help-10

[hidden compass]

Bring one cos to the other side

And then transform one of them by shifting the argument by π

Unless you have already tried something else

[wet moss]

there is a nice trick you can use, and it is pretty easy to derive with the compound angle formula

[fossil spoke]

oh ok

[hidden compass]

You mean prostapheresis?

[wet moss]

yeah

[hidden compass]

Mmh do they help here? I would need to try

[wet moss]

yeah it really helps

[hidden compass]

Oh Yeah you're right

Then you just make one of the factors = 0 and you're done

Nice tip👍

[wet moss]

however its only useful is OP knows the compound angle formula

[fossil spoke]

how to tranform exactly?

[obtuse pebble]

@fossil spoke Has your question been resolved?

[obtuse pebble]

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[timid silo]

should be h just because it is the orthogonal projection of f onto the space of trignometric polynomials of order 2, right?

[hidden compass]

Using cos(π - a) = -cos(a)

[obtuse pebble]

@timid silo Has your question been resolved?

[fossil spoke]

am I looking for a quadratic when I expand out

[timid silo]

open your own help channel please

[obtuse pebble]

@timid silo Has your question been resolved?

[tardy epoch]

Your guess is right, but I don't think the reasoning is.

Closer usually implies some type of distance. I don't think orthogonal projection is a distance

Do you know Parseval's theorem

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

I do not

please do elaborate tho

[ornate oar]

Personally I think your original answer is ok as long as you specify why orthogonal projection leads to smallest distance

[tardy epoch]

Plug in f-g and f-h for f

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[opaque galleon]

I need help to find the restriction on phi

I can do everything else except that

how do I do that?

[onyx sentinel]

I do not know the answer to your question but what concepts are required ro solve this?

*to

[opaque galleon]

BRUH

[lethal sand]

you can either

1. Graph everything onto the xyz space
   or
2. Graph out a slice of the solid, more preferably, graph out the slice on xz or yz-plane

after graphing it out, you'll see a clear bound for what phi needs to be

[onyx sentinel]

Mb sorry

[opaque galleon]

yeah but

I need to show it algebraically

i know its pi/4

but

can I do it like this

[lethal sand]

hmm

[opaque galleon]

z=sqrt(x^2+y^2)

rho cos phi = sqrt(conver this thing)

[lethal sand]

in that case, I suppose then just change everything to spherical coordinates and solve for the intersection

[opaque galleon]

I got this:
cos phi = sin phi

[opaque galleon]

from this

[lethal sand]

yes, then just solve for phi

oh

you aren't supposed to cancel out the r

[opaque galleon]

what r

[lethal sand]

the rho that is

[opaque galleon]

oh why

[abstract flame]

so you know that it is pi/4 right

[lethal sand]

basically you are supposed to find the intersection between rho^2(cos(2phi))=0 and rho^2=1, this gives you the intersection between the cone and the sphere of radius 1

now, you can indeed cancel out the rho in rho^2(cos(2phi))=0

then, you would do the same thing for the sphere of radius, and you'll see that the bounds for phi is the same for both intersections

when you cancel out the rho in rho^2(cos(2phi))=0, what happened is you just simplified this down to show you a surface represented by rho!=0, theta is any angle and phi is any angle that satisfies cos(2phi)=0. It doesn't really give you much information about the intersection.

[opaque galleon]

wait

where did you get this

rho^2(cos(2phi))=0

[lethal sand]

rho^2(cos^2(phi))=rho^2(sin^2(phi))

just pull everything to one side and you'll get cos^2(phi)-sin^2(phi) which is just cos(2phi)

[opaque galleon]

oh ok

[obtuse pebble]

@opaque galleon Has your question been resolved?

[obtuse pebble]

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Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[coarse comet]

please ping me

[timid silo]

.status

Or something like that TT

[coarse comet]

?

[timid silo]

Like where are you stuck

[coarse comet]

huh lol

im new

I understnad the problem

[crisp forum]

!status

[obtuse pebble]

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[coarse comet]

ah

[timid silo]

Thanks 😭

[coarse comet]

I guess 1

[timid silo]

Hmm okay

[coarse comet]

I understand the problem

[timid silo]

So prerequisite knowledge

[coarse comet]

but dont know how to approach it

[timid silo]

Yeah

[coarse comet]

All the way up to CALC BC

tryna improve on my algebra foundation

[timid silo]

Can you find a relation between a and b?

