help-38

[gritty ether]

I'm stuck after I factor out r^2 and convert that to x^2+y^2, the 2theta is tripping me out

Thanks in advance to whomever takes the time to look at this! :)

[lament reef]

sin(2 ϴ) = 2 sin ϴ cos ϴ

that's an identity

[gritty ether]

🤔

lemme try

dang

u are right

ok

still trying it out

[obtuse nexus]

Isn’t that just a circle

[gritty ether]

Yeah I'm still kinda stuck

Yeah its a circle, I'm just struggling with converting it to rectangular form.

[obtuse nexus]

I would graph it onto a polar grid and then simply find out the equation as a circle

Or

U might be able to just substitute x^2 + y^2 for r^2

[gritty ether]

The part where I'm struggling is to get it in forms of just x and y, but I end up getting stuck with the 2theta and having issues with getting rid of it.

[obtuse nexus]

You don’t need to do that

Just remember that r is sqrt(x^2+y^2)

And also the thingy for theta

U don’t need to change the trig yet

Into that

U might need some inverse trig for the theta part iirc

[gritty ether]

Hmmm

Yeah cause ultimately I'm still left with the sin2theta that I have absolutely no idea how to work with 😅

It's like no matter what I do I get stuck with that

[obtuse nexus]

Well sintheta is y rt

[wraith hinge]

well sintheta=y/r and costheta = x/r

[gritty ether]

dude

youre right

hold on

lemme try that

Highly doubt this is right but imma fuck around and find out

yeah nope 💀

Yeah I'm mad stuck

[wraith hinge]

ur not?

except for the last line

and maybe u could skip some steps if u didnt replace r^2 with x^2+y^2 immediately in line 1

[gritty ether]

😭

2yx=38 is my resultttt

Fried

[gritty ether]

The format is supposed to be in x^2 + y^2 = r^2

[wraith hinge]

send the full question pls

[trim jolt]

@gritty ether Has your question been resolved?

[gritty ether]

ye

[wraith hinge]

why you think it should be a circle?

[gritty ether]

im gonna be honest, i dont know.

all we've done is problems 😭

[wraith hinge]

ok so its not

its y=19/x

[gritty ether]

Huh lemme try dat

i was definitely going insane for not resting enough. it really was that easy huh

💀

villegas, thank you so much

um how do i close haha

.close

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Use .reopen if this was a mistake.

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[tough moth]

Where did I mess up?

[wary basin]

I'm a bit confused about your second step

where did the denominator go

[tough moth]

Denominator is the bottom

(x^2-1)

[wary basin]

well yes

but how did you go to x^2/1 + x/1 -2/1

[tough moth]

I thought that's all you can do :o

Since x alone is 1x ... I thought...

[exotic tundra]

that's you factoring the numerator right?

[wary basin]

you should factor numerator and denominator

both of them

[tough moth]

Ima be super fr with you

[wary basin]

there should be a common factor to cancel out

[tough moth]

I have 0 clue what numerator is

Or denominator

Only top and bottom

[exotic tundra]

numerator is the top part of the fraction

[tough moth]

Okay okay thank you

[exotic tundra]

I found it confusing too lol

[tough moth]

X^2

X-2/-1?

[exotic tundra]

your bottom factoring is wrong

bc (x-2)(x+1) ≠ x^2 -1

[tough moth]

Hmmm

Confusing guhh

Step by step?

[exotic tundra]

so, what's the factorization of x^2 + x - 2?

[naive swallow]

You can't separate a variable from a number like that. They go together like a set. You can only factor like that. If they cancel out with the numerator, they always cancel out together (in a parenthasis).

But keep going. I was just mentioning that general rule.

[tough moth]

I think I get it

[naive swallow]

You got this. We're here until the end with you.

[tough moth]

Good setup so far?...

[tough moth]

Might be a bit

I suck at math @.@

[naive swallow]

That's okay with me.

So back to the mighty Sepdron. What's the factorization of x^2 + x - 2?

[tough moth]

I'm avoiding the question cuz I have no clue ;~;

[naive swallow]

Do you know how to check once your done? What the four letter acronym is you need to know to check?

[tough moth]

I know

Pemdas

And

That's it

[naive swallow]

You use it when you BBQ sometimes...

[tough moth]

Gril

Uh

[naive swallow]

It's silver and shiny

[tough moth]

Tong

[naive swallow]

It starts with an F

[tough moth]

Ffff

[naive swallow]

It also means to [mess up] a plan

[tough moth]

Fact

@.@

[naive swallow]

FOIL. First. Outer. Inner. Last.

[tough moth]

[naive swallow]

It's how you multiply when you multiply two binomials. So (x + a) (x + b)

Try following foil to solve it

what are the two first letters of each?

[tough moth]

FO

[naive swallow]

First:
What are the two First letters of (x + a) and (x + b)?

[tough moth]

Idk ;~;

[naive swallow]

Does that help?

