help-17

[bleak prawn]

and they gave me cos(theta) and u • v so they want me to use it somehow

But I dont really know what else to do

[untold pike]

U can compute magnitude of v vector from given information

[bleak prawn]

Oh wtf how did I not notice

💀

[untold pike]

After u do so to compute |u+v| u just square it as square of a vector is equal to its magnitude

[bleak prawn]

IIvII is -12

[bleak prawn]

I dont follow sry

[untold pike]

|u+v|² = |u|² + |v|² + 2*u•v

[bleak prawn]

Oh

Is that a rule?

[untold pike]

Yes

[bleak prawn]

Or did you work it out?

Ah ok

So its just something to remember, you didnt get it algebraicly or something?

[untold pike]

Careful in 2*u•v it's a dot product not simple multiplication

[bleak prawn]

Ah true

[untold pike]

Yes u remember it

There are proofs but I can't go into them

[bleak prawn]

No worries haha

Thanks for your help 😇

❤️

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[lilac plaza]

Hello

[lilac plaza]

Is it possible to have the third integral boundaries as functions also?

Then it would give us a function not a volume though

[bitter pilot]

From what I know the inner integral has bounds that depends on the outer integration

So I'd say basically no

[lilac plaza]

Would that mean it is also not possible to integrate a sphere?

[lethal plank]

That doesn't need the outer bounds to be a function

The idea here is you are essentially parametrizing your volume

So at some point you need something independent to describe the shape

Since the shape is fixed

You expect to get a number out of the integral

If the outer bounds were a function

You would get the result as a function of the variable which the out bound is a function of

@lilac plaza

[lilac plaza]

Hmmm

Make sense

Thank you guys

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[lethal plank]

Np

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[bleak prawn]

y = mx + c

-1 = m(0) + c

-1 = c

y = 7x/6 - 1 is the equation of the line

Ah I see

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[terse fiber]

quick question about fractions

[terse fiber]

when working with unknown variables such as 6a/6

it becomes a

but 6/6a becomes 1/a

what's the general rule for this?

is it because the denominator in the second fraction is a variable so it will always be a fraction?

[sly glade]

6a/6 is $\frac{6a}{6}$

[twin meteor]

Jafar / جعفر

[sly glade]

Which can be turned into $\frac{6}{6}a$

[twin meteor]

Jafar / جعفر

[sly glade]

So you can see the 6/6 is gone and it turns to just a

[terse fiber]

i see, so in the second fraction it can't be turned into a as the denominator is a variable?

[mental falcon]

a / 1 = a

[swift bane]

yeah

i mean

where do yuo want the variable to go

what's your expected output

it's in the denominator so it remains in the denominator

[terse fiber]

there is not really an expected output, im just learning on why 6a/a = a, but 6/6a = 1/a

[swift bane]

6a/a = 6 no?

a cancels out, they are seperated by multiplication

[mental falcon]

like Jafar pointed out, in the first one a is in the numerator, in the second one a is the denominator, it doesn't change position with either cancellation when you cancel out the 6

[terse fiber]

i meant 6a/6, my bad

[mental falcon]

*if a is nonzero

but that isn't what they're asking

[terse fiber]

To get it straight
6a/6 = a
6/6a = 1/a

the second one is where the denominator is a variable, so therefore the outcome is a fraction?

[mental falcon]

$\f{6a}{6}=\f{a}{1}=a$

[twin meteor]

∫oosh (lemonsaurus appreciator)

Jafar / جعفر

[sly glade]

It doesn't matter whether it's a variable or not, it's simply what happens when you have numbers in the numerator and denominator like that

[terse fiber]

ah alright

[swift bane]

plug in like

4 for example

for a

[terse fiber]

so a/1 is a simplified as it doesn't change the value of a

[swift bane]

i replaced a with 4

[terse fiber]

ah, so there is no logic behind it, it's just how it simplifies

the same would count in 4a/a = 4 because we can cancel out the a's?

[swift bane]

yep

but a cant be equal to zero as mentioned above

[terse fiber]

yeah alright, thank you 🙂

[swift bane]

just plug in a number to experiment

5s cancel out

[terse fiber]

that's probably a good help line indeed

thanks!

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[swift bane]

np 👍

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[hot flare]

The question says:
Determine the real numbers alpha and beta knowing that the function is continuous in R

[tawny hollow]

!status

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1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
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5. I have a question about someone else's work/solution.
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7. None of the above

[hot flare]

I really don't know where to begin 😭

[tawny hollow]

In that case, do you recall what it means for a function to be continuous?

for example do we allow sudden jumps?

[hot flare]

No

[tawny hollow]

Would you say that each case on its own is continuous?

i.e alpha^2 x + 3 etc

[hot flare]

The problem is not how to work with the continuous function but the problem is this type of questions if yk what I mean

[tawny hollow]

Uh sure, thats what im trying to help you with; the above is a crucial step

Hello? @hot flare

[hot flare]

I'm trying to understand what do u meant here

[tawny hollow]

im asking whether each piece of the function is continuous, take the first piece for example alpha^2 x + 3

would that be a continous function on its own?

[hot flare]

Yes
Isn't the lim x>1 equal to f(1)?

[tawny hollow]

sure, but im also taking about all other x's

This is crucial i think because what you might notice is that we have 3 pieces of continuous functions, and what would make them all toghter continuous you think in g?

Since g is basically just gluing these together, where do you think we would need to focus our attention on to ensure continuity of g?

[hot flare]

When X is between 1 and 3 ??

[tawny hollow]

Well if x is in-between 1 and 3 we have a continuous function

namely (8-beta^2 x)/(2x - 1)

wouldn't you agree that this piece is continuous?

[hot flare]

Wait if it's not the case where we should focus our attention then ??

[tawny hollow]

Maybe the points where we go from one piece to the other?

[hot flare]

the jump occurred in 3 and 1 right?