[coarse comet]

a+b=180 I believe yes

[timid silo]

Yep

Now

Do you know that property where

An exterior angle of a triangle is equal to the sum of opposite two angles

[coarse comet]

oh I did not know that, thats a geometry concept no? The year I had geometry covid scuffed it up

[timid silo]

Yep

Lemme draw it for you

[coarse comet]

thats why im tryna build up my geometry and algebra foundation

please and thank you!

[timid silo]

Angle 1 = Sum of Angle 2 and 3

[coarse comet]

oh I see

ok

[timid silo]

You can also prove it using Angle sum property of triangle , to find the angle besides Angle 1 on teh line, then use Angl 1 + that angle = 180

Will give you the same result

[coarse comet]

hold up sorry

trying to undersatnd hahah

[timid silo]

Can you just remind me what its called when to angle makes 180° on a line, I frogot the name

[timid silo]

No worries

[coarse comet]

uh

supplementary

idk

i think thats it

[timid silo]

Aye aye

[coarse comet]

kk!

[timid silo]

So may I go again on that

Exterior angle thing of triangle?

[coarse comet]

ya please do

[timid silo]

It's been like 3 years since I last studied it so I have forgotten the exact names

I'm sorry

[coarse comet]

just so u dont have to scroll up

nah ur all good

[timid silo]

I remeber the question by heart

Dw

[coarse comet]

💀 thank lol

my memory is so bad

honestly its more for me thatn you

[timid silo]

So this one, can you similarly apply to our question?

[coarse comet]

yes

that means

[timid silo]

Trynna see the similarity somewhere

Yeah

Proceed

[coarse comet]

c + the other angle = a

[timid silo]

Correct

Now what's the other angle?

In relation to something given?

[coarse comet]

<2?

[timid silo]

I mean

[coarse comet]

or you talking about D

[timid silo]

Yes

[coarse comet]

oh so

[timid silo]

Cuz making a forumal with C wouldn't work well

[coarse comet]

D = B+C?

[timid silo]

So you see Angle D and try something with it

[coarse comet]

since b is a vertical angle

[timid silo]

Yeahhhhhh

You're far too good

Yep

[coarse comet]

haha lol

[timid silo]

And also

Now something between b , c and the new angle you found?

Angle sum property

[coarse comet]

let me search that up

rq

[timid silo]

Like ykr All angles of triangle adds up to 180°

[coarse comet]

ok

ya

[timid silo]

Yeah

Do it

With b , c and b+c , you just found

See they are all interior angles of a triangle??

[coarse comet]

yes

[timid silo]

So what do they add up to?

[coarse comet]

one second lol

[timid silo]

Aye aye

Where did you get stuck?

[coarse comet]

wait

you want me to add

angles

B and C?

[timid silo]

Nah

See

Just stop me when I'm not making sense

See

[coarse comet]

kk

[timid silo]

Angles inside a triangle

Add up to 180°

Right?

[coarse comet]

yes

[timid silo]

Now

You already know by that vertically opposite angle

[coarse comet]

yes

[timid silo]

The angle b can be brought INSIDE the triangle?

Like to the right side of a

Correct?

[coarse comet]

yes

[timid silo]

Now what did you get

Lemme draw it

[coarse comet]

no I got it

that part

OH

WAIT

[timid silo]

So this will be your new question

[coarse comet]

hold up

[timid silo]

Yeah

[coarse comet]

wait give me exactly 1 min

[timid silo]

Sure

Take 5

I'll solve teh question too now like the calculation part

[coarse comet]

would you just do this

C-15+C+C=180?

[timid silo]

Lemme check

[coarse comet]

kk

[timid silo]

How did you come to that?

[coarse comet]

so

in the problem

it says a= 2c

and using the angle thingy

from the top + angle c we counclude thats a

[timid silo]

It's no property it's just given

[coarse comet]

so thats C

ya

but thats how I got c on the top

and in the problem it also said b=c-15

so instead of putting b in the triangle i put c-15

so I would get

c-15+c+c=180 for the inside

of the triangel

in the bottom

[timid silo]

Good

[coarse comet]

am I right

idk

if i am lol

[timid silo]

That's the correct method man 😭

[coarse comet]

lol

alr

[timid silo]

Like there are multiple ways

And I would have never seen that coming

Nvm

Now

You get C as?

[coarse comet]

65

[timid silo]

65°

Yeah

[coarse comet]

ya

[timid silo]

Now put it in teh given eqn to get B

What do you get?