[tough moth]

Idk what Foil is ;~;~;~;

[naive swallow]

This

[naive swallow]

I'm walking you through the example

One step at a time

I'm here

Just remember to breathe

[tough moth]

O.o

Okeh

Uhm

A and b are both... X

x^2 and x

[naive swallow]

a and b are a and b. They are their own letters. They aren't equal to anyone else (just yet).

But the first two letters are x, right?

They are the first ones you read for each parenthesis. So you multiply them first. And they do multiply to x^2.

That's the F in FOIL.

The Outer is the outer numbers. So what are the Outer numbers of (x + a)(x +b)?.

[tough moth]

Uhm...

[naive swallow]

(or the outside sides)

[tough moth]

-2

That's the 3rd parenthesis

c

Heh

a*c to get theeeee

Number

Sooo 1 (x^2=1) * -2

Which is still -2

ALRIGHT

I might have it...

[naive swallow]

I'm sorry, you lost me.

[tough moth]

I lost myself tbh ;~;

[naive swallow]

Lets use your homework. (1)x^2 (+1)x (-2) right? This is skipping forward a bit but you add the inner and outer numbers together in the end of FOIL.

What two pairs of numbers multiply to get 1?

[tough moth]

1*1...

Right?

[naive swallow]

That's one pair. What's the other one?

[tough moth]

The only other one I think is... -2?

[naive swallow]

Nope. -1 * -1.

But since we are talking about the first set of numbers we dont usually use the double -1s. It's important to remember that for the other three numbers so remember that a negative times a negative is a positive.

How can you get -2?

[tough moth]

-1*2! c:

[naive swallow]

And?

[tough moth]

And?..

[naive swallow]

(1 times anything is itself....)

[tough moth]

I thought that was how you got -2 tho :o

1*-2

[naive swallow]

YES

So that says the numbers that make First and Last are either of those two pairs. And we never really use -1 * -1 so we can just put 1 * 1.

It's either (1x - 1) (1 x + 2) or (1x + 1) (1 x - 2).

Now we use the center number of x^2 + 1x - 2 to figure out which one it is

Since you add the Outer and Inner numbers, you just have to figure out which one results in a result of + 1

[naive swallow]

Here's a reminder of what Outer is.

[tough moth]

I

Confused

I dunno this Foil stuff ;~;

I barely just learned all this simplified stuff

[naive swallow]

If you've ever played Jenga, you are missing the blocks that connect the tower to the table. You need to learn FOIL and multiplying binomials before you can answer this question. You can't go backwards (simplifying) before you learn how to go forwards (multiplying, FOIL). Do you know how to multiply a(b+c)?

[tough moth]

Yes

Ab and ac right?

[naive swallow]

And I've been a tutor for twenty years at this point (jeez never make me say this again). I've done it and seen it done. It just takes more time than one night. It's the only reason why I'm asking.

Great!

You aren't that far behind then.

You need to practice FOIL. This question will feel a lot more natural and may feel so much easier once you learn it.

[tough moth]

Bwahhh alright alright

So

Alright

[naive swallow]

You can go back to the easy example or back to your homework. First one is easy mode. Second one is harder but is quicker if you get it (not quicker if you don't).

[tough moth]

I try homework

I'm already late on this lols

[naive swallow]

That's here

So the first and last of an equation is the A and C, right? And the B is adding the outer and inner together.

[tough moth]

Maths sucks

[naive swallow]

(x + a) ( x + b)
(x * x) + (x * b) + (a * x) + (a * b)
F + O + I + L
x^2 + bx + ax + ab
x^2 + (a+b)x + ab

Basically

Factoring is figuring it out backwards

What makes (x*x)?

What makes (a*b)?

Now it's what could (a+b)x?

And it's narrowing down the options like a detective

we got the (1x*1x) already

Now it could be (-1 times 2) or (1 times -2)

[tough moth]

Whatever it is tho

[naive swallow]

Both make (a*b)

[tough moth]

Doesn't it gotta equal b?

If you add it

[naive swallow]

That's the (a+b)x part!

Exactly!

[tough moth]

If you have -2 as a*c

And uhm

1 as b

It'd be

-1 * 2

Cuz that equals -2

But -1+2 will equal

[naive swallow]

So when you put it formally what is it?

[tough moth]

1, for b

[tough moth]

Something something...

(X-1)(x+2)

[naive swallow]

Yup

That's the numerator

[tough moth]

And the denominator...

[naive swallow]

Same thing. What's A, B, C?

[tough moth]

X^2-1 tho...

Could you square root it to get rid of the ^2?...

[naive swallow]

First rule

[tough moth]

Crap

[naive swallow]

If that pesky -1 wasn't there then sure but it is

[tough moth]

Bwahhhhhh

A=1 b= -1?

[naive swallow]

Close. B is the number before x

So that would make it x^2-x

[tough moth]

Hmmmmmm

So then what's c?

[naive swallow]

C is the one that has no x at all. The constant.

[tough moth]

Basic numbahhh

[naive swallow]

Mmhmm

So what is A, B, and C in X^2 - 1?

