help-28

[devout valley]

Was digging for your formulas

[thick hedge]

huh charbit you really love emotes, don't you?

[devout valley]

(also saves having to type out responses sometimes lol )

[next basalt]

I'm not sure how to solve

$\int_0^{\pi/4} \frac{x}{cos^2(x)}dx= [xtan(x)]_0^{\pi/4} - \int(tan(x)dx$

[thick hedge]

uh, the the second part should simply be $\int(tan(x)dx$

[glossy valve]

Why am. I here

[next basalt]

yeah.. sorry hard to keep track

[glossy valve]

Merineth

[next basalt]

ah idk aynmore.

aåpdklwa

[devout valley]

hint: tan(x) = sin(x)/cos(x)

[thick hedge]

try writing tan(x) as sin(x)/cos(x)

then use a u sub

[next basalt]

[thick hedge]

can you do it from here?

[next basalt]

y redoing it

1 sec

$[xtan(x) + ln(cosx)]_0^{pi/4}$

[glossy valve]

Merineth

[thick hedge]

looks right to me

[next basalt]

thank god...

I would've solved it if i recognized that tan could be rewritten :c

[thick hedge]

btw, is this dutch ?

[devout valley]

it's @next basalt's, posted from before

[next basalt]

It's Swedish

[thick hedge]

ah, ok

[next basalt]

Honestly based on the amount of time i spend on this server recently i'll end up with my own channel lmfao

[devout valley]

Some people have like their own threads, in the topic-specific channels, you never know

[thick hedge]

You can create your own in help forum

[devout valley]

Help forum is more for specific questions, basically like these channels (and "eventually" the channels here will be phased out)

[next basalt]

I usually always get help quickly here so I don't need one

[devout valley]

There's something like this thread https://discord.com/channels/268882317391429632/1195800467880558672 and there are others

[next basalt]

i'm also learning linear algebra..

this is way worse tho

I'm desperate for food now tho... so i'll take a 30 min break :)

I'll be back later

.close

[full forum]

Channel closed

Closed by @next basalt

Use .reopen if this was a mistake.

[next basalt]

[devout valley]

Wow you started like three hours ago, definitely well deserved break I'd say! Enjoy the food

[thick hedge]

Enjoy your break! You deserve it after working so hard!

[full forum]

Available help channel!

Send your question here to claim the channel.

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>.
• Type the command .close to free the channel when you're done.
• 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.

[hasty python]

anyone know how to do this the answer is 4 congruent triangles but i just dont know the steps to get to that solution

[waxen geyser]

I think there's no solution needed no? Just say 4 congruent triangles and you're set

[limpid junco]

I see 8 but ok

[limpid junco]

8 of these

[hasty python]

the thing is if i js say that my teacher wont accept it

[waxen geyser]

Just say 8

[hasty python]

ok

[waxen geyser]

well make it a sentence

[hasty python]

also i meant 4 pairs of congruent triangles

my bad

[waxen geyser]

what kinda teacher is that

[rugged void]

wait hang on

unique triangles or...?

if unique there's 1

and it says PAIRS of congruent triangles

that's why the answer's 4

[hasty python]

i think i figured it out

its mostly just reflexive property then vertical angles then by SAS they're congruent

[rugged void]

yep

[hasty python]

.close

[full forum]

Channel closed

Closed by @hasty python

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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>.
• Type the command .close to free the channel when you're done.
• 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.

[next basalt]

$\int_0^{\infty} \frac{cosx}{3+sinx} dx = [ln|3+sinx|]_0^{\infty}$

[next basalt]

How do i proceed from here?

[static bramble]

$\infty$

[glossy valve]

Steakanator

[thick hedge]

Enjoyed your break ?

[glossy valve]

Merineth

[rugged void]

DNE i think

[next basalt]

Haha yeah i got foodcoma !

The answer says divergent but i don't know what that means?

[rugged void]

the limit DNE

[next basalt]

That it is infinit large?

[rugged void]

no

not in this case

basically sin (inf) is undefined since it oscillates infinitely

so ln of sin inf is undefined as well

thus this integral should, theoretically, b undefined

[next basalt]

[rugged void]

,w calculate int from 0 to inf of cos x/(3 + sin x)

[glossy valve]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by renegade3192

[rugged void]

see

[next basalt]

Weird

[rugged void]

this is the graph.

it repeats

so at certain points the int is 0

from 0 as in

at others 1

etc

since it'll cancel out depending on the endpoint

inf... isn't defined so

[next basalt]

So how do i deal with integrals that deal with infinity?

