help-10

[cold thorn]

wait im confused

lemme try again

[ (d/dx sin(ax+b)) * cos (cx+d) - d/dx(cos (cx+d) * sin(ax+b) ] / cos (cx+d)^2

@timid silo this okay?

[timid silo]

I think yes

[cold thorn]

and what to do after that?

[timid silo]

Solve

[cold thorn]

apply chain rule when?

[timid silo]

Right @high lily ?

[high lily]

yes, that's a correct application of quotient rule

[cold thorn]

i was just confuse cuz i dont know if consider sin(ax+b) as one or two

[high lily]

chain rule to differentiate the
d/dx sin(ax+b)

[timid silo]

But i have no idea how will u solve this now

[cold thorn]

Ohhh

I see now i understand

first we apply quotient to simplify

then we use chain rule inside for derive

I try and give answer here pls wait

a cos (ax+b) cos (cx+d) + c sin (cx+d) (sin ax+b)

Divided By

cos(cx+d)^2

@high lily

[high lily]

would prefer that you put
(cos(cx+d))^2
for the denominator but yeh, that loks ok

[cold thorn]

yes i meant divided by like that

I not know how to use @warm shale

but yes thanks guys

.close

[obtuse pebble]

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[obtuse pebble]

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[prisma anvil]

i did variable saperable, reached on eq. logy= kx + c

[obtuse pebble]

@prisma anvil Has your question been resolved?

[prisma anvil]

<@&286206848099549185>

[vocal grove]

what function basically stays the same after differentiating?

[spiral maple]

yeah that's right, now apply the initial condition

[prisma anvil]

So we take x= 0?

[spiral maple]

yes

x=0, y=c.

$\ln(y)=kx+c\iff y=Ae^{kx}, A:=e^c$

[warm shale]

Mosh

[prisma anvil]

I think next step will be logy = c

[spiral maple]

also given c is defined in the question, you shouldn't have used c as the constant

[prisma anvil]

What's A?

[spiral maple]

I defined it

A is e^c

[prisma anvil]

So if logy = kx + c

Y= logkx + logc?

[spiral maple]

?????

what

$\ln(y)=kx+c \ y=e^{kx+c}=e^ce^{kx}=Ae^{kx}$

[warm shale]

Mosh

[prisma anvil]

I'm not getting how you get on y= Ae^kx

[spiral maple]

definition of logs, and exponent rules.

[prisma anvil]

Ohh now i got it

I'm dumb

[spiral maple]

then you just solve for A.

[prisma anvil]

@spiral maple i want to practice differential equation. From scratch. Easy questions. From where can i practice it

[spiral maple]

read your textbook then.

[prisma anvil]

They are high yield questions. I want easy

[spiral maple]

They have easy problems.

It's a textbook.

[obtuse pebble]

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[obtuse pebble]

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[timid silo]

The temperature of a point $(x,y)$ in the plane is given by the expression $x^2 + y^2 - 4x + 2y$. What is the temperature of the coldest point in the plane?

[warm shale]

d-static

[timid silo]

help please i dont know where to start

[high lily]

completing the square

[timid silo]

for x^2-4x and y^2+2y?

[high lily]

yes

[timid silo]

x^2+y^2-4x+2y+4-4+1-1

there

$(x-2)^2+(y+1)^2$

[warm shale]

d-static

[high lily]

wheres the rest of it

that isn't equivalent to what you started with

[timid silo]

ah

right

$(x-2)^2+(y+1)^2-3$

there

[warm shale]

d-static

[high lily]

no

is still wrong

[timid silo]

no?

[high lily]

how are you getting -3 on the end

[timid silo]

we had a +4-1

[high lily]

you had +4-1 "somewhere" in there...
that doesn't mean +4-1 will give you the constant you want on the end
nor does +4-1 equal -3

show all your steps clearly, (clearly show whats happening with all your values)
don't skip any steps

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

ok

i did completing the square here

$x^2+y^2-4x+2y+4-4+1-1$

[warm shale]

d-static

[timid silo]

is this correct

[high lily]

so far, yes

[timid silo]

ok

then i form the squares

$(x+2)^2+(y+1)^2-4-1$

how about now

[warm shale]

d-static

[high lily]

is wrong

check your signs

[timid silo]

$(x-2)^2+(y+1)^2+4-1$??

[warm shale]

d-static

[high lily]

is wroing

check your signs

don't skip steps

[timid silo]

$(x-2)^2+(y+1)^2-4-1$??

@high lily

[warm shale]

d-static

[high lily]

yes

[timid silo]

ok

is the answer

(2, -1)?

[high lily]

that isn't quite what the question is asking for

that's the location of the point with the lowest temp
they want the value of that temp

[timid silo]

isnt that what i am gibving?

[high lily]

reread what i said

[timid silo]

how do i give the value

@high lily

how do i calculate the value from the point

the distance formula, from the roigin???

are you there

[high lily]

don't overthink it

(2, -1)?
that's the location of the point with the lowest temp
the lowest temp would simply be the temp at that point

just sub it in and/or consider the min value of a square@timid silo

[timid silo]

Sub it in where

[high lily]

into the function for temperature

don't overthink it

[timid silo]

What is the function for temperature

[high lily]

you were given it in the question

don't overthink this

and reread what i've said

you pretty much already have everything you need

[timid silo]

Is the answer 11

[high lily]

no

how the fk are you getting 11

consider how you even got the point (2,-1)

the temperature is given by whatever you have

original expression, completed the square expression use whatever

you also found the location of the point with the lowest temperature

simply sub that point in to get the value of the lowest temperature

which is what the question is asing for

[timid silo]

I meant -5

[high lily]

yes

[timid silo]

Ok

.close

[obtuse pebble]

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[marble idol]

How do i know if a function is positive or negative ?