[tawny hollow]

Possibly, for certain alpha and beta yes, so if by assumption g is to be continuous, this (might) give us a way to determine alpha and beta

We want to ensure that the limit exists at x=1 and x=3

and is equal to whatever g(1) and g(3) is resp.

[hot flare]

But first how by just knowing these points
We could determine alpha and beta?

[tawny hollow]

What do you mean?

[hot flare]

I meant that if we ensure that the lim exists at x Equal to 1 and x Equal to 2 and it is continuous how is that going to help with the determination of alpha and beta??

[tawny hollow]

This puts a constraint on what alpha and beta could be, possibly so much that theyre unique

[hot flare]

AHH kkk

[tawny hollow]

However, nothing is in general stopping there from existing multiple alpha and beta, the question only assumes g to be continuous

so if we happen to get multiple alpha and beta, this doesnt mean we're in trouble. That just means there exists more than 1 solution

[hot flare]

Ok I get it

[tawny hollow]

Cool!

so try and see if you can make the limits at x = 1,3 exist

one trick is to make the left and right limits be equal

[hot flare]

Kk let me try

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@hot flare Has your question been resolved?

[hot flare]

Can u explain it more??

Do u mean the lim of the right of 1 and the left of 1

[tawny hollow]

Yeah

In most cases that trick suffices

you want them to be equal eachother

[hot flare]

So then the function is going to be continuous on 1 right

At this point I'm just doing nothing 😔

[tawny hollow]

So what you can do is just input 1 on both pieces

the one which are next to eachother that is

and set them equal

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[hot flare]

I think I got it

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[white scaffold]

I thought that if you rewrite the denominator as (s+3)^2 + 1, you could just use the inverse of the first translation theorem. Why is that wrong?

[gaunt sparrow]

Because the numerator would have to be s+3 as well to use the shifting property

$\frac{s}{(s+3)^2 + 1}$ is not in that form

[twin meteor]

Azyrashacorki

[white scaffold]

I see that makes sense, thanks

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[honest tinsel]

so the final answer is b = 4 because you use the period of pi/2 to get the answer, what im not understanding is idk how to explain it but the whole calculation at the bottom portion below "now, solve for b" like where did they get b(pi) from, if someone could explain it would be greatly appreciated!

[honest tinsel]

im so confused as to how pi/2 = 2pi/b becomes b(pi) = 4pi

[tranquil trellis]

cross multiplication

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[honest tinsel]

Meaning? Could you explain

[tranquil trellis]

If $\frac{a}{b} = \frac{c}{d}$, then $ad=bc$.

[twin meteor]

Ari

[honest tinsel]

So I just multiply ad and I get bc?

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[tawny hatch]

@graceful ibex

[mellow oyster]

!15mins

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[mellow oyster]

!original

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Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[tawny hatch]

This is the original formula. The questions are on page.

I got it from a YT tutorial.

My main question is what is significant difference between this version?

[regal bane]

This version, and which other version?

[tawny hatch]

There is version where denominator is combined as one

Standard deviation is SIG FIG different from N how so?

If you affirm this version as best, can you share a YT tutorial or khan academy recommendations for statistics?

In other words, what does a 135 class for communication major will need to prep study?

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[tawny hatch]

@regal bane

@graceful ibex

@high abyss

@high abyss

<@&286206848099549185>

[gentle thicket]

!15min

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[long jackal]

Here I only need to change variables and set the limits

[long jackal]

I know another way to solve this

I just wanna know if the way I solved it here is correct or not

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@long jackal Has your question been resolved?

[long jackal]

<@&286206848099549185>

[halcyon ice]

I believe it is correct yes

[long jackal]

I tried it on another problem

and it didn't work

so I donno

[halcyon ice]

what was the original problem asking exactly/originally

[long jackal]

this one

look

[halcyon ice]

yeah i mean what was the original question

what integral was it asking you to evaluate

[long jackal]

It was only asking to change variables and limits

to polar coordinates

so what I'm asking is

[long jackal]

is it ok to think of this circle as (r goes from 0 to 1) and (theta goes from -pi/2 to pi/2) even though the circle is off centre

[halcyon ice]

its ok to think theta goes from -pi/2 to pi/2 , but now that i think about it i believe the bounds for r are incorrect
lets start off with the circle equation
x^2 + (y-1)^2 = 1
convert it into polar coordinates
(rcostheta)^2 + (rsintheta - 1)^2 = 1
Expanding it out it becomes
r^2cos^2 theta + r^2sin^2theta + 1 - 2rsintheta = 1
since sin^2 + cos^2 = 1 it becomes
r^2 + 1 - 2rsintheta = 1
so
r^2 - 2rsintheta = 0
factoring out r it becomes
r(r - 2sintheta) = 0
so the solutions are either r = 0 or r = 2sintheta
r = 0 corresponds to the origin of the circle
and r = 2 sintheta corresponds to the boundary of the circle

so the inner integral should actually be going from 0 to 2*sin(theta) , and the outer integral should still go from -pi/2 to pi/2

im not 100% sure of this but i THINK its correct

oh and btw actually it should go from 0 to pi/2
not -pi/2 to pi/2
since it only covers the first quadrant

[long jackal]

aha

[halcyon ice]

so the polar integral should now look something like this

[long jackal]

ok so always work from the origin to be the same ?

[long jackal]

yup yup

[halcyon ice]

yeah in polar coordinates you always start r and theta at the origin

[long jackal]

alright

thnx bro

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[marble marlin]

I'm trying to solve for p and there seems like an infinite p

[marble marlin]

There is 1, 3, 9, 13, 27, 39, 81, etc

How does one continue?