Replace infinity with N?

[rugged void]

no use lims

oh wait yea

i get what ur tryna say

usually that's what we do

:)

[next basalt]

Do i have to type lim every time ._.

[rugged void]

i mean

yea?

[next basalt]

[rugged void]

💀

[devout valley]

You could do something like omit the limits but then "add them in later"

So like $\int_a^N (\ldots) \dd x = \ldots = \int_a^N (\ldots) \dd x \to \ldots$

[glossy valve]

@devout valley

[next basalt]

$\int_0^{\infty} \frac{x}{1+x^2} dx = \lim_{N\to\infty} \int_0^N \frac{x}{1+x^2}dx$

[rugged void]

wait latex doesnt have limits? damn

[devout valley]

it does, $\lim_{N\to\infty}$

[rugged void]

oh i see

[glossy valve]

Merineth

[next basalt]

:( i haven't solved it yet

[rugged void]

...

i am a fkn idiot

TEXIT delete the message

AAAAA

,texit purge

no?

[devout valley]

click the bin

[rugged void]

cant for texit

OH i see

im dum

[next basalt]

iirc

i did one similar earlier

don't i u sub denominator

[rugged void]

yes

try subbing x^2 = u

[devout valley]

[whole denom is a tiny bit easier, but just a tiny bit!]

[rugged void]

it won't make any difference, will it?
@devout valley (ur bio says to ping when replying)

[devout valley]

It doesn't in the end, but it may look nicer to use the integral you get with u = 1 + x^2 as opposed to u = x^2

[next basalt]

$[ln|1+x^2|]_0^N$

[glossy valve]

Merineth

[rugged void]

good point

no

close

very close

the derivative of x^2 + 1 is 2x, not x

what does that tell you?

[next basalt]

that i factor out 1/2

[rugged void]

yes

gj

[next basalt]

:)

[rugged void]

so $0.5 \cdot \ [ln(1+x^2)]_0^N$

[next basalt]

Once again the answer is divergent

[rugged void]

i think that's how you do it?

[glossy valve]

ren

[rugged void]

yes it is

[next basalt]

Why exactly is it underfined?

It has something ot do with ln?

[rugged void]

no

actually yes

since ln 1 + n^2 as n tends to inf is ln 1 + inf = ln inf

that's just inf

so yea

if it were smth else like x^-2 it'd be 0 instead of infinity

[next basalt]

Ah i see, since ln goes to infinity so will the integral

[devout valley]

[but at the other endpoint, 0, you may have problems ]

[rugged void]

i meant ln 1+x^-2

so yea
but they're not wrong either

u done @next basalt ?

[rugged void]

OH GOD

WHY

LIKE SERIOUSLY WHY

YOU KNOW IT'S UNDEFINED AND DIVERGENT

[next basalt]

ignore (a), why is (b) = 1

$\int_0^{\infty}xe^{-x}dx$

[rugged void]

hold up

[glossy valve]

Merineth

[devout valley]

"exponentials beat powers", so -Ne^{-N} goes to zero, and e^{-N} should clearly go to zero as N goes to infinity

[rugged void]

where did half that stuff cm from

also you left out dx (irrelevant here ik but my OCD is kicking in even with ur awesome, i might add, handwriting)

[devout valley]

By parts and evaluating

[rugged void]

id see half the parts

oh they just evaluated

isee

*i see

i thought that was the expansion of parts and was confused as hell

[next basalt]

Which part are you comparing the exponentials beat powers?

[rugged void]

,w xe^-x int from 0 to inf

[glossy valve]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by renegade3192

[devout valley]

"so $-Ne^{-N}$ goes to zero"

[glossy valve]

@devout valley

[devout valley]

You can see that as $-\frac{N}{e^N}$ and the denominator gets larger much faster than the numerator ever will

[glossy valve]

@devout valley

[next basalt]

ahh right

Since it's trending towards 0 the larger the N becomes

[rugged void]

yep

[next basalt]

$-\frac{1}{e^N}$

[glossy valve]

Merineth

[next basalt]

And same goes for this

[rugged void]

and for 0, it multiplies

and the e^-x turns to 1

happy ending

[next basalt]

[rugged void]

done?