[tough moth]

Oh

A= 1 B= 1 C= -1?

[naive swallow]

Closer! That is x^2+x-1

[tough moth]

Hmmmmmm

So then what's b?

[naive swallow]

Is there one?

[tough moth]

No...

So... 0?

[naive swallow]

Yup!

[tough moth]

1*-1 is -1

1*-1 equals -1

1+-1 equals

0

[naive swallow]

Exactly!

[tough moth]

It's x^ so...

[naive swallow]

So how do we write it formally?

[tough moth]

(X-1) (x+1)

[naive swallow]

Yes!

Amazing!

So write the fraction

[tough moth]

Wait tho

If I have (x-1) (x+2) up top

And (x-1) (x+1) on bottom

Typo sorry

Would the two (x-1)'s cancel each other out?

[naive swallow]

They would!

[tough moth]

So

X+2/x+1

:o

[naive swallow]

Wonderful!

See. All you had to do was do it and you got it so quickly.

[tough moth]

Small second of happiness before I realized I got about... 8 questions left ;~;

[naive swallow]

You're actually not bad at all. You just syke yourself out.

[tough moth]

Thank you thooo..

[naive swallow]

You definitely got this.

[tough moth]

Blehhhh I fell eep in class

[naive swallow]

Well now you understand better.

If anything, come back here!

[tough moth]

Thank you ;~;

[naive swallow]

Practice FOIL!

[tough moth]

Foillll okehhh...

8 questions left...

Gon be a long night

[naive swallow]

You got it down

[tough moth]

Bwahhh bet

Ima leave this open in case

[trim jolt]

@tough moth Has your question been resolved?

[tough moth]

Nah yet not u.u

Nah yet.

[tough moth]

What makes -25

And also makes zero if you add it

5 @.@

[naive swallow]

What ma--And?

[tough moth]

Mathand

[naive swallow]

5 and?

[tough moth]

-5! c:

[naive swallow]

yup

[tough moth]

I feel like I can just

Glide through this now

[tough moth]

Sorry to ping

But

[naive swallow]

See!

[tough moth]

If I have (x+4)(x+4) as my nominators

Do I leave them like that orrr?

(X+4)^2

[naive swallow]

Either way but you may be able to cancel something later

[tough moth]

Hmmm it's final solution

[naive swallow]

I'd leave it expanded because it's easier to write

[tough moth]

1 :o

I get one more guess guh...

I'm DUMB

The greatest multiple of 12 that could equal 8 wasn't 4*4

It was 6*2 ;~;

So

[naive swallow]

?

[tough moth]

I can divide by x

To get rid of 42x

And make it 42

To do that whole

a*c stuff

Right?

[naive swallow]

You can factor x out of both equations, right? Because a(b+c) = ab+ac?

[tough moth]

:O I think so

[naive swallow]

So if there's an x without a number on top and on bottom, then that rule doesn't apply!

[tough moth]

So I can't factor x out even if they have it in common?

[naive swallow]

No you can factor it out because there's no pesky single number

so you CAN do it

[tough moth]

OHS

Sorry sorry

1:26 rip

[naive swallow]

Its okay it's 03:26 for me. I understand. Lol

[tough moth]

Multiple of 42 that equalssss

13

Hmm...

7*6!!

[naive swallow]

Exactly~

[tough moth]

Is that the biggest tho?

Denominator is

(X^2+6x)

A= 1 b= 6 c= 0?

[naive swallow]

There's a batter trick for that one

but also yes

[tough moth]

Better trick?

[naive swallow]

What can we do when C=0?

[tough moth]

When c=0? I dunno :o

[naive swallow]

What does it mean when C=0?

[tough moth]

Meansss it equals nothing?...

[naive swallow]

What does?

[tough moth]

C

[naive swallow]

What's C in the equation?

[tough moth]

Uhms

I wanna say 0 but I fear the repitition is annoying @.@

[naive swallow]

Actually that's technically the right answer

Because C is the constant. The number of the equation. So C is the O of the equation.

[tough moth]

Yah yah yahhh

Do I still doooo

A*c?

Then -1*1 for the variables..

[naive swallow]

So what can you do when you don't have a number in an equation?

[tough moth]

Substitute it?

Orrr

Well

Yah...

[naive swallow]

You're allowed to scroll up

[tough moth]

I dunno ;~;

Just 0..

[naive swallow]

here

[tough moth]

Just put an x in C's spot?...

[naive swallow]

no need

[tough moth]

Then wot do?

Cuz I would just

Square root the x^2 and 6x and see if I fail lol

[naive swallow]

you can factor an x out whenever there is a collection that doesnt have a pesky number (our any other letters)

[tough moth]

OH

(X+6) would be the variable then!

Right??

Or...

[naive swallow]

mmhmm

[tough moth]

LET'S GO XD

That means

(X+7) is only left

[naive swallow]

Was there an x on top too?

[tough moth]

On top of what?