[oak magnet]

so when she replaces p by 1/p you get 1, 1/3, 1/9, 1/13, 1/27, 1/39, 1/81

maybe a pattern with numbers in the denominator

[marble marlin]

Yeh there is but I don't know how to generalize it

Oh wait I think I know it, it's any square of 39 and 13 and 3

Prime powers

Wait i'll try

367/228

P = 1 is possible

The rest is just 1/2, 1/38, and 1/ 12

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[wide briar]

Hay, may someone halp with physics a bit? I have mechanics, and I forgot how to calculate some vectors

[wide briar]

A block of mass M1 is placed between two blocks of mass M2. The entire system is placed on a smooth table with a hole so that the triangular block protrudes slightly. The two blocks are pushed with forces F1 and F2 so that the entire system is at rest.

No friction

We need to mach the forces

Here is how Ive done it, but I didn't calculate the un named ones

And Im not really sure Ive done it correctly at all

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[short pilot]

Fr

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[tidal dock]

you can cancel 3r_1

$\frac{3r_1 \cdot 2r_2}{6r_2^2} = \frac{3r_1 \cdot 2r_2}{3r_1 \cdot 2r_1}$\
$= \frac{\cancel{3r_1} \cdot 2r_2}{\cancel{3r_1} \cdot 2r_1} = \frac{2r_2}{2r_1} = \frac{r_2}{r_1}$

[twin meteor]

artemetra

[tidal dock]

so your entire LHS is just r2/r1

also i don't think the right solution is 2*r1

should be -2r1

it should

$r_2 = -2r_1$ indeed

[twin meteor]

artemetra

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[untold arrow]

$\int_{0}^{+\infty} \frac{1}{(1+x^2)(1+x^a)}dx(a\neq0)$

[twin meteor]

riyob

[untold arrow]

hints?

[bitter pilot]

maybe try partial fraction decomposition

[untold arrow]

Could you elaborate a bit?

[bitter pilot]

First of all, what is this integral really about?

Do you have to calculate it, or investigate convergence/divergence in terms of a?

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[untold arrow]

Calculate it

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[gray fractal]

1-tanx=x then x ? how can i solve this question ?

[hybrid flicker]

$1 - \cot(x) = x$ has infinite solutions and aren't expressed nicely

[twin meteor]

rafilou2003

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[gray fractal]

it is a multiple-choice question how i know the range of x ?

[gray fractal]

sorry it is tan not cot

[outer warren]

can you show the full original question

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[gray fractal]

1-tanx=x then x ? a) [pi/4,pi/6] b) [pi/3,pi/4] c)[pi/3,pi/6] d)[0,pi/6]

[outer warren]

is that how the values are given?

they don't look valid with the larger value coming first

if you're given intervals, you don't have to actually solve the equation

you can try applying ivt

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[woeful hull]

Hi, I was kind of confused on something, that being that I thought subarrays were practically the elements of the powerset of a set (array), but when doing that I am generating the wrong output for the code problem I am doing

[woeful hull]

For example, I thought that for the array [1,2,3] the subarray for this would be [1], [2], [3], [1, 2], [1, 3], [2,3], [1, 2, 3]

But the elements must be continguous?

[hybrid flicker]

subarray elements must be consecutive

to get a subarray, you choose "all elements between a and b"

subsequences and subarrays are different

[woeful hull]

I see

[woeful hull]

So the subarray for [1, 2, 3] is the same as what I wrote excluding [1, 3] right?

.close

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[unreal solar]

Not sure where I went wrong here

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[unreal solar]

<@&286206848099549185>

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@unreal solar Has your question been resolved?

[void sleet]

928.64<μ<1009.36

[unreal solar]

Wait how did you get that?

[vocal sleet]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[void sleet]

First, we need to find the t-score for a 90% confidence level with
𝑛−1=22−1=21 degrees of freedom. Using a t-distribution table or calculator, the t-score for a 90% confidence interval and 21 degrees of freedom is proximately 1.721.

[void sleet]

Got it, thank you

[unreal solar]

Ohhh okay

Got it. Thanks

.close

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[dire harbor]

is this solution sufficient

[dire harbor]

it seems extremely sus to me cause i made a bunch of oversimplifications (about how some terms are greater as n approaches inf) but i have no idea how else to solve this one

[pallid zenith]

I'm not sure that the root test is appropriate here

also, i think the sum converges

[dire harbor]

are you sure it does

wolfram says no

[pallid zenith]

idk eyeballing it

i guess im talking about the second sum there

i have doubts about the first

no, actually, that should be fine

[dire harbor]

what is wrong with the second one then

[pallid zenith]

i dont understand what youve done there

😳

oh, you know what

i get it

wont we end up like $\lim a_n \approx \qty( \frac{2}{\sqrt 3})^n$

,calc 2/sqrt(3)

[twin meteor]

Result:

1.1547005383793

jan Niku

[pallid zenith]

so fails ntt

[dire harbor]

whats ntt

[pallid zenith]

nth term test

[dire harbor]

what is that

[pallid zenith]

but idk i have a tendency to be wrong with your questions

[dire harbor]

ohhh do you remember me

[pallid zenith]

the nth term of a series has to limit towards 0

[dire harbor]

oh

[pallid zenith]

only because i helped you a handful of times that day all the airports got shut down

i was right i think ... only one time

[dire harbor]

oh i see lol

[dire harbor]

yes i agree with this

[pallid zenith]

im not sure how you connect to this loose argument

the limit does not look fun

[dire harbor]

its what i was trying to say for the second sum but i didnt know how to convey it well

[pallid zenith]

well its true overall

i dont know that you need to split it

[dire harbor]

but like if you do root test over that limit you get 2/√3 which is still more than 1

so it divergers

[pallid zenith]

root test isnt appropriate here

[dire harbor]

why

[pallid zenith]

root test is for series that are raised to the nth power

not for tests involving a root

unless

i guess i see what youre trying to do

[dire harbor]

but the limit is literally raised to n?