[next basalt]

Nop

[rugged void]

ope

we have more

it's almost midnight here so i gtg soon

[next basalt]

||a lot more||

[rugged void]

but this is genuinely good practice for me too

and it's fun

LOL

THAT FEELING AFTER THE SPOILER

i love that sticker

[next basalt]

I sit here between 13-21 CET

[rugged void]

CET?

idk what that is

and idrc

next question cmon

im hungry

[next basalt]

I tend to attempt them before posting :3

[devout valley]

,ti --at CET

[glossy valve]

The time in CET is 07:30 PM (CET) on Sat, 03/02/2024.
chartbit is 1 hour behind, at 06:30 PM (GMT) on Sat, 03/02/2024.

[rugged void]

damn

what's at?

like australian time?

that a thing?

[devout valley]

Command to tell you the time at a particular time zone, such as

,ti --at pst

[glossy valve]

The time in PST8PDT is 10:30 AM (PST) on Sat, 03/02/2024.
chartbit is 8 hours ahead, at 06:30 PM (GMT) on Sat, 03/02/2024.

[rugged void]

i am hammered aren't i

[next basalt]

||OKAY NOW I CAN NARROW DOWN WHERE CHARTBIT IS LOCATED BASED ON THE TIMEZONE. I'M COMING IN HOT ||

[rugged void]

at is just at

[rugged void]

LOL

https://tenor.com/view/rocket-shi-rocket-bomb-gif-26885344

Merineth lifting off to find chartbit

here we see a wild Merineth in it's natural habitat, stalking a mathematics DC server helper

it is primed to narrow down the exact locations of targets that warrant its interest, often soliciting a nuclear or orbital strike

@next basalt next one??

[next basalt]

Solving, one moment

[rugged void]

aaaaaaaaaaaa

im HUNGRY

and i dont even have chartbits to stalk like you

[devout valley]

Grab food and then come back and lurk

[rugged void]

https://tenor.com/view/sulk-mad-angry-hmp-cat-gif-17529533

[rugged void]

no im hungry for maths

nom nom nom

waits

https://tenor.com/view/hang-on-gif-27524073

https://tenor.com/view/waiting-titanic-gif-25999423

MERINETH IS BACK

FINALLY

WHERE IS MY SLAMS HANDS MATHS

[next basalt]

$\int_2^{\infty} \frac{x}{x^4 -1}dx$

[rugged void]

hmm

[glossy valve]

Merineth

[next basalt]

My answer is

[rugged void]

hang on hang on

i need to evaluate the integral

while absolutely hammered

oh partial fractions? no?

im dum

eh i give up

im horrible at PFD anyways

[next basalt]

$\frac14 \ln[x^4-\sqrt{x^4}]$

[rugged void]

...

u sure that's what u meant

[glossy valve]

Merineth

[next basalt]

the sqrt should be 4th grade

so ig just x?

[rugged void]

yea

x^3/x^4 - x...?

no wait

am dum

[next basalt]

I did not use partial fraction, i used u sub

[rugged void]

basically what im tryna say

is that the derivative of that is (x^3 - 0.25 / x^4 - x)

so doesnt seem to be correct

@devout valley lil help

[queen crater]

What u did you take?

[next basalt]

x^4-1

[rugged void]

that... doesn't work

bc du is 4x^3 dx

[queen crater]

Ye you need to be a little more clever about it

[devout valley]

(or at least not very nicely )

[rugged void]

no it straight up doesn't work

[rugged void]

it's 00:16 AM for me rn, i think i'll go in about 3-5 minutes

why did i write AM

it was obvious

see im hammered

[next basalt]

I think i forgot to cancel an x in denom

[queen crater]

That's just not right

[rugged void]

where did u get last step

it doesnt make any sense

please tell me im making sense

https://tenor.com/view/facepalm-really-stressed-mad-angry-gif-16109475

i cant tell anymore

[next basalt]

Hold on i see what i did wrong

Ur right nel

[rugged void]

waits

can i spoiler a WA request

imma test it

appears not

okay so it doesn;t

@next basalt done? i'd like to finish this b4 i sleep

[next basalt]