Nuh

Just x+7

[naive swallow]

So how did you cross out the (x)?

[tough moth]

I had (x+6) on top and (x+6) on bottom :O

[naive swallow]

mmhmm

[tough moth]

So thoseeee did the thing!

[naive swallow]

mhmm mhmm

[tough moth]

Multiple of -12

That can add to 1

-3*4

[naive swallow]

yup!

[tough moth]

I hate

How some problems switch the letter

I'm so used to x or y

And my brain goes on autopilot XD

[naive swallow]

you can always solve it using x and then answer it in the right letter. Just remember to write it in the right substitution

[tough moth]

Lolll I never remember

Best of luck to me

Frick

[naive swallow]

?

[tough moth]

Forgor to multiply

[naive swallow]

or to simplify

[tough moth]

Yeh..

Can I multiply 5 by x?

[naive swallow]

Nope

Cant separate them

[tough moth]

Okay

Good

I got that..

[naive swallow]

Nope. You were closer in the screenshot

[tough moth]

BWAH

That's after multiplying

Haven't simplified yet

Yahhh?

[naive swallow]

which one?

[tough moth]

This

[naive swallow]

Nope

[tough moth]

;~;

I dunno wot do

[naive swallow]

write the equation with x^2-25 factored out

[tough moth]

Like that?

[naive swallow]

yeah. Now you can cross out now or make it one big one and then cross out

[tough moth]

Bam?

[naive swallow]

Nope. You either leave the bottom two as parenthesis or you... FOIL

[tough moth]

Bam?

[naive swallow]

try it!

[tough moth]

Submitting it?

[naive swallow]

yeah

[tough moth]

Woah!

It worked!

Thank you!

Got it with your helpppp aaaaa

[naive swallow]

[naive swallow]

Did you put in your final answer?

[tough moth]

X+3 mm...

Was wrong ~^~

[naive swallow]

Where is the x+3?

[tough moth]

Bottom

Denominator

[naive swallow]

So if 2 was in the denominator what would have been the answer?

[tough moth]

Yes

If it was alone...

[naive swallow]

if two is the denominator of a number what is the number?

[tough moth]

Uhm...

I dunnos

@.@

[naive swallow]

1/2

[tough moth]

Half hm...

What does that mean for x+3 then?

1/x+3?

[naive swallow]

yes. BUT we usually put (x+3) in parenthesis

[tough moth]

True cuz then then it's 1/x +3

Right?

[naive swallow]

mmhmm

[tough moth]

Guess wut...

Wait nvm

One question left...

Heck....

Help requested ;~;

[naive swallow]

I'm here

[tough moth]

Wait nevahmind

I read chat

Was wrong

Wait

PH

OH

IS IT CUZ

X-2 on bottom cancels out x-2 on top??

[naive swallow]

[tough moth]

Making it 1/(x+2)?

Whaaa nu

Still wronk...

[naive swallow]

Write it all out with all the factors before you cross them out

Then cross them out

so when you do something wrong you can backtrack

[tough moth]

All I got is this

[naive swallow]

How did you get the top?

[tough moth]

2x^2 - 4x

Divided both by 2x

Got 1x-2

[naive swallow]

Well you factored out 2x

[tough moth]

Yahhhh

[naive swallow]

So it should be out there but not cancelled out

[tough moth]

Greatest factor

[naive swallow]

You would need another 2x somewhere to cancel it out

So it disappearing is bad

[tough moth]

Oof...

So

Cancel out by

X?

To make it x-4

2x-4

[naive swallow]

It's like a see-saw. There has to be equal parts on either side for to work. a 2x on the top is the only thing to cancel out a 2x on the bottom. If not, then the 2x stays.

(2x)(x - 2)

[naive swallow]

It's factored out so it stays outside

[tough moth]

I canceled out here :o

[naive swallow]

oh man my eyes cannot see that lol

[tough moth]

Mine too..

Fixed heh...

[naive swallow]

nope. B is the one with the x

[tough moth]

Double fixed blahhh...

[tough moth]

That's how I got rid of the 2x tho

You don't do that?

[naive swallow]

I amended to x + 2. But that's how you factor out the 2x. But you can only cancel out the 2x if there is a 2x in the denominator.

[tough moth]

Oooh I get it

2x/x+2?

[naive swallow]

try it

[tough moth]

3 2 1

3

2

1

Yes

You keep what you divide by

I just remembered that

Thank you

Guess

What

[naive swallow]

What?

[tough moth]

3.98

I so happi

[naive swallow]

[tough moth]

THANK YOU

[naive swallow]

Good job

[tough moth]

I STARTED THIS STUFF

AT 10

IS 3 A.M NOW

[naive swallow]

good night!

[tough moth]

Bababababababa goodnight to you too

I hope you eep wells

Thank you tons ;~;

[tough moth]

Was all you

[naive swallow]

Not at all

I never told you an answer

[tough moth]

Whoops

I honestly sucks

I gotta wake up in 3 hours

Whatever I did it to myself ~^~

[tough moth]

Woah

At some points you did :o

Then I hardly knew anything so you walked me through it closely lol

Anywho, thank you!