[pallid zenith]

our loose asympotic one yea

not the summand

[dire harbor]

well yes

[pallid zenith]

you need to connect to the loose approximation, presumably

this argument isnt gonna hold water to your prof

[dire harbor]

yes thats what im worried about

but like i dont know if there even is a perfectly mathematically sound solutioin here without making approximations

this is the posted solution btw

[pallid zenith]

okay, yea

[dire harbor]

so they also used (2/√3)^n

[pallid zenith]

so thisk is fine

[dire harbor]

yay!

[pallid zenith]

do you understand the argument

[dire harbor]

also how exactly do you end up with this limit

[pallid zenith]

it just leading order

[dire harbor]

what is that

is it just arguing that functions like x^2 grow slower than 2^x

[pallid zenith]

like in theory you could expand out the summand into some other representation

of hopefully less and less significant terms

if we truncated this hypothetical expansion at a single term

it would be this one

(primary behavior for large parameter) + (correcting terms)

[pallid zenith]

yea, basically

getting that expansion is sometimes a bit of work it could be done here, youll need binomial theorem for non-int, i think

and you will need to split the series

and youl gett something like the second series is $\sum \frac{2^n}{ \sqrt{3^n} \qty( 1 + (\text{small}) ) }$

[dire harbor]

ermmmmmmmmmmmm

i dont believe will go forward with doing that

[twin meteor]

jan Niku

[dire harbor]

oh yeah

[pallid zenith]

well im not typing it super great

[dire harbor]

that makes more sense lol

thats what i was trying to convey

[pallid zenith]

but you rewrite $\sqrt{3^n+n^3} = \sqrt{3^n} \cdot \sqrt{1 + \frac{n^3}{3^n} }$

[twin meteor]

jan Niku

[pallid zenith]

the second term is small for large n

[dire harbor]

that n^3/2 is much smaller than √3^n at inf that you can ignore it

[pallid zenith]

we argue this is $\sqrt{3^n} \cdot \qty( 1 + small)$ by binomial expansion

[twin meteor]

jan Niku

[pallid zenith]

our expansion becomes approximated $\sqrt{3^n} + ( \text{small} ) \cdot \sqrt{3^n}$

[twin meteor]

jan Niku

[pallid zenith]

we argue the second term is small

[dire harbor]

[pallid zenith]

there is some power weirdness

[dire harbor]

honestly

[pallid zenith]

actually no

its fine

[dire harbor]

i think i wont bother with proving it too rigorously

[pallid zenith]

you can

[dire harbor]

the profs solution is even shittier so i dont think they care that much

[pallid zenith]

youll need an asymptotic expansion via binomial

and that will cinch it

but yea

your prof seems chill about it

[dire harbor]

i dont think that is wihtin the scope of a calc 2 course

but useful to know

[pallid zenith]

its not

[dire harbor]

its beyond me then

ok ty!

[pallid zenith]

i didnt learn this til grad school

np

[dire harbor]

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[keen osprey]

It asked me to determine the length of the fence required to enclose the pooland deck with fence

i got 118ft

just stuck on how to go about solving for this 1

nvm i figurerd itout

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[random rock]

How do I do c

[random rock]

The answers 12000

[slate dagger]

Can you explain your thought process for this solution?

[random rock]

2 hot dogs are sold for 1 dollar and
2 hot dogs are produced for 80 cents

So u make 20 cents

[loud walrus]

Yes

[slate dagger]

Yes, you make 20 cents for a pair of hot dogs

How many pairs of hot dogs do you need to sell to get $1200 then?

[random rock]

6000

[slate dagger]

Yes. You need 6000 pairs of hot dogs

What does the question ask?

[random rock]

Ah

How many individual hotdogs

[slate dagger]

yep

[random rock]

Ty

[slate dagger]

Alternatively, you could compute it as one hot dog giving you $0.1 of profit. But that gets tricky if the numbers don't divide evenly (e.g. if you sell 3 hot dogs for $1.50)

[random rock]

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[gilded peak]

Determine whether the piecewise function is differentiable at x=0

[vocal sleet]

Please don't occupy multiple help channels.

[gilded peak]

If someone could please assist me on where to start here

[void sleet]

I think you can do it with graph

[gilded peak]

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[gilded peak]

Haha I meant to post this problem

[torn glacier]

U need to check at 0

Check the function from left and right

Of 0

So 2(0) - 1 = - 1

(0)+ 0 +7 = 7

Since the left side is not equal to the right side the function is not differentiable

@gilded peak

[gilded peak]

Thought so, thanks man 🙏🏽 @torn glacier

How about the explanation behind this guys

[harsh canopy]

well cosx is always less than 1, so |1-cosx| = 1 - cosx

from that you can simplify this fraction to find the limit

[gilded peak]

You want cos (0)? Which is pi/2 @harsh canopy

[harsh canopy]

cos(0) isnt pi/2

[torn glacier]

Cos 0 is 1

[gilded peak]

OH OH

[harsh canopy]

what?

[gilded peak]

I was thinking , literally cos 0

I see what you mean > at it’s 0 value, it is 1

[harsh canopy]

in some cases you can just sub in the limit point (if its continuous there), but you cant here since the denominator is 0

[gilded peak]

Ok?

[harsh canopy]

ie you dont sub in 0

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[cloud harness]

is this correct

[vocal sleet]

@cloud harness Has your question been resolved?

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[modern ledge]

Hi, the entire question is quite long so i will just snip to the part where im struggling with

[modern ledge]

its a volume of object question through rotating around axises and for a portion of the object im realising that my trig substitution for the integral isn't working

im quite certain ive done it correctly but its currently

integrating from -a to a sqrt(a^2-x^2) dx

[hidden kelp]

What was your trig sub?