Hm i tried it again but this time i got $\frac14 \ln|x^2|$ which is most likely wrong

[glossy valve]

Merineth

[next basalt]

Partial fraction seems to be the only way afaik. Thought i could get away by using u substitution

[queen crater]

u-sub is the way to go

Just not the same u

[rugged void]

what's the u sub

spoiler it

i'd like to know

[queen crater]

When you're dealing with an x^4, always check if (x^2)^2 is easier to work with

[rugged void]

bc i gtg soon

||so it's 0.5 int of du/u^2 - 1?||

[queen crater]

||yes||

[rugged void]

i see

and then PDR?

appears so

[next basalt]

So my mistake was taking the entire denominator with u substitution when i should've just taken x^2?

[rugged void]

cool ty @queen crater

[rugged void]

yep

[next basalt]

How in the world do i recognize that i should take x^2 and not x^4 -1?

[rugged void]

||i thought about u^2 first but wasn't sure about the later int, thanks for helping||

for nel

[queen crater]

You should recognize that the derivative you want should cancel the numerator

[rugged void]

always try it

jic

[next basalt]

Yes, that i did with x^4

[rugged void]

chartbit's still here??

[queen crater]

The numerator is x, something that cancels it is x too, so x^2 seems like a good thing to at least try

[rugged void]

s(he)/they're like our audience

@next basalt solved?

[next basalt]

So the goal is to cancel the numerator but also at the same time make sure we aren't left with something worse after the cancelation? Since we can just factor out 1/2 after u subbing x^2?

[rugged void]

yes

[queen crater]

Well in general the goal is to get to a form for which you know a formula

[rugged void]

but that helps

[queen crater]

Here it would be 1/(x^2-1)

[rugged void]

in getting to nel's goal

[rugged void]

-1

not + 1

[queen crater]

Yes

[rugged void]

then factorize

via PFD

my god i cant type

alr guys i think imma b heading out

it's 00:34 (kinda offended you aren't stalking me Merineth, i've announced my time before. hmph)

okay jk, this was honestly a lotta fun

peace

[devout valley]

Have a good rest

[rugged void]

yep

[next basalt]

If i factor out 1/2

and then i apply partial fraction

do i multiply 1/2 infront of both the integrals?

[devout valley]

Yep, you'd need to make sure you apply it to both the integrals you find

[next basalt]

hm

I applied partial fraction and got

$-\frac14 \int_2^N \frac{1}{u-1} du + \frac14 \int_2^N \frac{1}{u+1}du$

[glossy valve]

Merineth

[next basalt]

the only way i see what i can do here is u subbing again?

so i get 1/t

Is it okay to sub while it's already subbed?

Is that equivalent to several chains in the chain rule?

[queen crater]

Of course

You may want to remember these simple cases though

[devout valley]

you may have the + and - the wrong way around I think

[queen crater]

Integral of 1/x is ln(x), so integral of 1/(x+a) is just ln(x+a)

Because the derivative of x+a is the same as the derivative of x

[next basalt]

I don't think so? Checking atm

I got A = 1/2 B = -1/2

[devout valley]

But what did A and B relate to originally?

the numbers are fine but their order isn't

[next basalt]

$\frac{1}{(u-1)(u+1)} = \frac{A}{(u-1)} + \frac{B}{(u+1)}$

[glossy valve]

Merineth

[next basalt]

That could be because you solved with the in the other order on the LHS?

[queen crater]

This plus A = 1/2 B = -1/2 gives you the other order

[next basalt]

$1 = A(u+1) + B(u-1)$

[glossy valve]

Merineth

[next basalt]

If u = 1 then
1 = 2A
A = 1/2

if u = -1 then
1 = -2B
B = -1/2

[devout valley]

We're happy with how you got those values

[devout valley]

put them in here: what does it look like?

[next basalt]

okay i see what you mean

i forgot to factor out the -?

$\frac14 \int_2^N \frac{1}{u-1} du - \frac14 \int_2^N \frac{1}{u+1}du$

[devout valley]

Yea, should have been the other way around, the thing that corresponded to your first term should have been +1/2 and the second term -1/2

[glossy valve]

Merineth

[next basalt]

I despise integrals

one tiny - wrong and gone goes 5 points on the exam

lmao

$\frac14 (ln|x^4-2| - ln|x^4|)$

[glossy valve]

Merineth

[next basalt]

wrong?

i give up..