[coarse comet]

65-15= 50

thats B

[timid silo]

Yep

And D is?

[coarse comet]

and C+D=180

c=65

so D= 115

[timid silo]

Yeah

So the answer is

[coarse comet]

115-50

is

uh

[timid silo]

Good work

[coarse comet]

65

oh

[timid silo]

😂😂

[coarse comet]

can u help me with more

i got a couple more

[timid silo]

Sure

[coarse comet]

say less let me take some quick pic

[timid silo]

Okay so in these questions , key word - MUST be true

[coarse comet]

ya

[timid silo]

Cuz it can be , or can't be

Now Lemme just

Gimme a uhh 48 seconds

[coarse comet]

alr lol

[timid silo]

Yeah got it

Did I take more than 48 secs?

damn failed

[coarse comet]

idk 💀

[timid silo]

Anyway

[coarse comet]

probs not

[timid silo]

So let's start

[coarse comet]

alr

so what i know

is

<Q=<T

<V=<S

<R=<U

[timid silo]

Don't make the angle signs dw

[coarse comet]

kk

and all of them must add up to 360

[timid silo]

U *

[timid silo]

Yeah

[coarse comet]

and half of it adds up to 180

[timid silo]

Yeah

So what is given to you in the question first of all

Like except this

[coarse comet]

q+u=s+v

[timid silo]

Good

And
Relation between s and v you can probably use?

[coarse comet]

yes

[timid silo]

Wrote it down fir me

For*

[coarse comet]

kk

wait let me think

and ill write down my stuff

[timid silo]

Yep

Just tell me when you're ready

[coarse comet]

kk

[timid silo]

Yeah?

[coarse comet]

alr

not really sure if this is any use of all

[timid silo]

Dw geometry is all about that

You learn with practice

[coarse comet]

but if s+v= q+u does that mean s+v=t+r?

[timid silo]

Yeah

[coarse comet]

is that any use to us

[timid silo]

That's correct

Yep for the third one

But to develop a thinking you must

Like Trynna think FOR the quesntion and not randomly

[coarse comet]

alr

so for I

[timid silo]

Like see what's given and try to find something related to it

[coarse comet]

ok

q+u = s +v and thats a fact

[timid silo]

Yep

S= V that's a fact too

See

You wrote it

[coarse comet]

ya

so for I

[timid silo]

So now what would you write ?

[coarse comet]

oh

[coarse comet]

wait

wym

[timid silo]

Like substitute s with v, as we already know s=v

[coarse comet]

kk

so if 2s = q+ u

[timid silo]

See

1. Trynna make things in less variables
   2.Trynna substitute some variable with values given in question, like straight line is there make it 180
2. Sometimes just keep things as they are for future use,

[timid silo]

Yes

Now

You want something from q and u

Right?

[coarse comet]

ya

[timid silo]

So if I told you

I gave you some value of v , can you find it?

[coarse comet]

if you give me a value of v

i would be able to find s

and from that

[timid silo]

Yes

Goood

[coarse comet]

yes i would be able to find it

[timid silo]

Yeah

[coarse comet]

if you give me v

[timid silo]

Now you have to find v

Now

This one is not so easy to think, you'll get it with practice

But yeah

In teh question can you see q+u+v

What is it equal to ?

[coarse comet]

180

[timid silo]

Good

Now what do we have?

q+ u = 2v

Yeah?

[coarse comet]

wiat

[timid silo]

Sure

[coarse comet]

ok

this is lowkey stressing idek why

[timid silo]

Dw

Where did you get stuck?

[coarse comet]

could you just explain from the top you did that last time and I got it super easily lol

it lowkey helps a lot

[timid silo]

From the top?

[coarse comet]

ya ig

last time u did it

i figured it

haha

[timid silo]

Yeah

So I can start all over again if you want

[coarse comet]

if you want

you can skip

[timid silo]

Yeah

[coarse comet]

the simple stuff

like the stuff i alrdy knew

[timid silo]

Nvm

Yet I'll recap that too

[coarse comet]

kk

lol

[timid silo]

Except the angle you wrote equal

That's correct

Also we are given q + u = s + v

Now you knew, s = v

[coarse comet]

ya

[timid silo]

Can I write v anywhere s appears

[coarse comet]

ya

[timid silo]

Liek if I had 1 + s

J could have written it as 1 + v

So here

[coarse comet]

yep

[timid silo]

q + u = 2v

Sure?