[desert hornet]

this integral can simply be solved by considering the area of a circle

[modern ledge]

well i think its more so of a technical issue

im aware that it is solvable via that way

but ive actually yieled 5pi a^2/2 instead of pi^2/2

*pi a^2/2

[hidden kelp]

Ok, but what was the trig sub you did, so that I can see wether it‘s right or not

[modern ledge]

acos theta

well its just the bounds really

im yielding different answers depending what bounds i choose to pick

which i hadn't really thought about before

[hidden kelp]

if x = acos(theta), if x = -a we have -a = acos(theta). Solve for theta

[modern ledge]

my mistakes on forgetting to explicit the problem im facing but its just the fact im yielding different answers

say i choose to pick the theta from the next priod

*period

i would yield a different solution

as for why i have done that, i have a bunch of objects of similar computations and i thought it would have been easier if i tried to shift everything alongside each other so that im not losing track of which one i have done and which one i havent

[hidden kelp]

Of course

[hidden kelp]

Of course to this I mean

Usually you take the first period for substitutions. Because the base function (sqrt(a^2-x^2)) is not periodic

[modern ledge]

is this a general restriction with trignometric substitution?

i've never really thought about it so it was kinda confusing now that im thinking about it

[hidden kelp]

I‘ll try something one moment

Yeah I‘m not sure. Trying to visualize on desmos but idrk.

[modern ledge]

im kinda taking shots in the dark here since im only at highschool maths rn but i've done a question before where it makes u prove that the substitution only works if its invertible on the integrating bounds

im not sure if thats related here but

but its the only thing i can recall ive seen that ever mentions on trig subs not working

[modern ledge]

although the context for this question was the last question of a past paper and it gave u a fair amount of structure to take u through the proof

[hidden kelp]

In my head, the only thing that makes sense to me is that if you look at the graph of the sqrt function, it‘s only a function on [-a,a], so if you pick some theta > a, it no longer makes sense to integrate there

,w plot sqrt(1-x^2)

[twin meteor]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by imtyp0

[hidden kelp]

Like if you pick theta = 3pi or whatever, it doesn‘t make sense to integrate that function over there

[modern ledge]

yeah that seems reasonable

might just keep this in mind then, math at highschool lvl is just not very strict and im just assuming i missed important theories on substitutions

thx for the response

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[hidden kelp]

To be honest, I think math at highschool level is "do this and don‘t ask why", so it‘s cool you‘re experimenting and asking 🙃

[modern ledge]

its also kinda motivated by the fact that sometimes it comes to haunt me in an exam setting;

the syllabus here lets u pick what level of math u want to do and im doing the hardest thats offered here and sometimes the examiner kinda goes out of syllabus

there has been occassions where they hit u with rather obscure properties or borderline out of the scope of teachers haven't taught

so it always feels a bit more reliable to overprepare

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[hollow panther]

Finally wrapped my head around reflecting a vector across a normal (I think), could someone check this working please? Sorry it's an image but I tend to work on paper

[hollow panther]

Main thing I am unsure about is the last step of doubling and subtracting the original vector, as well as my understanding of the dot product

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@hollow panther Has your question been resolved?

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@hollow panther Has your question been resolved?

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@hollow panther Has your question been resolved?

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[prisma parcel]

Hello chat. Our question today is quite simple. Can someone please provide a geometric explanation of why $\vec{a} \cross \vec{b} = -\vec{b} \cross \vec{a}$?

I've seen the ways of expanding it algebraically and such but I don't think I really understand what's going on here. Like I know that crossing 2 vectors gives a vector perpendicular to both, so does the anticommutativity mean that the vector is pointing the other way or....idk

[twin meteor]

The Cat Collective

[grim lotus]

make your thumb, index finger, and middle finger on your right hand perpendicular. Let a be in the direction of your thumb and b your index finger. Then a x b is in the direction of your middle finger.

swap a and b and see what happens

[prisma parcel]

Like this?

[grim lotus]

yes

[gentle thicket]

$\vec{A}{\times}\vec{B}=|A|\cdot|B|sin(\theta)\hat{n}$ where $\theta$ is the angle measured from the Direction of $\vec{A}$ to $\vec{B}$ and $\hat{n}$ is the unit vector in the direction of resultant. \
Now here, $\vec{B}{\times}\vec{A}=|B|\cdot|A|sin(-\theta)\hat{n}$ the angle will be measured from the Direction of $\vec{B}$ to $\vec{A}$. And since we assumed that angle from the Direction of $\vec{A}$ to $\vec{B}$ is $\theta$ then the angle between $\vec{B}$ to $\vec{A}$ will be of same magnitude but in opposite direction(that's where negative sign came from). \
SO, $\vec{B}{\times}\vec{A}=|B|\cdot|A|sin(-\theta)\hat{n}$ $\implies$ $\vec{B}{\times}\vec{A}=|B|\cdot|A|sin(\theta)(\hat{-n})$ \
and hence $\vec{a} \cross \vec{b} = -\vec{b} \cross \vec{a}$.

[twin meteor]

77²

[gentle thicket]

a mathematical proof if you want

[prisma parcel]

Tysm, will read in a sec

[grim lotus]

that implies you're already choosing a direction for n somehow

[grim lotus]

are you a physicist by any chance?

[gentle thicket]

bruh no
just completed schools a few months ago

[grim lotus]

mhm alr
i was just asking cuz it sound like smth a physicist would say

[gentle thicket]

I want to be someday

[grim lotus]

i meant that more as an insult but okay

do you want to major in physics

[gentle thicket]

I don't take anything negatively

[gentle thicket]

yeah

[grim lotus]

you'll fit right in

good luck!

[gentle thicket]

tysm

[prisma parcel]

Thank you both for your help

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[bleak prawn]

Not really sure what to do here

I know that getting the tangent involves getting the derivative

But the fact that the equations are factorized leads me to think that I am supposed to solve for t?