[marble idol]

E.g. :

[zinc flax]

do u mind if I explain for this prblm ?

[marble idol]

Yeah please

[high lily]

completing the square and/or expressing parts as squares usually help

[marble idol]

Can you explain that further please? @high lily

[high lily]

can you see that the denominator is a perfect square?

[obtuse pebble]

@marble idol Has your question been resolved?

[marble idol]

Yes

[high lily]

express it as a perfect square

also note that 2x^2 + 8 on the numerator is always positive

[marble idol]

So then it should be positive

[high lily]

yes

[marble idol]

But then why is it concave down?

If it is positive

[high lily]

,W graph (2x^2+8)/(x^4 -8x^2+16)

it's not concave down

[warm shale]

Results provided by WolframAlpha

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Requested by ramonov#0921

[marble idol]

Maybe if i showed you the original question it would make more sense. Because i have the same answer as you have but the teacher wrote that it was concave down

[high lily]

yes. show the original

[high lily]

you have different signs for the. second derivative

[obtuse pebble]

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[high lily]

.reopen

[obtuse pebble]

✅

@marble idol Has your question been resolved?

[marble idol]

I dont understand :/ so sorry @high lily

[obtuse pebble]

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[sinful grove]

Anyone know how i'd go about this? I tried that but it was incorrect

[obtuse pebble]

@sinful grove Has your question been resolved?

[obtuse pebble]

@sinful grove Has your question been resolved?

[obtuse pebble]

@sinful grove Has your question been resolved?

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[rain moss]

help pls

[timid silo]

ok

so

the first thing to do when you get a question like this

is think

"what do we know?"

meaning

waht numbers do I know and what do I need to get

so what numbers do you have here

[obtuse pebble]

@rain moss Has your question been resolved?

[rain moss]

uhh

3

@timid silo

[timid silo]

yeah but what are they

[obtuse pebble]

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[timid silo]

.reopen

[rain moss]

per gigabyte?

[timid silo]

what numbers do you have

list them out

[obtuse pebble]

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[amber fjord]

i am working on this lab where i must obtain the maximum wet area exposed, i understand that i need to take the derivative of the area and set it equal to 0 to find thee maximum area, the part where im struggling is finding a good equation to use for the area exposed, i believe that it is the area of the whole wheel (pi r^2) minus the area of the dry region (pi h^2) minus the area of the unexposed region, i am mostly confused on finding the area of the unexposed region.

[amber fjord]

also my professor gave the hint to think of integrals that affect area as well as to think of the fundamental theorem of calculus, i also think im supposed to use the area of a semi ellipses for the area unexposed, lastly the end of goal of the lab is to determine the length h ( radius of the dry region) that leads to the greatest exposed wet area.

[rain moss]

help

pls

[amber fjord]

bruh this channel isnt open

@rain moss delete your messages

[rain moss]

yes mam

[amber fjord]

<@&286206848099549185>

[obtuse pebble]

@amber fjord Has your question been resolved?

[amber fjord]

<@&286206848099549185>

[obtuse pebble]

@amber fjord Has your question been resolved?

[amber fjord]

<@&286206848099549185>

[obtuse pebble]

@amber fjord Has your question been resolved?

[amber fjord]

<@&286206848099549185>

[amber fjord]

please ping me if you can help

<@&286206848099549185> plz

[obtuse pebble]

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[obtuse pebble]

@lament patio Has your question been resolved?

[lament patio]

.close

[obtuse pebble]

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[sterile spire]

Anyone know how to do question three

I know that we must set the slope value as 0 but not sure what to do after that

[short spruce]

solve for x, make sure it's a turning point

though it's a quadratic, so you can just use the equation for it's vertex

which is -b/2a

[sterile spire]

ok ill try that and see if it works

so i got my x valyes

ohh nvm i read what you said wrong

ok i got it thanks

.close

[obtuse pebble]

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[celest condor]

hi can someone help me with derivatives

[celest condor]

<@&286206848099549185>

[nocturne minnow]

You have to wait at least 15 minutes before pinging helpers

[short spruce]

also nobody can help you if you don't post a question

[nocturne minnow]

Also, if you have a specific question too. Derivatives is a large concept

[timid silo]

take the derivative of deez nuts

[nocturne minnow]

Don't troll

[timid silo]

mb

[celest condor]

i just need to know for a polynomial of degree n, how many derivatives does it take to reach 0

[short spruce]

n+1, no?

[celest condor]

i thought n

[short spruce]

just take the derivative of some polynomial

[nocturne minnow]

It's n + 1

[short spruce]

take derivatives of x^2 until you get to 0

[timid silo]

can yall help me when u done with him

[celest condor]

it takes 2 derivatives of x^2 to reach to 0?

[short spruce]

no

what's the derivative of x^2

[celest condor]

2x

[short spruce]

and the derivative of that?

[celest condor]

2

[short spruce]

and again?

[celest condor]

0

[short spruce]

and that's 3 derivatives

[celest condor]

my niga ✊

one more question

[timid silo]

yeah don't be saying that

[celest condor]

h(x)=1/x does not have finite derivatives right

?

[short spruce]

i do not believe so

you can just keep using the power rule infinitely many times

[celest condor]

aightttt

[obtuse pebble]

@celest condor Has your question been resolved?

[obtuse pebble]

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[bitter sphinx]

Can someone help with homework?

[autumn adder]

Hi? What's your homework

[bitter sphinx]

Quadratic functions and equations

[obtuse pebble]

@bitter sphinx Has your question been resolved?

[timid silo]

hard to resolve a question without a question

[bitter sphinx]

He just disappeared

[obtuse pebble]

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[rugged hill]

So I'm trying to solve for t in terms of r and B.

Circle with radius r (C) is touching the rectangle of thickness t, and it is tangential with both circles of radius q (A & B), which are tangential with each other. Both A & B are (25.4-t) away from the rectangle.