[coarse comet]

ya

[timid silo]

Rn were solving for question 1

Which wants us to get something in terms of q and u

So now , how do you get q and u from this ?

[coarse comet]

ya i dont know, is it a type of rule or smt

?

[timid silo]

Just wait a sec I'll be back

[coarse comet]

kk

[timid silo]

Had to attend a call

Sorry

[timid silo]

No

[coarse comet]

nah you good

[timid silo]

I told you it would be hard to think of as a beginner

Uhh

Yeah

q + u = 2v

Sure?

[coarse comet]

wait could you explain that

[timid silo]

And also can you make out some straight line from q and u ??

In the question

[coarse comet]

on why q + u = 2v

[timid silo]

Okay

q + u = s + v

s = v

Therefore

q + u = v + v

q + u = 2v

Understood?

[coarse comet]

ya

[timid silo]

For sure?

Or just, ya

[coarse comet]

i got it

100%

[timid silo]

Hmm good

Now find for me

Some relation between q and u

Like find some straight line or something

[coarse comet]

line C?

[timid silo]

Can you make something on line C?

Try a better one

[coarse comet]

its cutting q and v

so

oh

[timid silo]

Umhmm

[coarse comet]

wait

[timid silo]

Yeah

[coarse comet]

hold up my brain is thinking rn

[timid silo]

Ikr

[coarse comet]

oh my goodness

so

would r and u be equal to 90?

idk if my brain is thinking right

[timid silo]

Hwo did you get to that?

[coarse comet]

so i imagined there not being a middle line

[timid silo]

just for convenience

[coarse comet]

so just lines D and A

[timid silo]

Umhm

[coarse comet]

wait

nah i dont think my brain was thinking right

hold up

[timid silo]

Yeah

[coarse comet]

qv = st

yes

was that the relationship

u were talkign about

[timid silo]

Resend the que again please

It got lost

Not qv = st

Wouldn't it be q + v = s + t

[coarse comet]

ya

thats what i mean lol

[timid silo]

And that's something we already knew

Cuz q = t , v = s

Innit?

[coarse comet]

ya im just totally lost on this question

[timid silo]

😂

Dw

[coarse comet]

lol its probs cuz its 1 am

my midn is lowkey fried

[timid silo]

😂

Dw

May I ?

Proceed?

[coarse comet]

ya

i wont sotp you

[timid silo]

Yeah so

Again

Send me the que one sec

q + u = 2v

[timid silo]

We still good on that!

[coarse comet]

YA

ya

[timid silo]

Yay

Now what's q + u + v

[coarse comet]

180

[timid silo]

Good

So can I not just add v , on both sides of q + u = 2v

To get

q + u + v = 2v + v

180 = 3v

v = 60

[coarse comet]

wait why

[timid silo]

Any questions?

[coarse comet]

oh

i mean it makes sense

but doesnt

[timid silo]

You'll have to think that ways

Okay

[coarse comet]

oh

[timid silo]

From where?

[coarse comet]

like

the math makes sense 100%

its jsut hard

[timid silo]

Yeah

[coarse comet]

i would have never thought of that

[timid silo]

Don't Trynna imagine it btw like you prolly can

But it'll be hard

[coarse comet]

ya

[timid silo]

See I asked you to find a relation between q and u

If you had done that on Line B

You would have thought of that

You see?

[coarse comet]

wait

hoenstly i probs still would not have

ugh

[timid silo]

VC?

Like fingers hurt

Or nvm

[coarse comet]

if u want lol

[timid silo]

I'll write it's 1 am you say

[coarse comet]

nah

its not that i dont midn lol

[timid silo]

You can ?

[coarse comet]

i can hop i nvc

ya

stay up till ike 5

any way somedays

[timid silo]

Lmao

So

[coarse comet]

is there a vc

in ehre

[timid silo]

Nah

[coarse comet]

dam

thats unfortunate

lol

its fine

ya but i think i can solve the problem

from there

since i can find eveyr value

[timid silo]

Accept friend req

[coarse comet]

cant

[timid silo]

Aye

[coarse comet]

sorry

lol

[timid silo]

No problem

Okay so

you know rigth why we added v on both sides?

[coarse comet]

ya

[timid silo]

so that we get a value

180*

[coarse comet]

ya

[timid silo]

q + u + v = 180°

[coarse comet]

ya

[timid silo]

We just had q + u

And in the other side v

Now if you had written , q + u = 180 - v

That would have been correct too

[coarse comet]

ya but i wouldbt have thought

of 180 being 3v

[timid silo]

And also we knew q + u = 2v

[timid silo]

Yeah but now you know?