I know that they want it at the origin so im supposed to set them equal to 0

[lone linden]

The factorisation it mainly to make the value of t such that you have a pt at the origin more obvious

[bleak prawn]

Whats a pt sry

point?

[lone linden]

yeah

just me being lazy and abbreviating

[bleak prawn]

Ah sry haha

Which point tho?

At (0,0)?

[lone linden]

I mean the question wants the tangent at the origin, so yeah

[bleak prawn]

Ok, that makes sense, but what is the factorization showing us

[lone linden]

The factorisation it mainly to make the value of t such that you have a pt at the origin more obvious

it makes solving for t much easier

than if you had a polynomial that was expanded out

[bleak prawn]

Why do we want to solve for t?

Sry if im asking dumb questions

[lone linden]

because you want the tangent line

at the origin?

how else are you supposed to find the value of t to take the derivative at

[bleak prawn]

So I set them equal to 0, then take the derivative? Not the other way around?

[lone linden]

do it in whatever order you want

the only important thing is that you need to find the value of t to take the derivative at before you do said calculation

but that's kinda forced by the laws of nature anyway

[bleak prawn]

How sry?

[lone linden]

find where x(t)=0 and y(t)=0 so that you're at the origin

[bleak prawn]

for x(t) we have -1/7, 4/5 and 2/5

for y(t) we have -1/7, 5/3 and 2

[bleak prawn]

What do I do now sry

[lone linden]

you need both coordinates to simultaneously be zero

take that as you will

[bleak prawn]

But theyre both zero at 3 each

I can clearly tell which ones im supposed to pick but I just want to understand the reasoning behind it

[lone linden]

you need both coordinates to simultaneously be zero

this means that your value of t needs to make both x(t) and y(t) equal to zero

[bleak prawn]

Ah ok I get you

So do I literally just expand the brackets

Take the derivative

Sub in -1/7

And thats the answer?

Oh wait no

Equation of a line

So thats x and y

[lone linden]

you could do that or do triple product rule

[bleak prawn]

Whats that

[lone linden]

product rule but for a product of three functions instead of two

you basically do the same thing tho

$(fgh)'=(fg)(h')+h(fg)'$

[twin meteor]

Civil Service Pigeon

[bleak prawn]

Oh cool

I need to learn that one

[lone linden]

what?

no I'm showing you how to derive it

that's not the finished version

we just applied the product rule where one function is fg and the other is h

[bleak prawn]

Haha yeah I searched it on google

But something im still confused about

What do we actually get when we do this

So do I literally just expand the brackets
Take the derivative
Sub in -1/7

We get two coordinates

Are they just random coordinates on the line?

[lone linden]

wait a minute

yk how parametric differentiation works ... right?

[bleak prawn]

ummm just differentiate x(t) and y(t) no?

I didnt realize it was different 👀

[lone linden]

$\frac{dy}{dx}=\frac{y'(t)}{x'(t)} \implies \frac{dy}{dx} |_{t=\frac{1}{7}}=\frac{y'(1/7)}{x'(1/7)}$

[twin meteor]

Civil Service Pigeon

[bleak prawn]

We get a fraction?

[lone linden]

a fraction for what?

[bleak prawn]

y'(t)/x'(t) ?

[lone linden]

yeah, that's the formula

[bleak prawn]

I got it! Thanks

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[lone linden]

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[gilded heath]

can someone help me show why x_n > \alpha for all n in this

[oak magnet]

Well show that $x_{n+1} - x_n < 0$

[twin meteor]

YakuBros

[gilded heath]

yeah i feel like i can't really do that without assuming x_n > \alpha

if you assume x_n > \alpha then it follows immediately

[hybrid flicker]

don't assume it, prove it

[gilded heath]

wait yeah

that's what i'm asking help with

i looked up a few solutions to this exercise and all of them assumed this is true

but didn't show why

which is like insanely sus

[oak magnet]

Ok

[hybrid flicker]

you can show it by computing x_n+1^2 - alpha

[oak magnet]

x_1 > sqrt(a) > 0
So (x_1)^2 > a > 0 ...

[gilded heath]

wait wdym

[hybrid flicker]

suppose x_n > sqrt(alpha)

and compute x_n+1^2 - alpha

[gilded heath]

yeah i understand that being positive is equivalent but like

i did that earlier and the expression i got was this

\frac{1}{4}(x_k^2 - 2\alpha + \frac{\alpha^2}{x_k^2})

which is like not a useful expression

[hybrid flicker]

it is... (a-b)^2

[gilded heath]

OH FUCK

i was looking at that expression

and i didn't notice

thank you

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[fair gorge]

Solving a rational equation, wondering whats the best way to go about it and want to make sure im right.

[fair gorge]

i assume the best way is to multiply each by 7t and 5?

wondering if there is an easier way, if i can ignore the 5 or this is the correct wa

[gaunt sparrow]

You could just multiply through by t, but the useful thing about mutliplying by t, 5 and 7 is that you get rid of fractions.

This means less messing around with common denominators, and in the long run, fewer mistakes :p

[fair gorge]

ok sweet thanks!

.close

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[stoic trout]

i'm unsure what to do here because the null space is just 0?

[simple mason]

Why do you need the null space though?

[stoic trout]

isn't that all of the solutions for the subspace?

[gaunt sparrow]

The subspace V is just made up of vectors that get mapped to themselves.
The nullspace doesn't have much to do with that. Was that something discussed in a)?

[stoic trout]

Part A was just proving that V was a subspace

i might be overthinking this

could i just use the equations in the third or second row as the equation?

[simple mason]

Isn't that solvable by getting the eigenvalue of the matrix

Then we just put it in the matrix then get the null space

[stoic trout]

i think you were right that the null space was irrelevant

we just need a linear equation that represents the subspace

[gaunt sparrow]

Yeah it ends up being equivalent to finding the eigenvalues, but you can also pretty easily write out the equations you need to get, and it turns out you end up with 3 equivalent equations for the plane.