I tried to solve this a few different ways but they mostly revolved around assuming that the line passing between the centre of C and the centre of either A or B would meet the rectangle at a point that would also meet the tangent of circle C.

Turns out it entirely doesn't and I have no idea how to solve this now. I also don't actually know if this is pre-u or above, because as far as I'm aware this is mostly just geometry but I could be wrong.

[This is literally to solve an ongoing argument in a game, and doesn't really matter much except that it's driving me insane now that I've spent 3 hours trying to figure it out.]

[obtuse pebble]

@rugged hill Has your question been resolved?

[rugged hill]

<@&286206848099549185>

[obtuse pebble]

@rugged hill Has your question been resolved?

[visual kelp]

I recommend drawing it out with desmos to solve it yourself

[autumn adder]

Do we know anything about the length of the rectangle

[rugged hill]

No, it's irrelevant; only the thickness matters

[autumn adder]

Hm, okay lemme see

We got to find out t?

[rugged hill]

Image is a segment of basically an infinite line of circles in a line with A & B and circles of C

Yea, in terms of q and r

[obtuse pebble]

@rugged hill Has your question been resolved?

[autumn adder]

Is 25.4

An exact value

@rugged hill

[rugged hill]

Yes

[autumn adder]

Oh

[rugged hill]

It's representing one inch in mm, because this game uses both imperial and metric for some reason

@autumn adderIf the solution is actually extremely complex then it's alright, really.

[autumn adder]

Yeah I know

I just wanna arrive at it now

[rugged hill]

I was just trying to figure it out with what I knew (which is around high school/A-Level) and then got stuck so I figured I could ask

[rugged hill]

basically why I was asking in the first place, I spent 3 hours on it and almost fried my very much not-mathematics-inclined-anymore brain

[autumn adder]

No issues

Lemme solve

I felt interested in this

Hey umm

Looks like Im arriving at the conclusion

Finally

[rugged hill]

Nice

[autumn adder]

Oooh need to do the equations again

Phew

Well Im enjoying it

I think its much easier yay]

Ill be back after lunch

And Imma do it fully

Dont close the channel

[rugged hill]

Got it

[autumn adder]

Yeah im here

@rugged hill

I think we got the solution

I dont know if its all correct

But the method is correct

This took half of my day

[warm shale]

Muhammad Hussaini

[autumn adder]

@rugged hill

[rugged hill]

Ooo nice

[autumn adder]

Hahahaha

Now I need better role in the server hehe

Come over in a VC after 5 min

Imma show u how I did this

It took around 5 pages

@rugged hill if you gotta join, you can. Im in the mathematics channel

[rugged hill]

I'm uh on the train

I'll be home in about 15

[autumn adder]

Haha, okay

[rugged hill]

I'll hop in then

[high lily]

can't you do this from pythag and segment addition/subtraction

[autumn adder]

It was just Pythagoras Theorem

[high lily]

it feels like the stuff under your root isn't quite right

[autumn adder]

Well should I send all of it

Its the pythagorean theorem

And just comparing with a quadratic equation

$at^{2}+bt+c=0$

[warm shale]

Muhammad Hussaini

[rugged hill]

I am finally home but I ran some of the numbers and I keep getting an exceptionally high value for t

[narrow nexus]

im stuck plz help 😦

[narrow nexus]

A number a is randomly selected from (0,1) a number b is randomly selected from (0,2) and a number c is randomly selected from (0.3). what is the probability that a<b<c ?

[autumn adder]

@narrow nexus

Please got to #❓how-to-get-help

[autumn adder]

Thats what I did

[high lily]

should only take like 3 lines of work in total

[autumn adder]

No it wont

Because you arrive at a quadratic result

[high lily]

what exactly were you doing

[autumn adder]

Use the quadratic formula

You got to find t

[high lily]

why do you even need the QF

after pythag to find the blue line, you can apply segment addition.
not much else needed

[obtuse pebble]

@rugged hill Has your question been resolved?

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[obtuse pebble]

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[timid silo]

im really stuck on this

[timid silo]

Peter is stuffing 5 letters into 5 envelopes. In how many ways can he do this so that exactly
1 letter is in the wrong envelope,
2 letters are in the wrong envelope,
3 letters are in the wrong envelope?

do they even give enough information?

[west sierra]

How many wrong envelopes are there

[karmic whale]

i dont get it in this
1 letter is in the wrong envelope,
2 letters are in the wrong envelope,
3 letters are in the wrong envelope?

[west sierra]

They say wrong envelope so I'm guessing there is a right envelope for each of the 5 letters

[jolly ginkgo]

1 letter have 1 right envelope

[west sierra]

Oh

[jolly ginkgo]

Each letter is assigned to some envelope

[plain owl]

So well, exactly 1 letter in wrong envelope isn't possible i suppose

[timid silo]

it isn't?

[west sierra]

There are 5 envelopes so ofcourse 1 letter can go to wrong envelope

[jolly ginkgo]

Yeah

It can be more than 1 in the case

That's right

[plain owl]

Only 1 letter is in the wrong envelope means the 4 others are in correct respective envelopes but that isn't possible

[west sierra]

Yeah

[jolly ginkgo]

So there's no way for part a I think

0

[karmic whale]

yeah?

[plain owl]

Two wrong envelopes would be equivalent to choosing two and swapping those

[west sierra]

So there r 5 letters whose envelopes We gonna swap individually and the number of possible swaps would be number of ways

[timid silo]

i dont really understand

the other 4 could be in wrong envelopes also

[high lily]

that exactly
exactly 1

(different to at least)

[timid silo]

ah

[plain owl]

That's why I edited it to "exactly"

[west sierra]

3 others*

First one was only 1 wrong envelope which is not possible, and in second question we take 2 wrong so 3 would be right

[timid silo]

ah

[plain owl]

Do you mean "odd" number of letters? If yes then that is incorrect.