[coarse comet]

ya

lol

[timid silo]

Good

v = 60 right

[coarse comet]

ya

i got that

and ik

[timid silo]

Still this isn't what the question 1 asked right

[coarse comet]

ya

[timid silo]

So subbing that

[coarse comet]

but i could just solve for everything now

tho

no?

[timid silo]

Yep

[coarse comet]

ya

[timid silo]

Tbh yeah

Try it

[coarse comet]

ok

wait can I?

[timid silo]

Yeah you can

[coarse comet]

alr

wait

my brain must be really fried or smt

give me asecond

[timid silo]

No worries

[coarse comet]

is it 60 for all of them

[timid silo]

No

Only s and v

For all you can say

You can't prove they're all 60

[coarse comet]

i see

[timid silo]

They can be, but can't say for sure

[coarse comet]

is there a way to actually solve for all the values of the angles

with only knowing

[timid silo]

Nah

[coarse comet]

v is 60

[timid silo]

Nah

[coarse comet]

ok

for some reaosn

[timid silo]

You can't get the values , only relations

[coarse comet]

i thought

ya

i realzied

[timid silo]

Yeah

May I

Help?

😂😂

[coarse comet]

is

it c

[timid silo]

Yessss

Good

[coarse comet]

ya ok

im done 💀

[timid silo]

😂😂😂

Go sleep

[coarse comet]

ya most def

gn

[timid silo]

😂😂, cya

[coarse comet]

cya

[timid silo]

It's like 10 am here so

Goodnight 😂😂

[coarse comet]

goodngiht ill probs be on here tmr

lol

pce

[timid silo]

.close

Do it meapmeep

@coarse comet use .close before you go

[coarse comet]

.close

[obtuse pebble]

Channel closed

Closed by @coarse comet

Use .reopen if this was a mistake.

[coarse comet]

kk

[coarse comet]

done!

gn

[timid silo]

Gn

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
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[sage dagger]

I know they used the chain rule here but how did they get e^x(ln a) from e^(ln a)x

[hollow dove]

multiplication is commutative

[wooden cipher]

${(a^b)}^c=a^{bc}$

[warm shale]

GarlicBredFries

[sage dagger]

oh

How did they get (ln a)e^x•ln a

How did they get ln(a) at the front

[wooden cipher]

d/dx(e^f(x))=f'(x)e^(f(x))

f(x)=x(ln(a))

[sage dagger]

could you texxit that

[wooden cipher]

f'(x)=ln(a)

ok hold on i takr forever texing

[sage dagger]

The derivative of e^x is e^x

[wooden cipher]

$\frac{d(e^{f(x)})}{dx}=f'(x)e^{f(x)}$

frick

[warm shale]

GarlicBredFries

[wooden cipher]

By chain rule

[sage dagger]

Shouldn’t this be the final answer then

[frank monolith]

whats wrong in the last one

[sage dagger]

It doesn’t make sense why does the power of ln a disappear and why does it be rearranged to a^x where did e^x•ln(a) go

[wooden cipher]

$(e^{ln(a)})^x=a^x$

[warm shale]

GarlicBredFries

[wooden cipher]

Use this again

But backwards

[sage dagger]

But how

Why does e and ln(a) just disappear

[wooden cipher]

$x^{log_x(y)}=y$

[warm shale]

GarlicBredFries

[wooden cipher]

$ln=log_e$

[warm shale]

GarlicBredFries

[sage dagger]

Ok I sorta get it

[royal basin]

$\ln$ and $\log_e$, y'all.

[warm shale]

Ann

[obtuse pebble]

@sage dagger Has your question been resolved?

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[karmic hedge]

If I have 1/(3x^2 + 1) I should make u = (√3)x but what do I do if I have 1/(x^2 + 3)

[karmic hedge]

Cause I can take out 1/3 and put it in front of the integral, but then I need to multiply by 3 because of du, right? Which goes in front of the integral too, but that cancels out the 1/3? This doesnt seem right....