[simple mason]

Oh choosing the eigenvalue = 1

[gaunt sparrow]

3y-2z = x and so on. They will all give the same equation, so you don't need to overthink it.

[simple mason]

We just assume the eigenvalue = 1 then we just try to find the equations

[stoic trout]

wait how do i find eigenvalues my prof has mentioned it like once before

[simple mason]

By the way solving for the eigenvalues of the matrix gives 3 identical solution which are all 1

We need to determine the values of h such that
det(A-Ih) = 0

@stoic trout ^

[stoic trout]

h?

oh

[simple mason]

I wrote it as h but most of the text books use lambda

[stoic trout]

is the eigenvalue the d value here in the linear equation

[simple mason]

No

You can determine it from the matrix

Just minus lambda from the diagonal of the matrix like that then take the determinate of the matrix and set it equal to zero then solve for lambda

Like that

Kinda

[stoic trout]

i dont understand why that works conceptually

[stoic trout]

wait why is it equal to x?

[gaunt sparrow]

Do matrix mulitplication on the first component, you want that to be equal to x, since T(v) = v.

So vector (x,y,z) gets sent to vector (3y-2z, x - 2y + 2z, 2x - 6y + 5z) = (x,y,z).

[simple mason]

Eigenvalues are the values that for a given eigenvector x it give
Ax=hx
We want to solve for h here so we say
Ax-hx = 0
Ax-Ihx = 0
(A-Ih)x = 0
Here for a matrix A-Ih to make any vector X multiplied by it equal to zero it needs to have a zero determinant so the solution to that equation is
det(A-Ih) = 0 where we solve for h

After solving for eigenvalues it would result that h=1 is repeated 3 times
Which is excepted to get 1 as a solution as it is the only value of eigenvalues that would keep Ax = 1x = x
So we will apply that to A-Ih
Where h = 1 so A-I
Then we solve for its null space to get the solution

Like this

The x - 3y + 2z = 0 is the final solution

[stoic trout]

tysm, I will go through and parse through all of this to make sure I fully understand

🙏

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[simple mason]

I think what I said there is confusing though

So don't worry if you don't understand

Actually an easier approach is the approach Azyrasha suggested

Which is take any line from the matrix
Say line2
And set it to
1x -2y + 2z = y
Then simplify it

And it will be the answer

We just set Ax = x

Then take any equation from the equations and it will be the answer

Better @stoic trout ?

[stoic trout]

yeah i was just experimenting with stuff like that rn

[simple mason]

Ok

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[hidden anchor]

what does the arrow and + symbol mean here

[hybrid flicker]

context?

[hybrid flicker]

arrow is most likely to say it's a vector

and the text says that Z+ is a "direction"

so it's a unit vector

oriented "positively"

[hidden anchor]

ah okay, thanks a bunch

.close

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[copper otter]

i need help

[pale perch]

any idea

[copper otter]

um

i did getvthis one right but i 'm confuse by the decmal one

[pale perch]

decimal one?

[half trail]

should i just solve it

[copper otter]

no

i wil give u he deciaml one

give me s sec

this one

one

[pale perch]

whats troubling about it?

[half trail]

just multiply all numbers by 10

and then divide ur result by 10

gets rid of the decimals when u dont want em and puts em back in when u need em

[copper otter]

the culition it give me is weired

so 19.2 times 10 12.9 times 10 9.6 times 10 8.8 times 10?

[half trail]

yup

so ur left w a square of 192 * 129 km in size

rectangle*

[copper otter]

so 192 ,129 96 88 right?

what now?

[half trail]

yup

do what u did with the first one

[half trail]

here

[copper otter]

k

24,768 and 8,448

[half trail]

now divide by 100

cuz u multiplied 2 numbers by ten

[copper otter]

anwer is 16,320

[half trail]

so u have to 10²

which is 100

[copper otter]

163.20?

[half trail]

im bad at explaining

[half trail]

lemme check rq

[copper otter]

k

[half trail]

yup

u have successfully solved ur homework

[copper otter]

ty

[half trail]

also take a look at ur dms and remember to .close this ticket

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@copper otter Has your question been resolved?

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[hidden kelp]

,, \int_{0}^{\frac{\pi}{6}}\sqrt{\tan\theta +\sec\theta}\sec^2\theta\dd\theta

[twin meteor]

𝔧𝔪𝓣𝛾𝜑𝜽

[hidden kelp]

I feel like this is elementary, but my guess is I'm completely missing some trig identity

[vocal sleet]

@hidden kelp Has your question been resolved?

[hoary cypress]

i feel llike taking tantheta as x might work

but then u m8 have to simpllfy it further

im not sure

[hidden kelp]

that gives what I originally started with, namely $\int_{0}^{\frac{1}{\sqrt{3}}}\sqrt{x + \sqrt{1+x^2}} \dd{x}$

[twin meteor]

𝔧𝔪𝓣𝛾𝜑𝜽

[hidden kelp]

You can probably see why I wanted x = tan(theta)

<@&286206848099549185> pin time

[static socket]

sec² is dtan/dO btw

[static socket]

Oh, it ends as that too

[hoary cypress]

rationalize it?

nvm

[static socket]

You can do a sum by its sub substitution by multiplying inside the integral by sqrt(x-sqrt(1+x²))

Then it becomes
sqrt(x²-1-x²)/sqrt(x-sqrt(1+x²))

Oh, thats not easier

[hidden kelp]

yeah I was thinking doing it the trig way would be better cause we get identities

[dry orchid]

u could simplify using formulae for submultiple angles?

not sure

[hidden kelp]

Meaning? Like sin(x/2) and whatnot?

[dry orchid]

yeah

wait

hm

u grt 1+tan T/2 / 1-tanT/2

from doing that

[hidden kelp]

From doing what?