[timid silo]

what property do you mean?

because of pigeonhole?

im going to try and write it up

a) This would be impossible. If all the 4 other letters were in the correct envelopes, then the last has to be in the correct envelope.

i understand it

probably a terrible explanation

[plain owl]

This is a good explanation (perfectly fine even)

[timid silo]

ah, much beter

revised

a) This would be impossible. If all the 4 other letters were in the correct envelopes, then the last has to be in the correct envelope, which contradicts the statement of one letter is in the wrong envelope.

not many changes

right

let me make it a proof by contradiction

 Let’s say that 1 letter is in the wrong envelope. This means that the other 4 letters are in the correct envelopes, but then this would mean that the letter that is in the wrong envelope is in the correct envelope since there are 5 total envelopes. Therefore, we have a contradiction, proving this impossible.```

tada

what property do you mean>

this too

[plain owl]

3 is possible.

[timid silo]

i agree

wait

i dont

its impossible

parity?

does it have to do with odds and evens?

[plain owl]

I tried to warn earlier

[timid silo]

i missed that

im confused

there are like two opinions

it seems impossible to me

[high lily]

consider a simple game like rock-paper-scissors

[plain owl]

Then what property were you insinuating? Only 1 wrong envelope isn't possible. 2,3,4,5 all others have non-zero arrangements.

[plain owl]

Let each letter and it's corresponding envelope be assigned the same number from 1,2,3,4,5.
Assume that letters 4,5 are correctly put in envelopes 4,5 respectively
Then 1->2, 2->3, 3->1 is a way to rearrange letters 1,2,3 such that only those three are assigned three wrong envelopes.

[timid silo]

i see

what would the number of possible arrangements be for that

2^3?

[plain owl]

Split our actions into two parts. First, we choose the three envelopes from 5 we want to swap the letters of. Then, the swapping itself looks like a "rotation" of sorts which can be done two different ways

[timid silo]

why is it not 3 different way?

three different rotation

[plain owl]

Consider we chose envelopes 1,2,3

Letter from 1 can be placed in either 2 or 3. This fixes the location letters 2,3 can be in.

[timid silo]

right......

so the number of combinations is 6

[plain owl]

It is 2*(5C3) for overall 3 wrong envelopes case

[timid silo]

hmm

sounds a lot like a problem i did today

[plain owl]

5C3 for choosing the three envelopes

[timid silo]

ok

[plain owl]

Yep

[timid silo]

the *2

why 2 again

the way the rotations work doesn't really make sense for me @plain owl

how would i explain 2

[plain owl]

Didn’t this work for you?

Are you familiar with “derangements”?

[timid silo]

not realy

[timid silo]

wdym by "fixes"

[plain owl]

If letter 1 is placed in envelope 2, letter 2 has to be in envelope 3 and letter 3 has to be in envelope 1. Similarly for letter 1 placed in envelope 3

[timid silo]

i see

it makes sense

let me write it up

[plain owl]

These are the only two cases possible

[timid silo]

how do i write 5 choose 3 in latex

$5 \choose 3$

[warm shale]

d-static

[timid silo]

lol

[plain owl]

$^5C_3$ or $_5C_3$ or $\binom{5}{3}$

[timid silo]

@timid silo@plain owl ```
The way we can choose 3 letters out of 5 is 53 (5 choose 3). We then have to multiply that by 2. For example. (if we corresponded the correct letter and envelope with the same number), letter 1 could be placed in envelope 2 or 3. This fixes the locations in which letters 2 and 3 can be in. The number of fixed positions is two, which is why we multiply by 2. Therefore, our answer is 2* 53.

[plain owl]

How is 5 choose 3=53??

[timid silo]

it didn't render

[plain owl]

Ok

[timid silo]

2 letters are in the wrong envelope, the answer should be pretty similar for this one also, right?

[plain owl]

We choose two envelopes and swap the letters between them

But there is only one way to swap here

[plain owl]

Derangements are very useful (read it later). From that, we can say 2 comes from !3=2

[timid silo]

so the answer is 5 choose 2

[plain owl]

Yep!

[timid silo]

but how is the answer i wrote

#help-10 message

[plain owl]

Probably include what 1,2,3 represent. Assuming the three letters we choose are numbered 1,2,3 or like that

[timid silo]

i did

my full answer for all three parts

a) Let’s say that 1 letter is in the wrong envelope. This means that the other 4 letters are in the correct envelopes, but then this would mean that the letter that is in the wrong envelope is in the correct envelope since there are 5 total envelopes. Therefore, we have a contradiction, proving this impossible.

b) The way we can choose 2 letters out of 5 is  52 (5 choose 2). We then have to multiply that by 1. For example. (if we corresponded the correct letter and envelope with the same number), letter 1 could be placed in envelope 2. This fixes the locations in which letter 2 can be in. The number of fixed positions is one, which is why we multiply by 1. Therefore, our answer is 1* 52. 

c) The way we can choose 3 letters out of 5 is  53 (5 choose 3). We then have to multiply that by 2. For example. (if we corresponded the correct letter and envelope with the same number), letter 1 could be placed in envelope 2 or 3. This fixes the locations in which letters 2 and 3 can be in. The number of fixed positions is two, which is why we multiply by 2. Therefore, our answer is 2* 53. 

[plain owl]

I will come back after a while (looks correct at first glance)

[timid silo]

oof

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

<@&286206848099549185>

[plain owl]

Why did b) have to be this long?

[timid silo]

but don't i have to explain why its just 5 choose 3?