I eep now

Then... Coffee in morning

.close

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[lyric basalt]

How do I find the local min? I already found that the global max is at E, and the global min is at T, I found the local max's which are C,E,S

[steep patio]

the method for finding local minima will be very similar to what you did to find local maxima

[lyric basalt]

I know that T has to be apart of the group since the global also counts for the local, and R is also a local, but I'm unsure about A,B,D

[steep patio]

as we have allowed C as a local maximum, I think we can use the same logic to say that both B and D are local minima

As they "sit above" all of the points near them

If we are calling T a local minimum, do you think we should get the analogous result that A is a local maximum?

And similarly for C

[lyric basalt]

Based on the fact that E counted as a local max, I made the assumption that T can be considered a local min, given that A wasn't considered a local max I think it's safe to say that it won't be considered for a local min either

<@&286206848099549185>

<@&286206848099549185>

<@&286206848099549185>

<@&286206848099549185>

<@&286206848099549185>

.close

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[green palm]

I'm clueless please help

[green palm]

<@&286206848099549185>

[trim jolt]

@green palm Has your question been resolved?

[past cargo]

what do you know about triangle AED

[pseudo dawn]

Even this is wrongly simplified

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[dusk ivy]

hi

[trim jolt]

Please don't occupy multiple help channels.

[dusk ivy]

how do i solve this

[meager bloom]

You already have one channel open , I would suggest close one

[dusk ivy]

i cant find the other one thats why i opened this

[acoustic topaz]

bro it's 2 channels down

[dusk ivy]

.close

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[marsh forum]

Prove that if $W_1$ is any subspace of a finite dimensional vector space $V$, then there exists a subspace $W_2$ of $V$ , such that $W_1 \oplus W_2= V$

[solid kiln]

A dense set

[rancid helm]

why spam ping

[round imp]

it was separated by 10 min intervals

not spam

[marsh forum]

We propose that $\beta_1$ is the basis of $W_1$ and that $W_1$ is a subspace of $V$. We also know that the basis of a subspace, can be extended to the basis of the larger vector space. We call this basis $\beta$. We now propose that the basis of a $W_2$ that satisfies this condition, would be $ \beta \setminus \beta_1 = \beta_2$. It follows that $\beta_1 \cap \beta_2 = \varnothing$.We have also previously proven that if $\beta_1; \beta_2$ be disjoint basis for subspaces $W_1$ and $W_2$ respectively.Then that if $\beta_1 \cup \beta_2$ is a basis for $V$, then $V = W_1 \oplus W_2$. From this it follows that if $W_1$ is any subspace of a finite dimensional vector space $V$, then there exists a subspace $W_2$ of $V$ , such that $W_1 \oplus W_2= V$

[solid kiln]

A dense set

[marble wharf]

first line is not true

well ok second line corrects it

but it should not be written that way

wdym by "appropriately"

[marsh forum]

Like say I'm considering the direct sum of the xy cordinate system, and the z-axis

[marble wharf]

not examples

[marsh forum]

Well, then how do I write it

[marble wharf]

well thats the question of the proof

how can you choose beta and beta_1 in such a way

[marsh forum]

I chose $\beta$ and $\beta_1$ such that $\beta_1 \subseteq \beta$

[solid kiln]

A dense set

[marsh forum]

We propose that $\beta_1$ is the basis of $W_1$ and that $W_1$ is a subspace of $V$. We also know that the basis of a subspace, can be extended to the basis of the larger vector space. We call this basis $\beta$. We now propose that the basis of a $W_2$ that satisfies this condition, would be $ \beta \setminus \beta_1 = \beta_2$. It follows that $\beta_1 \cap \beta_2 = \varnothing$.We have also previously proven that if $\beta_1; \beta_2$ be disjoint basis for subspaces $W_1$ and $W_2$ respectively.Then that if $\beta_1 \cup \beta_2$ is a basis for $V$, then $V = W_1 \oplus W_2$. From this it follows that if $W_1$ is any subspace of a finite dimensional vector space $V$, then there exists a subspace $W_2$ of $V$ , such that $W_1 \oplus W_2= V$

[solid kiln]

A dense set

[marble wharf]

basis can be extended
thats what I wanted to hear

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[marsh forum]

Thanks

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[terse portal]

if

[terse portal]

i want to put cos^3x

into dx/dt

do i need to seprate them?

so i get cos^2x*cos x?

or is there a diffrent way to do it?

u guys mean chain rule?

[dapper swift]

yes

note that $\frac{dx}{dt} = \frac{dx}{dy} \frac{dy}{dt} = \frac{1}{dy/dx} \frac{dy}{dt}$

[solid kiln]

south's secret twin brother

[terse portal]

yes i understand that but to get there

do i need to use u and du?