[dry orchid]

subtituting tan T and sec T

tan T becomes 2tan T/2 / 1-tan^2 T/2

sec T is just 1/cosT

thats 1+ tan^2 T/2 / 1-tan^2 T/2

u could write that in terms of sin and cos

then turn the numberstor and denominator sin terms

[static socket]

what if you integrate by parts first

[dry orchid]

on top would be sin(pi/4 + T), down would be sin (pi/4 - T)

[static socket]

what happens with the root when diff.

[dry orchid]

i'll be honest idrk integration

ik trig

just tryna simplify the thing

oh got it

u can write the term inside the root as tan(pi/4 + T/2)

try it for urself

but again idk if that would help u solve the problem

oh i took extra unnecessary steps

[dry orchid]

thus can immediatey be written as tan(pi/4 + T/2)

this just complicates it

[hidden kelp]

Wait what does it give, you said you didn't know much integration but maybe reducing it as that helps?

[short pilot]

Yo

@hidden kelp

Here to help guyz

[hidden kelp]

okay?

[short pilot]

Ok

Give the quotion

[hidden kelp]

↑

[short pilot]

Fr bro has Tau gamma phi and theta in his name

Fr

[dry orchid]

the inside term can be simplified into tan(pi/4 + theta/2)

[short pilot]

You know sec square t-tan square t=1

Aaah

[dry orchid]

inside the square root, that is

[short pilot]

it can be factored as (sect-tant)(sect+tant)=1

So we write sect+tant as 1/(sect-tant)

@hidden kelp

[dry orchid]

would that be simpler to integrate?

[short pilot]

Yes

Fr

It will get in denominator

So we will have sec^2t dt/√(sect-tant)

[dry orchid]

itll turn into ln[ abs( sec ....) ]?

[short pilot]

Wait bro

I did not integrate

Imma do it now

[hidden kelp]

Let me know your findings :)

[hoary cypress]

that doesnt make it any better r8?

[short pilot]

Wait imma try differently

[bitter pilot]

,,u=\tan\theta + \sec\theta

[twin meteor]

𝔸dωn𝓲²s

[bitter pilot]

,,\dd \theta = \frac{1}{\sec^2{\theta} +\tan\theta \sec \theta} \dd u

[twin meteor]

𝔸dωn𝓲²s

[bitter pilot]

,,\int_1^{\sqrt{3}} \sqrt{u} \frac{\sec^2\theta} {\sec \theta \cdot u} : \dd u = \int_1^{\sqrt{3}} \frac{\sec \theta}{\sqrt{u}} : \dd u

[twin meteor]

𝔸dωn𝓲²s

[hidden kelp]

I thought of that but I don't know how to have sec theta wrt to u

[bitter pilot]

,,\sec\theta = u - \tan\theta = u - \sqrt{\sec^2\theta - 1}

[twin meteor]

𝔸dωn𝓲²s

[bitter pilot]

Then

square both sides

nah

[hidden kelp]

Probably leads somewhere honestly

[bitter pilot]

,,\sqrt{\sec^2\theta -1} = u - \sec\theta \Rightarrow \sec^2\theta -1 = (u-\sec\theta)^2

[twin meteor]

𝔸dωn𝓲²s

[bitter pilot]

quadratic equation

then solve in terms of sectheta

x² - 1 = (u - x)²

,w x² - 1 = (u - x)²

[twin meteor]

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Requested by adoniszgks

[bitter pilot]

So

,,\sec{\theta} = \frac{u^2+1}{2u}

[twin meteor]

𝔸dωn𝓲²s

[bitter pilot]

,,\int_1^{\sqrt{3}} \frac{u^2+1}{2u\sqrt{u}} : \dd u = \frac{1}{2}\int_1^{\sqrt{3}} \frac{u^2}{u^{\frac{3}{2}}} + u^{-\frac{3}{2}} : \dd u

[twin meteor]

𝔸dωn𝓲²s

[bitter pilot]

I am tired of this I give up

[short pilot]

What now it can be integrated

u^(2-3/2)

Yes it's integrable

Ow

Now*

@hidden kelp

Check answer

[bitter pilot]

how

prove it

[short pilot]

Bro

u^(2-3/2)

u^(1/2)

+u^(-3/2/

u^(1/2) integrates to

2/3u^(3/2)

,w integrate u^(1/2)

[twin meteor]

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Requested by oppailol.

[hidden kelp]

Makes sense

[short pilot]

,w integrate u^(-3/2)

[twin meteor]

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Requested by oppailol.

[short pilot]

Makes sense

Now just put the values

[bitter pilot]

ok bro

[hidden kelp]

Adonis

You might wanna go see a doctor

[short pilot]

But pls check

[bitter pilot]

why

[hidden kelp]

Your brain is abnormally huge

[hidden kelp]

The original answer was 2/3 iirc

[short pilot]

No I mean check the answer after finding

Completing it

What

Just 2/3

[bitter pilot]

,av oppailol.

[twin meteor]

oppailol.'s Avatar

Click here to view the image.

[short pilot]

Lol

[hidden kelp]

Yeah ik, pretty unexciting answer

[bitter pilot]

it's also empty

[hidden kelp]

I've got better ones up my sleeve if you want a challenge oppailol

[short pilot]

Na bro's brain filled with maths

[bitter pilot]

oppa i was playing with you

[short pilot]

We solving the Riemann hypothesis in his brain fr 🗣️🙊

[torn timber]

Math=not real thus empty

[short pilot]

@hidden kelp ok let's do more but atleast check the answer

Lord convergence

I bow to the majesty

[hidden kelp]

I'm gonna give you one I know the answer and is really sick

[short pilot]

🫡

[hidden kelp]

just give me a moment

[short pilot]

Check this one bro

I don't think this just becomes 2/3

[bitter pilot]

was this your channels or oppas

[short pilot]

Fr

Does it become 2/3 fr?