[plain owl]

Well, choosing 2 envelopes can be done in 5C2 ways and swapping to ensure both those letters are wrongly placed can be done only in 1 way. Giving 1*5C2

[timid silo]

OK

how about now

a) Let’s say that 1 letter is in the wrong envelope. This means that the other 4 letters are in the correct envelopes, but then this would mean that the letter that is in the wrong envelope is in the correct envelope since there are 5 total envelopes. Therefore, we have a contradiction, proving this impossible.
b) The way we can choose 2 letters out of 5 is  52 (5 choose 2). Well, choosing 2 letters can be done in 5 choose 2 ways and swapping to ensure both those letters are wrongly placed can be done only in 1 way. Giving 1* 52.
c) The way we can choose 3 letters out of 5 is  53 (5 choose 3). We then have to multiply that by 2. For example. (if we corresponded the correct letter and envelope with the same number), letter 1 could be placed in envelope 2 or 3. This fixes the locations in which letters 2 and 3 can be in. The number of fixed positions is two, which is why we multiply by 2. Therefore, our answer is 2* 53. 

[plain owl]

Looks... good

[timid silo]

ok

[timid silo]

you sound a bit unsure

[plain owl]

Looking for someone else's inputs

[timid silo]

ok

ill wait

[final plaza]

Your solutions look good but technically you didn't give an answer for part a

It asked "how many ways" and you said "contradiction"

That's not an answer

Finish it off properly

[plain owl]

Yes you should have concluded by saying thus the number of permutations satisfying this condition is 0

[timid silo]

oh, they just dont get rendered

should i close this

[final plaza]

Your solutions look right (once you fix part a) so you're probably done

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

ok

a) Let’s say that 1 letter is in the wrong envelope. This means that the other 4 letters are in the correct envelopes, but then this would mean that the letter that is in the wrong envelope is in the correct envelope since there are 5 total envelopes. Therefore, we have a contradiction, resulting in the number of permutations satisfying this problem to be 0.
b) The way we can choose 2 letters out of 5 is  52 (5 choose 2). Well, choosing 2 letters can be done in 5 choose 2 ways and swapping to ensure both those letters are wrongly placed can be done only in 1 way. Giving 1* 52.
c) The way we can choose 3 letters out of 5 is  53 (5 choose 3). We then have to multiply that by 2. For example. (if we corresponded the correct letter and envelope with the same number), letter 1 could be placed in envelope 2 or 3. This fixes the locations in which letters 2 and 3 can be in. The number of fixed positions is two, which is why we multiply by 2. Therefore, our answer is 2* 53. 

last time (i hope)

[final plaza]

Uep

Good

[obtuse pebble]

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[spiral maple]

what have you tried?

[bleak jackal]

I integrated it

[spiral maple]

ok

[bleak jackal]

idk if thats right tho lmao

[spiral maple]

it is

[bleak jackal]

and idk wat to do from there

oh nice

do i sub in x value?

[spiral maple]

you're solving that DE

you find the constant of integration, yes

[bleak jackal]

what do I do after I sub in x?

[spiral maple]

solve for the constant of integration

you plug in the point and get the constant.

[bleak jackal]

ohh ok

ty

.close

[obtuse pebble]

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[wanton perch]

How do I find the coefficients of a Fourier series?

[forest sinew]

integration, usually

applying condition which allows you to set it equal to a function of some variable

[wanton perch]

Can you give an example?

[forest sinew]

there are lots of examples online, im a student in this as well so im not sure how much you want typed or what youre confused about

there are a lot of resources for these though

youll need to be more specific

these are typically memorized

youll need eigenvalues for your frequencies

[obtuse pebble]

@wanton perch Has your question been resolved?

[obtuse pebble]

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[past hull]

Can anyone help with this homework problem dealing with parametric equations

[past hull]

<@&286206848099549185>

[obtuse pebble]

@past hull Has your question been resolved?

[past hull]

<@&286206848099549185>

[soft sail]

what have you tried already? in (a) for instance

[past hull]

I tried finding the non parametric equation

Since x = f(t)
Y = g(t) = t
Would the Cartesian equation just be f(t) but with in f(x) form @soft sail

Hello?

[obtuse pebble]

@past hull Has your question been resolved?

[past hull]

<@&286206848099549185>

<@&286206848099549185>

[past hull]

<@&286206848099549185>

[frigid nexus]

δαμν\

[obtuse pebble]

@past hull Has your question been resolved?

[past hull]

No

<@&286206848099549185>

[obtuse pebble]

@past hull Has your question been resolved?

[jolly ginkgo]

Still need help?

[past hull]

Yes @jolly ginkgo

[jolly ginkgo]

Ok

Do you know what you have to find for a?

(f(t),g(t))

[past hull]

Would it be (16,8)
And dist= sqrt of 16^2+8^2

[jolly ginkgo]

How did you get 16?

That's correct

[past hull]

From plugging t = 40 into ft

[jolly ginkgo]

Wait for distance

Distance is right too

[past hull]

I'm not sure what to do for b

Or c

Really the rest

[warm shale]

it's Sam

[past hull]

Well g(t) derivative is 0 at 10-20 and 30

[jolly ginkgo]

Ok

And what about f(t)

[past hull]

Not sure

[jolly ginkgo]

Differentiate f

[past hull]

2/5 + 8/pi(pi/10)cos(pi/10 t)

[jolly ginkgo]

Equate it to 0 and simplify

[past hull]

20/3

[jolly ginkgo]

Is that the only solution

[past hull]

I believe so

[jolly ginkgo]

No there are more

Tell me what you get cos =

[obtuse pebble]

@past hull Has your question been resolved?

[languid stirrup]

Does anyone know why we log a time series? Just curious

[obtuse pebble]

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[proven orbit]

If I have a curve, and I have to find the region in which it is increasing, do I also include the critical point? For example, for a curve y=f(x),
I observe that it is increasing in -inf to 2, then decreasing from 2 to 3, and then increasing from 3 to -inf.
So should my answer be:

1. (-inf, 2) U (3, inf)
   Or
2. (-inf, 2] U [3, inf)

The answer key for the exam says its 2. but I don't get it why. Wouldn't it be parallel to x axis at 2 and 3? Why include them?

