ (A U B)^C

this feels weird to prove something like this

is this valid?

[vernal briar]

Yeah but you need both directions

Now prove that if x is in (AUB)^c then it's in A^c intersection B^c

[solar parcel]

yeye

still though

that last jump from
x ∈/ A and x ∈/ B
to
x is in neither A or B

is this really a valid jump?

coz it's purely common sense

[solar parcel]

also a lot of this is iff not just if then

so would this be a full proof:

x ∈ (A^C ^ B^C) 
<=> x ∈ A^C and x ∈ B^C
<=> x ∈/ A and x ∈/ B
<=> x is in neither A or B
<=> x ∈/ A U B <=> x ∈ (A U B)^C

[vernal briar]

x isn't in A and x isn't in B implies x isn't in AUB

That's a valid jump

But yes you can make this iff

That works

[solar parcel]

I see

thanks brother!

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[split chasm]

find the sum then divide by 2

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[clear topaz]

help

[clear topaz]

So at the top I have a curve that I am trying to find the critical points + extrema of, but I am having trouble in trying to parameterize the equation. It's not written on the paper, but the boundary of the curve is (x^2)<=y<=4

I just feel that I am having an algebra error in trying to parameterize this thing. I should be able to get x and y in terms of t as a dummy variable... but can't figure it out

wait lol its the boundary i get it now

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[upbeat pike]

can someone please explain why is the case

[upbeat pike]

when you do |y| = f(x) it flips it about the x axis and only includes for x > 0

[sullen ore]

well the function is negative for x<0

so |y| cant be negaitve

[upbeat pike]

ohh that makes sense

and then it becomes +- f(x) if you remove the sign?

[sullen ore]

yes

|y|=f(x) => y = +-f(x)

when f(x)>0

[upbeat pike]

yea

thank you

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[atomic totem]

This is 864, right? I’m fairly certain but also I am utter dogshit at combinatorics so

[lusty delta]

seems legit

[atomic totem]

Ok thank god

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[wraith hinge]

it says this isn't continuous?

it meets all conditions of continuity though

at x=3

[ionic pendant]

what are the conditions of continuity?

[wraith hinge]

f(a) exists, limf(x) exists, f(a)=limf(x)

[burnt spade]

Does f(3) exist?

[wraith hinge]

I think I made the mistake of simplifying

and finding a hole

hole = discontinuous

I see, thanks

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[low gazelle]

Nah u were right

[wraith hinge]

.reopen

[burnt spade]

You're right

[low gazelle]

Nvm I lied

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✅

[low gazelle]

Lmao

[wraith hinge]

wmy

wym

[burnt spade]

Hole does mean discontinuity

[wraith hinge]

yeah

[sullen ore]

moral of story: do not simplify when taking f(a)

[burnt spade]

But u can't simplify the fraction here when evaluating at a point where denominator = 0

[wraith hinge]

when do you simplify though

[sullen ore]

when taking the limit

[burnt spade]

Evaluating as a linit

Limit*

[low gazelle]

You can simplify as long as u state x=3 isnt in the domain

[wraith hinge]

because for the limit i just used the original function f(x)

[low gazelle]

You just have to always keep that in mind

[wraith hinge]

I assume you simplify for convinence

[burnt spade]

You can simplify when evaluating as a limit

But when evaluating as a function it isn't always possible

[wraith hinge]

yeah

[sullen ore]

when it's indeterminate

[burnt spade]

And what makes it even more confusing is desmos doesn't tell u it's undefined until you hover over the point

[sullen ore]

you do black magic algebra

[wraith hinge]

lol

[wraith hinge]

yeah it shouldn't do that

anyways thanks guys

[burnt spade]

I think it does that for all removable discontinuities

[wraith hinge]

mm yea I checked the graph thinking I was right

so i came here lol

until I hovered over 3

😂

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[sullen ore]

moral of the story: dont use graph

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[elder steppe]

I was wondering if

I let A = this matrix

then -1 <= det(A) <= 1

using that formula of u o (v x w)

in the event where v x w is perpendicular, we get the bounds of -1, 0, and 1

but then if they arent

we can get a value in the middle of this

either negative or positive

[edgy willow]

yea with the u o (v x w) formula you can write it with a cos() or sin() from the dot/cross product formulas and then just go through all the angles theta

it might be weird like you have to choose angles between u and (vxw)

[elder steppe]

could I involve it with inequalities

[edgy willow]

I'm not sure how to use inequalities to show that something acquires all values between two things, I'd just pick an example with a cos(theta) in it and say you can vary theta to get everything from -1 to 1

[kind forum]

This is fairly easy problem, if you know how to interpret matrices

Where do the basis vectors get sent

[sullen ore]

Cant do that here

We have vectors as elements

[kind forum]

What do the lines mean?

I interpreted this question as meaning that the vectors filled up the columns

[sullen ore]

Wait

Oh didnt see the lines

Mb

[kind forum]

Like col 1 is u, col 2 is v and col 3 is w

Yes this problem is not too bad, the determinant is a measure of change in volume

You're bounded by a unit transformation, the rest can only flatten or deform the unit paralleleliped

Therefore the det is bounded by -1 and 1

[sullen ore]

First time where intuition beats computation

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[sullen ore]

Where did u go

[wraith hinge]

mb

[sullen ore]

I wanted to help:(

[wraith hinge]

.reopen

[sullen ore]

Wait open a new help channel

[wraith hinge]

ok

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[lost minnow]

need to simplify how to do this?

[burnt spade]

Difference of cubes formula, difference of quartics formula

[lost minnow]

oh ya

thanks

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[orchid wadi]

I don't understand the last line in this explanation

isn't the Rightsubdomain of {e1, e2, e3... ek, f1, f2, ... fi)
the {f1...fi) part? And isn't that part irrelevant?

so why are we doing the P function on ONLY the irrelevant parts??? THIS DOESN"T MAKE ANY SENSE

😭

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[mossy breach]

Hi,
Can anyone help me understand how the axiom of foundation prevents circular sets?
If I understand it correctly, the axiom of foundation says that there has to be at least one element in a set which is disjoint from it. Or ∀X ≠ ∅, ∃x ∈ X, such that X ∩ x = ∅. But how does that prevent a set like A = {1, A}? Wouldn’t 1 be disjoint from A, allowing the set to contain itself and be like {1,{1,{1,…}}}?

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[mossy breach]

<@&286206848099549185>

[tribal jetty]

https://proofwiki.org/wiki/Set_is_Not_Element_of_Itself

[dapper swift]

The reason why we have the rule 'a set can't be an element of itself' is because the set of all sets that don't contain themselves is a parado

[shrewd ridge]

i agree it's confusing

it's just that A = {1, A} implies that {A} is also a thing

[tribal jetty]

you can connect a = {b,a} with the proof I sent by stating a ∈ a already

see that the a thingy is in a, right to left

it won't matter what's b

you have b ∈ a and a ∈ a. If you have worked with line-by-line proofs you can separate the conjunction and still get the contradiction with a ∈ a

[shrewd ridge]

so it says "at least one", but it's not clear if it would be different if it said "every element is disjoint" instead

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[tribal jetty]

btw with this a ∈ a you are done

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[mossy breach]

.reopen

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✅

[mossy breach]

I'm not quite sure if I understand what disjoint means at this point. Is it correct that in the set B = {2, 3, {4, 5}} 2 and 3 are disjoint because the intersection of 2 or 3 with B is {} but {4, 5} would not be disjoint and yield an intersection {4, 5}?