[dapper swift]

no so you just find dy/dx from y = cos^3 x normally, using the chain rule

and then the dy/dt bit should mean you multiply at the end by dy/dt

don't forget it's 1/(dy/dx) also

[terse portal]

okay let me see if i solve it

[dapper swift]

https://math.stackexchange.com/questions/3165409/related-rates-calculus-confused-about-what-dx-dt-dy-dt-and-dx-dy-mean

is this related rates btw?

[terse portal]

i have a different question but its similar

i have both x=cos^3t and y=sin^3t

and i need to find the curve length

[terse portal]

i think this type of question is before uni right but somehow in this math course its half linear algebra and half integration

but calculus have integrations in it so yea

[dapper swift]

ah so then you need $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$, chain rule

that makes so much more sense now

[solid kiln]

south's secret twin brother

[dapper swift]

and then replace $dx = \frac{dx}{dt} dt$

[solid kiln]

south's secret twin brother

[terse portal]

no

[dapper swift]

that's how you find the arc length of a parametric function

[terse portal]

the rules is ds= sqrt of (dx/dt)^2 + (dy/dt)^2 dt

so i have done it like this

i let u= cos t

and then x= u^3

so dx/du = 3u^2

and du/dt = -sin t

[dapper swift]

yeah that's equivalent to what I did

[terse portal]

and then simplify everything and i got -3cos^2t sin t

[terse portal]

uhhh okay i see

[dapper swift]

yeah so $\frac{d}{dt} \cos^3 t = 3 \cos^2 t \cdot (-\sin t)$

[solid kiln]

south's secret twin brother

[dapper swift]

that's correct

now do the same for the other one

[terse portal]

yea

[dapper swift]

y = sin^3 t

[terse portal]

should be the same thing but other wise

so 3 sin^2t * cos t

but what i wonder

in this type of question

is the chain rule always relevant?

or can there be other rule that need to be applied ?

[terse portal]

or i am wrong?

[dapper swift]

yep that's correct

[dapper swift]

yeah, so whenever you have a composition of two functions, you need the chain rule

so y = sin^3 t is f(g(t)) where f(t) = t^3, g(t) = sin t

basically the chain rule is the most frequently applied rule

it's just everywhere

[terse portal]

well not for linear algebra xD

[still rose]

xd

[terse portal]

i have a another question which i wanna understand fast

[dapper swift]

oh lol, but that's a totally different area of maths

[terse portal]

yea in my opinion its much more fun and easier to understand

anyways

i have issue with divergence

i dont get the grasp of it

the 5 b is easy but a) in the other hand

i dont understand how to think

the only thing i understand is i need to seprate it

so i use A and B

[dapper swift]

oh well 1/(x^2 + 2x) < 1/x^2, and the integral of 1/x^2 from 1 to infinity is convergent

wait

there's the 0 to 1 part

[terse portal]

no?

[dapper swift]

yeah so 5a) also diverges, can't explain why

[terse portal]

uhh okay

[dapper swift]

if you really wanted to, use partial fractions to integrate the indefinite integral

you get $\frac{1}{2} \left(\ln(x) - \ln(x + 2) \right)$

[solid kiln]

south's secret twin brother

[dapper swift]

but then ln 0 is undefined, so diverges

yeah it says to find the value if it is convergent

might as well integrate

[terse portal]

yea uhh u are swedish xD?

.close

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[dapper swift]

nah neighbouring languages are close enough

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[graceful egret]

Im a bit unsure if im doing it wrong

im struggeling with part a)Decide P(x is more or 3)

i dont know where to start

like do i look for the x value? if so then i must do P(X=1)+P(X=2)+P(X=3)=0.10+0.20+0.10=0.40

right?

[thorn sparrow]

yeah

youre right

[graceful egret]

oh alr cool

[thorn sparrow]

you need help with the second one?

[graceful egret]

im struggeling with this one too

yes please

i have to explain why this is a real(?) equation

[thorn sparrow]

can you translate?

[graceful egret]

let me translate the text theres a bit one sec

[thorn sparrow]

here the second one is probably asking for the value of a, so you need to sum up all the probabilities P(x=t) ones and it should be equal to 1

itll form a linear equation and on solving that you should get your answer

[graceful egret]

There is 7 books, 4 childrenbooks and 3 thrillers
Lise picks 3 books for her trainride randomly.
She gets told that each possibility of picking 2 childrenboooks and 1 thriller can be solved with this equation:

Explain why this is true.

[graceful egret]

yus i did this:

P(2 childrenbooks and 1 thriller)=....

[thorn sparrow]

im assuming K(4,2) is choosing 2 out of 4 aka 4C2

ok so you want 2 children books and there are 4 of them in total, so ways to get any two of those is 4c2

[graceful egret]

yes i think k(4,2) is 4 childrenbooks but i only need/want to pick 2

k is combinatorics

if im not mistaken

im struggeling with understand why i add k(4,2)*k(3,1)

i believe i devide with k(7,3) because i only need 3 books and the total of the books is 7 combined

[thorn sparrow]

mb device died

[graceful egret]

nws nws

[thorn sparrow]

you are correct there

you also need 1 thriller at the same time so you multiply ways with which you can select 1 thriller out of the 3 present

hence the 3c1

[graceful egret]

oh so im just multiplying my possibilites of picking the books with eachother to find the combined possibility of finding wanted books to then devide it by the total books needed?