[obtuse pebble]

@proven orbit Has your question been resolved?

[proven orbit]

<@&286206848099549185>

[nocturne marten]

The answer key seems wrong to me too

At x=2,3, the tangent should be parallel to the x axis as you said so they're technically not increasing

[proven orbit]

yeah

[obtuse pebble]

@proven orbit Has your question been resolved?

[proven orbit]

fuck it

.close

[obtuse pebble]

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[hallow glacier]

do we have any name for this inequality
(a1^n + b1^n)(a2^n + b2^n)...(an^n + bn^n) >= (a1a2...an + b1b2...bn)^n
with (a1,a2,...,an and b1,b2,..,bn > 0)

[obtuse pebble]

@hallow glacier Has your question been resolved?

[hallow glacier]

<@&286206848099549185>

[novel knoll]

Almost a variation of mahlers inequality

[hallow glacier]

hmmm

i think it is

thanks

.close

[obtuse pebble]

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[timid silo]

Consider all the points in the plane that solve the equation $x^2 + 2y^2 = 16.$ Find the maximum value of the product $xy$ on this graph.

(This graph is an example of an "ellipse".)

[warm shale]

d-static

[timid silo]

i know that i can use am-gm inequality, but how?

what should i do fist

the am-gm inequality doesn't help me find the maximum

<@&286206848099549185>

anyone please

please

<@&286206848099549185> <@&286206848099549185>

..........

come one

someone,

there are 35 helpers on

[lethal crest]

are you allowed to use multivariable calculus?

i.e. Lagrange multiplier?

[timid silo]

i dont know calculs

all i learn was am-gm inequality

[lethal crest]

...

like not even single variable?

ig you can do it with that

$\frac{x^2+2y^2}{2} \geq \sqrt{2x^2y^2} = \sqrt{2}|xy|$

[lethal crest]

can you handle the rest?

[timid silo]

i was thinking of something like that

[lethal crest]

then why didn't you do it?

[timid silo]

i didnt see hwo it would help

i still dont see how it would help

[lethal crest]

bruh

$|xy| \leq \frac{16}{2\sqrt{2}}$

[timid silo]

ah

i didn't see that-

wow-

[lethal crest]

-_-

literally one step later

[timid silo]

-_-

ok

thank you

i have to think sometimes.

.close

[obtuse pebble]

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[vestal lily]

⦁ A sphere of volume was melted into a cylinder of height 2cm
⦁ Calculate the radius of the cylinder (3 marks)
⦁ How many cones of height 3cm and radius 4cm could instead be made from the sphere if it was melted? (3 marks)

[timid silo]

@vestal lily

[vestal lily]

oh hy i need help

[timid silo]

what have you done so far?

[vestal lily]

i have done this V(cylinder) = πr^2 • h, V(sphere) = (4/3)πr^3

πr^2 • (2) = (4/3)πr^3

and this

@timid silo

[timid silo]

(this is wrong) now you can cancel out the r^2 on both sides, or more accurately divide both sides by r^2 to get r

wait

[timid silo]

a sphere of volume what?

did they give you the volume for the sphere?

[timid silo]

don't do this

[vestal lily]

il show you the message my teacher said it for the question

There is no diagram. If there was it would just be a cylinder with a height of 2cm marked on it and a radius of r so wouldn't really be that useful I think.
The volume you are given is equal to the volume of the cylinder, that's what melted means there.
Use the volume of a cylinder formula. You know h so they only unknown then is r. Rearrange to get that.

For part two work out the volume of one cone (you know r and h) and then see how many of those you could get from the original volume, the 288pi.

[timid silo]

Did they give you the volume for the original spehre?

[vestal lily]

no they did not

or i can show you the first question if its linked to that

[timid silo]

sure

[timid silo]

"A sphere of volume was melted into a cylinder of height 2cm" is kind of an incomplete sentence, like a sphere of volume 2? 3? what? lol also you need either the volume of the original sphere or the radius of the original sphere to solve it

[vestal lily]

⦁ Which area is larger: a circle of radius 3cm or a square of length 5cm?
(2 marks)

and the answer i got 9pi cm1

@timid silo

[timid silo]

are you sure they didn't give you the sphere's volume?

[vestal lily]

no they dont il show you a scx

[timid silo]

I'm pretty sure you can't solve without knowing either the radius or the volume of the original sphere, well you can if you assume the sphere and the cylinder have the same radius, but they didn't give that there

I'm assuming it's a mistake

[vestal lily]

There is no diagram. If there was it would just be a cylinder with a height of 2cm marked on it and a radius of r so wouldn't really be that useful I think.
The volume you are given is equal to the volume of the cylinder, that's what melted means there.
Use the volume of a cylinder formula. You know h so they only unknown then is r. Rearrange to get that.

For part two work out the volume of one cone (you know r and h) and then see how many of those you could get from the original volume, the 288pi.
this is what ny teacher said @timid silo or does it not make sense

[timid silo]

oh wait

your teacher did give you the original volume

288pi

@vestal lily

[vestal lily]

oh did he i didnt know i checked the email

[timid silo]

"from the original volume, the 288pi."

[vestal lily]

@junior inlet then what do i do to find out

[timid silo]

lol do u want me to help you or can that person help?

[vestal lily]

i want you to help me

because thats my teacher

[timid silo]

so they want you to find the cylinder's radius right?

[vestal lily]

yeah to calculate the radius

[timid silo]

and they said that the sphere was melted down to make the cylinder, which you already understand as both of them having the same volume

so just set 288pi = (pi)(r^2)h

h = 2

now solve for r

[vestal lily]

will that be 2

r2

[timid silo]

what?

how did you get r = 2??