[tardy epoch]

Factor out a 1/3 first

[karmic hedge]

Oke so 1/3 ∫ 1/((x^2)/3 + 1)

[tardy epoch]

What

3/3=1

Oh I got mixed up from your question

[karmic hedge]

Haha

What do I do next

[tardy epoch]

Which integral is your starting one again

Don't put two in the same question

[karmic hedge]

This one

∫ 1/(x^2 + 3)

[tardy epoch]

Yes then this is the right next step

Do u sub now

[karmic hedge]

1/3 ∫ 1/((x^2)/3 + 1)
u = x/3, du = 1/3, dx = 3
1/3 ∫ 1/(u^2 + 1)(3)
(1/3)(3) ∫ 1/(u^2 + 1)
∫ 1/(u^2 + 1)
arctan(u)
arctan(x/3)
Is this correct?

[tardy epoch]

x^2 / 3 does not become u^2 if u = x/3

[karmic hedge]

u = x/√3?

[tardy epoch]

Yes

[karmic hedge]

Ah oke

1/3 ∫ 1/((x^2)/3 + 1)
u = x/3, du = 1/3, dx = 3
1/3 ∫ 1/(u^2 + 1)(√3)
(1/3)(√3) ∫ 1/(u^2 + 1)
∫ 1/(u^2 + 1)
√3arctan(u)
√3arctan(x/√3)
Correct now?

[tardy epoch]

,w diff sqrt(3) arctan(x/sqrt(3))

[warm shale]

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[karmic hedge]

Nice

Oke it makes sense now, I see my mistakes

Thank you!

❤️

.close

[obtuse pebble]

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[obtuse pebble]

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[karmic hedge]

https://i.imgur.com/mQ8jjPq.png

[karmic hedge]

I know that when lissajous figures are in the form Asin(B(t+C))+D that I can take 2pi/B to get the period, but the periods are different, the period of x(t) is 2pi, the period of y(t) = 4pi. What do I do to get the period of them together?

[haughty coyote]

Think about it

What's the definition of the period?

[karmic hedge]

How often it repeats

[haughty coyote]

More rigorously

[karmic hedge]

(not me googling rigorously)

I dont really know haha its just how long the function goes before it repeats itself

[haughty coyote]

a function with domain D is said to be T-periodic iff for all x in D, x+T is in D and f(x+T) = f(x)

Then, "the" period of f is the smallest T such that f is T-periodic

(ignore the case where this min doesn't exist)

[karmic hedge]

1 moment I need to read this a few times to make sure I understand haha

[haughty coyote]

Here you have a function from R to R^2, f(t) = (x(t), y(t))

[karmic hedge]

By R^2 do you mean like the 2nd value of k?

[haughty coyote]

What's k ?

[karmic hedge]

like sin^-1(0) is pi(k)

that k idk what its called

[haughty coyote]

An integer

[karmic hedge]

oh

[haughty coyote]

sin^-1({0}) = pi Z

[haughty coyote]

Never seen Cartesian products ?

[karmic hedge]

Ah I thought k was to do with the period haha

[karmic hedge]

Maybe but I dont recognise the name I will google them

[haughty coyote]

Cause then the notation of powers makes a lot of sense

[karmic hedge]

No I dont think so

[haughty coyote]

But otherwise no, R^2 is just the set of all pairs of real numbers, i.e. 2D space

[karmic hedge]

I think I get what you mean by this tho

Like I just take the lowest common multiple of 2pi/B, right?

[haughty coyote]

Cause you don't say sin is 42pi periodic, even though it is

[karmic hedge]

https://i.imgur.com/G87Kwfh.png

Yeppp I get it now

Thats easy enough haha

[karmic hedge]

This trick saves so much time

[haughty coyote]

You want a period of x and y at the same time

[karmic hedge]

Yeye

[haughty coyote]

I wouldn't call that a trick

[karmic hedge]

I didnt understand how to get both together but now its kinda obvious haha

[karmic hedge]

oh

Well whatever it is its handy 😅

Thanks again!!

❤️

.close

[obtuse pebble]

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[still badger]

If $x+\frac{1}{x} = 4$ what is $x^6+\frac{1}{x^6}$?

[warm shale]

babario

[frank monolith]

use $a^3 + b^3 = (a+b)(a^2-ab+b^2)$

[warm shale]

Dissrupt

[still badger]

@timid silo

[timid silo]

oh right okay

okay i think

so

Uh, okay, we have this equation, like x plus one over x equals four. And we need to find x to the power of six plus one over x to the power of six. To get that, we use this expression with x squared plus one over x squared, which is, umm, (x plus one over x) squared minus two, I think. So, that's like 16 minus two, which is, umm, 14, I guess.