[thorn sparrow]

yeah

[graceful egret]

oh...

that is so simple xd

[thorn sparrow]

so whenever you need 2 things together (they use an and) you multiply

and whenever theres a choice like a or b you add

[graceful egret]

and the reason i write p(2 childrenbooks and 1 thriller) is because the 2c and 1t is essentially my x value i want to find?

oooh

thank you C:

[thorn sparrow]

yeah

theys why they used and there

if theyd used 2 children or 1 thriller then the answer wouldve been different

(if you wanna know what it would be just ping!)

[graceful egret]

imma bite im curious C: i would like to know

[thorn sparrow]

im gonna assume we choose 3 books in total still

oh wait nvm it wont work with 2 children or 1 thriller with the current data

[graceful egret]

oh alr XD thank you the help tho

[thorn sparrow]

!done

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[wraith hinge]

Small doubt

[wraith hinge]

Is this the proof for Euler's relation

?

[dapper swift]

yeah

[wraith hinge]

Okay,thankyou

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[undone sierra]

I have a proposal that might contradict Virgagnon's theorem.
Just a heads up: Virgagnon's theorem states that if you take the midpoints of the four sides of any convex or non-convex quadrilateral, they will always form a parallelogram.
I have this counterexample:
A non-convex quadrilateral ABCD such that AB = CD and (AB)//(CD). I chose, I, J, K and L the respective midpoints of [AB], [BC], [CD] and [DA].
Using the previous information, ABDC would be a parallelogram thus his diagonals [AD] and [BC] would intercept at their midpoints. If so, then J and L would be on top of each other. So, [BC] has J as midpoint.

[undone sierra]

We can prove then that K, J and A are colinear by midpoint theorem in triangles ABC and BCD and so, this cant be a parallelogram

is there anything that can be considered false in this?

[trim jolt]

@undone sierra Has your question been resolved?

[undone sierra]

<@&286206848099549185>

what about non convex quadrilaterals

ur statement wouldnt necessarily imply then

i think they call it a crossed quadrilateral

u said in ur world, when u call ABCD a quadrilateral

this means each side intersects the other side at one point

for example [AB] and [AD] intersect at A

is this what u meant?

in non convex quadrilaterals this is not a necessary case

concave quadrilaterals are not alqays in that form

always*

convexity and concavity are a bit too technical

heres a simple definition of a concave quadrilateral

it helps indicate them

if u take two points inside the quadrilateral and connect them with a line

if that line somehow ends up outside the quadrilateral

its concave

if the line never does, its convex

convexity and concavity was a whole huge lesson for us

i couldnt even keep up

thats not really a part of what defines a quadrilateral

whats a quadrilateral by ur definition

wdym do not start with a quadrilateral

literally it is a quadrilateral what i stated

some call it crossed quadrilateral too

the virgagnon's theorem works too with a lot of examples of the type of quadrilateral i proposed too

i js dont see why it works with the one i stated

indeed it is

in quadrilateral ABCD

angles are ABC, BCD, DAB, CDA

u cant really tell first off

also

js google it

Oh i think u misunderstood

when i mentioned abt angle ABC

i was talking abt its exterior

not interior

u can find everything online abt crossed quadrilaterals

youll understand urself

thanks for ur time though, man

i cant deny that ur right, this is very old and probably couldve been discussed before

i wouldnt be surprised if someone proved me wrong

but u dont seem to convince me where im at fault though

[final coyote]

sorry for wasting your time.

[undone sierra]

wouldnt call it a waste of time so thanks for this discussion

[undone sierra]

<@&286206848099549185> ?

[undone sierra]

<@&286206848099549185>

[wraith hinge]

yes

hello

[undone sierra]

hi

[undone sierra]

this

[trim jolt]

@undone sierra Has your question been resolved?

[undone sierra]

<@&286206848099549185>

[hexed dragon]

Reading your question

Your proposed counterexample to Virgagnon’s theorem, involving a non-convex quadrilateral ABCD with AB equal to CD and AB parallel to CD, seems to misinterpret the geometry of the situation. While you suggest that J and L, the midpoints of sides BC and DA, would coincide because the diagonals of the quadrilateral meet at their midpoints, this conclusion is incorrect. Even though ABCD may have special symmetry, the midpoints of the sides will still form a parallelogram according to Virgagnon’s theorem. The confusion likely comes from misunderstanding the positions of the midpoints, but this doesn’t provide a valid counterexample. Virgagnon’s theorem holds for both convex and non-convex quadrilaterals.

Virgagnons theorem still holds

@undone sierra

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[fair oracle]

is 2x√5 the same as 2√5 x?