[junior inlet]

uh
it's 22:50 here
i'm not in a mood to solve problems
sorry

[vestal lily]

Ohh really it’s okay

[timid silo]

no it won't

how did u get 2 for r?

[vestal lily]

i thought it will be pi 2

i was wornk

[timid silo]

do you know how to solve this for r?

[vestal lily]

no i dont how do you do i t

[timid silo]

divide both sides by pi

now you'll get 288 = (r^2)h

now divide both sides by h which is two

now you'll get 144 = r^2

[vestal lily]

oh okay

[timid silo]

take the square root on both sides to get r

[vestal lily]

which is 12

[timid silo]

yeah

[vestal lily]

really

thanks is that it or we have to do more

[timid silo]

yes

you found r

which is what they asked you to find

so that's for that question

[vestal lily]

so what will be the formula for that question would

[timid silo]

wdym?

[vestal lily]

sorry not the formula the working would it be 288 divided 2 =144 square root =12

[timid silo]

yeah

[vestal lily]

thank you very much so what would be this question How many cones of height 3cm and radius 4cm could instead be made from the sphere if it was melted? (3 marks)

[timid silo]

you do pretty much the same thing

but with the cone formula

[vestal lily]

yeah alright so will the cone formula be this πr^2 • (2) = (4/3)πr^3

[timid silo]

no dude

do you know what's the formula for the volume of a cone is?

[vestal lily]

no i havent learnt that personally

[timid silo]

bruh

[vestal lily]

V=πr2h
3

is it that

[timid silo]

you need it to solve that

[vestal lily]

V=1/3hπr².

[timid silo]

this

set that equal to the sphere's volume 288pi

oh wait

brb

[vestal lily]

alright and thank you for taking the time with me

[timid silo]

you do basically the same thing

[vestal lily]

yeah so divide

[timid silo]

288pi = n(formula for the volume of a cone)

you multiply it by n and solve for n

n is the number of cones

[vestal lily]

which is 3

conmdes

so 288 divided by 3

[timid silo]

288pi = n(1/3)(3)(4^2)(pi)

n =18

[vestal lily]

oh alright so is that that answer done

or is there more to do @timid silo

[timid silo]

nope

that's it

[vestal lily]

thank you can you also show me for the first question in units

would it be 288pi=n

@timid silo thank you soo much i appreciate you i was stuck on that question for awhile

.close

[obtuse pebble]

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[timid silo]

question 28 i could show my attempts but they seem to be wrong

[naive fossil]

how does +4 +2 (when you move the 2 over to the lhs of the equation) become 2?

[timid silo]

ignore that part that i had

[naive fossil]

Ok, What method are you using to solve?

[timid silo]

well c = (b/2)^2

[naive fossil]

You're trying to complete the square?

Ok, that's a method.

Well, setting up for cts I go about it as k^2 - 24k + ____ = -6 + ____
and the blanks would be (b/2)^2 -> 144

[timid silo]

tes

[timid silo]

continue

i’ve understood that part

[naive fossil]

Then the LHS becomes the perfect square -> (k-12)^2

and the rhs becomes 138

[timid silo]

how’d you get those and whats a LHS?

[naive fossil]

LHS / RHS = Left Hand Side / Right Hand Side

The left side becomes a perfect square. That's what completing the square does.

[timid silo]

continue but how’d you get (k-12 )^ 2?

[naive fossil]

Try to multiply it out - (k-12)^2 = k^2 -24k + 144

Which is what our LHS side was once we put (b/2)^2 in the blanks

[timid silo]

didnt it equal 144 ?

[naive fossil]

The blanks did, yes

[timid silo]

so you found its square root and also k’s square root and made them in ()^2

[naive fossil]

So, the LHS was k^2 - 24k + 144 -> perfect square -> (k-12)^2

[timid silo]

yeah but it doesn’t explain how you’ve got the (k-12)^2

[naive fossil]

It's the perfect square of the LHS - I factored k^2 - 24k +144 into a perfect square.

[timid silo]

well how’d you factor it exactly wouldn’t it be finding two numbers that add to 24 and multiple to 144?

yeah continue you got it right

[naive fossil]

A helpful thing with CTS is whatever (b/2) is will be your factorization. Here, half of -24 is -12...so that's how I knew it was (k - 12)^2. It's just a helpful thing that works. Or yes - adds to 24 and multiplies to 144. Same thing, different ways to get it.

[timid silo]

oh okay continue

[naive fossil]

Now you just solve.
You've got (k-12)^2 = 138

[timid silo]

thats it ?

can i find both of their square roots ?

like / (k-12)^2 , / 138?

[naive fossil]

? I don't know what you're trying to do here...divide them?

You would take the square root of both sides.

[timid silo]

just putting them under the root

[naive fossil]

Then yes, that lol

[timid silo]

oh okay so it would be k-12 ≈ 11.75

take the 12 to the other side

[naive fossil]

Yes, or just simplify the square root - it depends on how they want you to answer.

[timid silo]

11.75 + 12 = 23.75, ≈ k

[naive fossil]

Remember, it's a quadratic.

2 possible solutions.

[timid silo]

yeah we need to the other k now

[naive fossil]

+/- 11.75

[timid silo]

oh

yess

so this is how we got those two?