Now, we can use 14 in the main expression, which is x squared plus one over x squared whole cubed minus three times x squared plus one over x squared. Let me plug in 14 and see what we get. So, it's 14 cubed minus three times 14. Doing the math, it's, umm, 2,744 minus 42, maybe?

So, the final answer is, 2,744 minus 42, which might be, 2,702. Yeah, I think that's it, but not entirely sure.

took me time that

and dont lecture me that i cant explain well

i already know

[still badger]

you can do stuff like this here actually $x + \frac{1}{x}$

[warm shale]

babario

[still badger]

just make sure u wrap the latex round $$

[timid silo]

hmmm okay

[still badger]

but yea nvm ill read

wait a min

[timid silo]

yeah alr, i was unsure throughout as i dont usually write out my whole thought process

as i think i may actually be getting some parts wrong

[frank monolith]

your answer is correct

[timid silo]

@frank monolith i got it correct?

[still badger]

uhh so ur main expression is ${(x^2 + \frac{1}{x^2})}^3 - 3x^2 - \frac{1}{x^2} = x^6 + \frac{1}{x^6}$?

[frank monolith]

yes

2702 is correct

[warm shale]

babario

[frank monolith]

oh wait, whose question is this?

[still badger]

mine

[frank monolith]

confused screaming

[still badger]

this is correct right?

[timid silo]

hmmm im not sure

im pretty sure its

$x^6 + \frac{1}{x^6} = \left(x^2 + \frac{1}{x^2}\right)^2 - 2$

[warm shale]

Avaaaa

[frank monolith]

no

[timid silo]

ist not?

its

[still badger]

nah

[timid silo]

dam

[frank monolith]

you are missing a 3 with -1/x^2

[still badger]

u said cube previously

[still badger]

tats squared

[frank monolith]

but your expression wont be much useful

[still badger]

oh

[timid silo]

hmmm right

[frank monolith]

use this

where a is x^2

and b is 1/x^2

[still badger]

ah gotit

thats it?

hm wait

[timid silo]

so this then

$x^6 + \frac{1}{x^6} = \left(x^2 + \frac{1}{x^2}\right)^2 - 2$

[warm shale]

Avaaaa

[frank monolith]

nope

[timid silo]

wait no

i did the wrong thing again

dam i got it before

now its just

well i got it right the first time so, oh well

[still badger]

so in the end its $x^6 + \frac{1}{x^6} = (x^2 + \frac{1}{x^2})(x^4 + \frac{1}{x^4} - 1)$?

[warm shale]

babario

[frank monolith]

yeah

now convert them again

by using $a^2 + b^2 = (a+b)^2 - 2ab$

[warm shale]

Dissrupt

[still badger]

yea

so in the end its 14 * (194-1) right?

yea

thx mate(s)

.close

[obtuse pebble]

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[fossil spoke]

i expanded the LHS and also worked from RHS but got stuck

[tight thunder]

can you show your work

[obtuse pebble]

@fossil spoke Has your question been resolved?

[tight thunder]

FLASHBANG

[fossil spoke]

?

.close

[obtuse pebble]

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[novel grotto]

Square the sum and simplify, then noting that the original sum is positive you can take square root again

There is no need for trigonometry

[tight thunder]

you mustve done something wrong

1/sqrt(2)-1 is not equal to sqrt((1/sqrt(2)-1)^2)

@digital jay

[obtuse pebble]

@digital jay Has your question been resolved?

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[pastel sluice]

HELP can someone show me working out

[weary vortex]

Do you know basic trigonometry?

[pastel sluice]

uhhh yea no new

ok i got 26.78 but the answer is 49.32

[weary vortex]

Can you show your work? 🙂

[weary vortex]

its not 11.3

[pastel sluice]

OHHHH

I KNOW NOW

[weary vortex]

🙂

[pastel sluice]

ye im retade hahaha

well thanks anyawys

[weary vortex]

no worries! 🙂

[wind scaffold]

find the length of the adjacent first forgive my possible spelling error

[obtuse pebble]

@pastel sluice Has your question been resolved?

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[obtuse pebble]

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[steady lodge]

“graph the function y=4(cosX)^2 from -2(pi)<_X<_2(pi)

[steady lodge]

<_ is less than or equal to

I don’t even know where to start to be honest

[grizzled shore]

Usually we use <= and >=

[steady lodge]

I think it’s like the wavelength thing right?

ohhh smart thanks yeah

[grizzled shore]

Can you graph just y = cos(x) from -2pi to 2pi?