[left oriole]

sure, multiplication is commutative

[fair oracle]

ok thx

.close

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[rugged cairn]

https://youtu.be/vtv_VSw_nhg?si=QVYB9JbsSBUA_CNd

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[wraith hinge]

,rotate

[wraith hinge]

1-2 a

what do they mean by quotient of length?

[marsh forum]

$\frac{|z_1|}{|z_2|}$

I suppose?

wait

oop

[wraith hinge]

they already defined that z is z1/z2

[marsh forum]

opps

my bad

[solid kiln]

A dense set

[wraith hinge]

oh

like

we have to get z1/ z2 into the form a+jb and then find its magnitude

[wraith hinge]

and we have to show that it is equal to this ?

[marsh forum]

I suppose so?

[wraith hinge]

👍

[marsh forum]

I'm not too sure

[wraith hinge]

i guess that is what they meant

i thought by quotient we have to divide z1 by z2 and write down the "quotient"

which sounded dumb

so..

thank you

.close

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[wraith hinge]

.reopen

[trim jolt]

✅

[wraith hinge]

1-3

i dont understand what they are asking

[wraith hinge]

should i start by taking z as a+jb and eitheta as costheta + j sintheta

?

[marsh forum]

show that $ze^{i \theta}$ is equivalent geometrically to rotating a vector $z$ by theta degree

[little glen]

@wraith hinge

[solid kiln]

A dense set

[little glen]

say you have a vector z = re^i(beta)

[wraith hinge]

let me try xD

[little glen]

now multiply it by e^itheta and see what you get

[wraith hinge]

alr,did the same

thank you

thanks wai

.close

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[wraith hinge]

.reopen

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✅

[wraith hinge]

boy im not getting a single question today

1-4 a

"explain the meanig of this relation in a vector diagram"

[marsh forum]

if $dz$ stands for the differential of $z$ here ,I swear I'mma explode

[solid kiln]

A dense set

[wraith hinge]

i understand that jz means rotating the z vector 90 degrees in anticlock wise direction

[wraith hinge]

xD

it is

idk how to represent dz

[marsh forum]

I don't think I know enough to help, sorry

Are you sure $j$ isn't just a real number here

[solid kiln]

A dense set

[wraith hinge]

no it represents sqrt -1

and the way they showed that is quite funny

[marsh forum]

Bruh. Using j instead of i is diabolical

[wraith hinge]

engineering textbook

[marsh forum]

I feel like you could just differentiate both sides wrt theta to start

[wraith hinge]

i showed tht

that

but not sure how explain it usiing a vector diagram

[marsh forum]

Hmm, that's an interesting question

[wraith hinge]

dz

is that a small part of vector z ?

?

🤷‍♂️

[dry copper]

dy = f '(x) . dx means that when you have a point (x0, y0) and you apply a little change to x0, like you add a dx to x0

the new value of y will be automatically y0 + dy

and that amount of change in the function "dy" is related to the amount of change in the x "dx"

similar here, imagine you got a vector z0 with constant length A and making an angle θ0 with +x axis

if you rotate it counter-clockwise by a small angle dθ

automatically the vector will change

becomes z0 + dz

dz being amount of change due to the change in the angle "dθ"

and the relation says that z0 will change by exactly jz * amount of change in θ

[little glen]

i was trying this

it made no sense to me

[dry copper]

relation tells you that if you grab a cmplx number and you change its angle by a small amount

the new cmplx number will simply be the original + dz

dz is the change that occurs due to the small change in the angle

drawing is not to scale, dθ is very small, infinitesimal

the smaller, the more accurate the relation holds

[wraith hinge]

one thing

in the relation

dz = j z d (theta)

dz and d(theta) are vectors right ?

[dry copper]

no, dz is a vector only

dθ is scalar

[wraith hinge]

okay

[dry copper]

kinda difficult to explain lol

hope you get it

[wraith hinge]

so change is z is perpendicular to z

thats hard to visualise

i kinda get it

i will come back to this later

thank you

[dry copper]

you can do this

,, dz = jz d\theta = e^{j\frac{\pi}{2}} \cdot A , e^{i\theta} , \cdot d\theta = A , d\theta \cdot e^{j(\theta + \frac{\pi}{2})}

[solid kiln]

Milena

[little glen]

the angle is so small that the arc is almost a straight line

[dry copper]

so yes, perpendicular

[little glen]

kind of like those derivations in ray optics

cool question

Thanks @dry copper

[dry copper]

yw

[dry copper]

yw

yeah, i liked this question

[little glen]

anw gotta head out for school now

[dry copper]

i knew it, but i never actually tried to visualize it

i tried it just now

[wraith hinge]

exactly,that would make dtheta vector right

if that was the case,it would make sense to me

[little glen]

d(theta) is NOT a vector

it's a scalar qty

angle is not a vector qty

[dry copper]

the change in the angle is what made the vector change from z0 to z1

dθ scalar

[wraith hinge]

oh

mb

mb