[naive fossil]

So, 12 +/- sqrt(138)

Yes, because of that +/-

[timid silo]

oh fair got it

mind waiting for me to solve another question

[naive fossil]

I'm doing other things, so just ping me if you'd like

[timid silo]

sure i’ll ping u if i face any issues with this , thanks

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

hm

but what about 4x^2 - 20x + 25 = 0

its a perfect square

@naive fossil

i tookthe sqrt of 4x^2 and 25 and the sign of the middle term

(2x-5)^2 = 0

uh then i find the sqrt on the both sides again

[naive fossil]

Yes

[timid silo]

2x-5 = 0

take the 5 to the other side

2x = 5

divide by 2

x = 5/2

but that only gets us the first part

we need the 2nd solution

part where i'm confused

[naive fossil]

With quadratics, there is always the possibility of 2 solutions, not necessarily will always be 2 though

If you were to graph this, you would see why there is only one solution

[timid silo]

oh havent learn about grahping them yet , altho i think we did just didnt remember it well , will def relearn it

so my answer was correct ?

[naive fossil]

yes

[timid silo]

thanks !!!!!!!!

i really appericate it

you dont know how much this helped

thanks once again

.close thanksss

[obtuse pebble]

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[pastel fern]

Hi, I'm halfway through this exercise, I've calculated the rotational with general derivatives for f bc i don't know what the function is

I'm now trying to solve the integral by integrating rot(F)(S)*n

[high lily]

<@&268886789983436800>

[pastel fern]

i have no idea why i sent that

i'm sorry

[upbeat plinth]

ty

[pastel fern]

lol

so can anyone help?

i got (fy,fx,-2) as rot(F)

also lmk if you need help with the translation

thing is idk how to evaluate (2x,2y,-1) which is my normal vector on rot(F) since 2 of the coordinates are implicit derivatives

[obtuse pebble]

@pastel fern Has your question been resolved?

[pastel fern]

Somebody suggested me to use Stokes' theorem

[obtuse pebble]

@pastel fern Has your question been resolved?

[obtuse pebble]

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[woeful ibex]

how to prove that 6 has no inverse mod 10 ? I tried to prove that by using the fact that if 6 have an inverse (let s call it b) mod 10, we have 6b = 1 mod 10 (it s three bar not two) <=> 6b - 1 = 10k <=> b = (10k+1)/6... and because b needs to be in Z, i need to prove that 6 doesn t divide 10k+1

[woeful ibex]

how to do that ?

(if you don't want to read everything, how to prove that 6 doesn t divide 10k+1)

i mean, i could do it for k = q, k = q+1, k=q+2, etc... no ?

[obtuse pebble]

@woeful ibex Has your question been resolved?

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[timid silo]

Are 1 and 2 true?

[obtuse pebble]

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[timid silo]

nvm

[obtuse pebble]

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[lament sorrel]

How would I go about thinking out the minimum radius of convergence conceptually?

[lament sorrel]

Here is my work for this one, but I have a feeling that I'm overcomplicating it

<@&286206848099549185>

[obtuse pebble]

@lament sorrel Has your question been resolved?

[obtuse pebble]

@lament sorrel Has your question been resolved?

[lament sorrel]

.close

[obtuse pebble]

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[obtuse pebble]

@drifting ether Has your question been resolved?

[obtuse pebble]

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[crystal stream]

Hi there

[crystal stream]

I am back with another Calculus question 🙂

What I don't understand is: when a function is defined from R -- > R

and I need to prove a specific deleted neighbourhood

Doesn't it mean that it is not from R --> R

?

like

i dont get it

[spiral knot]

If you input a real number, the function output a real number

That’s all it says.

[crystal stream]

so

for instance

if I want to think of a counter example

I can choose whatever I want?

but it has to cover all R?

[spiral knot]

As long as you don’t use something like sqrt yeah

[crystal stream]

like

sqrt because it is not defined in x < 0?

[spiral knot]

because it’s not real in x<0 yes

[crystal stream]

So it means that for all functions in R --- > R

each and everyone of them

has a limit in x -- > -8

[spiral knot]

it also says increasing

[crystal stream]

which is not inf or -inf

[spiral knot]

yes

but it also says increasing so its derivative is never negative.

[spiral knot]

Wait no. There is a difference between -8 and 8-.

[crystal stream]

yeah yeah

I meant 8 -

its ok

[spiral knot]

ah ok

[crystal stream]

So if it is increasing

in every real number

doesnt it mean that it is defined for all numbers?

I am trying to construct a counter example

[spiral knot]

not necessarily.

[crystal stream]

but I can't think of a jump function

like

rationals and irrationals

[spiral knot]

wait I just looked some stuff up

It said it is R->R meaning it’s defined at all reals

it said that at the start

but yeah terrible wording. If this is work ask your teacher about what they mean by increasing and R->R

because there are some definitions that allow stuff like f(x)=1/(8-x) to be R->R and increasing and some don’t.

[crystal stream]

damn

I am so stuck at this...

Anyone can help?

<@&286206848099549185>

What they mean is that the function is defined in all x's in R @spiral knot

and that the output can be any number in R

[spiral knot]

ooooh ok thanks that clear so much up

[crystal stream]

I have to prove or disprove that the limit exist for all function that hold that

now

for first thought

[spiral knot]

well if it is defined for all R then it is defined for 8.

[crystal stream]

it looks correct

and since it is increasing

[spiral knot]

Does the function have to be continuous?

[crystal stream]

it cannot go up or down

because it has to come back and then it wont be increasing

[spiral knot]

because I can think of a function that is defined for all R but not lim(8-).

so does it have to be continuous?

[crystal stream]

Well

we actually learn what is continuous just today

and the homework is from last week

so I think it is not the point

but

I don't think it is incorrect

so I think anything will be accepted if it well defined

[spiral knot]

also another important question: does it have to be algebraic? (i.e. I can't just define a function as "f(x)=x for all x<=8, but for x>8 it is f(x)=1/(8-x)")

it has to be actual stuff

[crystal stream]

of course you can

[spiral knot]

I assume it does but idk

[crystal stream]

anything

We already gave tons of example of jumping functions

[spiral knot]

like?

or just the piecewise stuff as jumping?

[crystal stream]

piecewise yeah