[steady lodge]

tell me if this looks right let me try

so 4 is a stretch factor

does that effect magnitude?

[frank monolith]

yeah

stretches the graph along y axis so magnitude

[frank monolith]

can x^2 be negative?

also that is similar to sinx graph

[steady lodge]

no

yeah I only learned sinx last year I think

[frank monolith]

so can (cosx)^2 be negative?

[steady lodge]

I don’t think so

so it can’t dip below 0

[frank monolith]

no

[steady lodge]

so does it like bounce off 0 or something?

[frank monolith]

and what will be the maximum value of this function?

yeah

[steady lodge]

either 4 or 1

[frank monolith]

why 1?

i just asked maximum

[steady lodge]

yeah, like it only goes that high

isn’t that the default

but there is a 4

so it goes higher than default

perhaps, up to 4?

[frank monolith]

yeah

cos(x) has a maximum of 1, similarily cos^2(x) also has a max of 1

[steady lodge]

so would the equilibrium be 2?

[frank monolith]

so 4cos^(x) has max of 4*1 = 4

[steady lodge]

ohhh okay

that makes sense cause it is just stretching it vertically

i’m trying to think, if the middle line isn’t 0, is it just the average of 4 and 1?

so 5/2

[frank monolith]

why 1?

can cosx be zero?

[steady lodge]

0 sorry

so it bounces of 0 but never touches it

[frank monolith]

why?

[frank monolith]

.

[steady lodge]

I think so yes

let me think

no because that would mean a side length would be 0

and that’s not possible

but I feel like i’ve seen 0 in a calc after putting in cos(x)

[frank monolith]

,w plot cosx

[warm shale]

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Requested by old.guy

[frank monolith]

is cosx zero somewhere?

[steady lodge]

yes

[frank monolith]

so will cos^2(x) be zero?

[steady lodge]

yes

at 90* right?

[frank monolith]

yeah

and 270

[steady lodge]

why is that?

[frank monolith]

you havent studied cosx?

[steady lodge]

isn’t taking like cos(90) like not allowed?

because the opposite is the hypotenuse?

[frank monolith]

why is that

[steady lodge]

.

[frank monolith]

i dont think i understand that?

[steady lodge]

like

say you have a triangle right

if you put yourself in the 90* corner

the hypotenuse is always the greatest side of the triangle, and there id no greater angle than 90, therefore, the opposite side of the 90* angle would always be the hypotenuse

[frank monolith]

do you know the quadrants?

[steady lodge]

yes

[obtuse pebble]

@steady lodge Has your question been resolved?

[obtuse pebble]

@steady lodge Has your question been resolved?

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[inner spear]

in this video professor leonard posts about this "figure 8" shape and says that greens theorem is not applicable to this case because it is not a "simple curve". Why must the curve be simple?

[inner spear]

I know that green's theorem only works on "simple curves", but why?

the curve still encloses a region (2 regions)

[polar fossil]

it's because the curve is going around one region clockwise and one region counterclockwise

it's for morally the same reason that this integral is 0, even though it looks like the area enclosed is > 0

[obtuse pebble]

@inner spear Has your question been resolved?

[inner spear]

I'm a bit uncertain - do you mean that they will have opposite signs of magnitude?

[polar fossil]

yes

it's also similar to the reason that $\int_a^b f(x) = -\int_b^a f(x)$

[warm shale]

Hayley

[polar fossil]

you're effectively reversing your bounds

[inner spear]

ah I see

so by convention we take anticlockwise as being positive

thanks

.close

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[fervent stump]

is it possible to numerically/algebraically show that 1. as x increases y increases; 2. as x increases y increases at an increasing rate for the following equation. The domain is restricted to Pi/2

[fervent stump]

The domain is restricted to (0;Pi/2)

[polar fossil]

sure, take the derivative

[fervent stump]

i tried and got the following expression -3/8 (sinθ/θ)^((-11)/8) ( cos⁡(θ)θ-sin⁡(θ))/θ^2

[polar fossil]

yeah now show that's > 0

[fervent stump]

this looks better

[polar fossil]

for your domain

[fervent stump]

and you get as far as i got it x<tan x

and then i am stuck

and taking a second derivative of this hell doesn't look plausible

so perhaps there is a different way or am i missing something?

[polar fossil]

well

it might be possible to prove something about sinx/x

and apply that to your function

[polar fossil]

i would look more at x^-3/8

,w graph x^(-3/8)