[devout valley]

Oh wait a moment actually, forgot you subbed

[next basalt]

$\frac14 \ln|u-1| - \frac14 \ln|u+1|$

[glossy valve]

Merineth

[devout valley]

Yeeaaaa not this cause the sub was u = x^2

[next basalt]

oooh

i completely forgot

i thought i did the x^4 one lol

[devout valley]

Me too tbh

[next basalt]

but yeah u = x^2

[devout valley]

I'm gonna just make the excuse that you should have changed the limits and then we could happily have forgotten about the sub

[devout valley]

[similar don't forget to disinguish the limits here that they're wrt x, not u]

[full forum]

@next basalt Has your question been resolved?

[next basalt]

time for linear algebra.. :(

How do i determine a planes equation?

If the information i'm given are two lines?

I understand that a plane is

Ax + By + Cz = d

[queen crater]

Two lines can't always form a plane

Do they intersect? Are they parallel?

[next basalt]

Yes they intersect

It was the question from yesterday but b)

Denote with L1 the line through the points A = (2, −2, 0) and B = (7, −17, 10) and with L2 the line (x, y, z) =
(−3, −7, −14) + t(1, 2, 3), t ∈ R.
a) Show that L1 and L2 have a common point. Which?
b) State the equation for the plane that contains both L1 and L2.

I determined a point (1,1,-2) in a)

[queen crater]

Ok so an easy way to find an equation of the form ax+by+cz=d is to find a normal vector and then adjust d

[next basalt]

and the normal vector is determined by the cross product between both lines directional vector?

[queen crater]

Yes

(a normal vector, doesn't really matter which)

[next basalt]

Okay i got the n vector = (65, 5 -25)

[queen crater]

Ok, haven't checked, but that gives you a,b,c, now you need to adjust d so that all the points in the two lines are also in the plane

[next basalt]

Can i take the dot prodcut between the normal vector i just found and dot it with (r - P) where r and P are two points in the plane?

[queen crater]

That would just give 0

[next basalt]

oh right

[queen crater]

Just set up an equation

I mean, pick a point, use the incomplete equation of the plane that you have

It straight up gives you d

[next basalt]

oh just put in the cordinates of x y z of a point and solve for d?

[queen crater]

Yes

That's the point of the equation of the plane: every point (x,y,z) that is in the plane satisfies the equation

[next basalt]

Oki i got d =120

is there a way to verify my plane?

Ooh i can check if the dot product between the normalvector and the plane = 0

[queen crater]

Yeah

[next basalt]

I'm given a plane, and a line

(a) Determine the point P where the line L and the plane pi intercept

(Sorry if it's bright btw)

But I plugged in the lines equation into the planes equation to find out t, then i took that t into the lines equation and found the point (-3,3,2) seems ok?

[queen crater]

Yes

[next basalt]

(b) asks me to determine the ortogonal projection of L_1 and L on the plane pi

However they haven't given me the equation for L_1?

So how can i do the cross method

[queen crater]

I think you're not reading it correctly

They are calling the projection L_1

[next basalt]

"Enter the orthogonal projection L1 of L onto the plane π."

[queen crater]

Right

So L_1 is the projection of L onto pi

Find L_1

[next basalt]

oh so it's the cross product between the normal vector and L which gives me the vector V_l

[queen crater]

The cross product will give you a vector orthogonal to both

It will be orthogonal to L_1 too

[next basalt]

My suspicion is that i shall use the projection formula to somehow establish L_1

[queen crater]

You can

[next basalt]

Oh i think i have an idea

if i project v onto the plane?

[queen crater]

What's v?

[next basalt]

The vector of L

[queen crater]

Yeah that works

[next basalt]

Just to confirm

This formula means "projecting u onto v?

or is it v onto u

[queen crater]

Read it like a function

The function is called proj_v

So it's the projection of u

The u' indicates that too, it's a new vector that comes from u, it's its projection

[next basalt]

I figured out what my lambda is or rather the coefficient before my planes vector

do i just take it not and multiply it into my planes vector?

[queen crater]

You lost me

[next basalt]

Basically i'm projecting v towards the plane by using the projection formula.

$\frac{v \cdot n}{|n|^2}$

[glossy valve]

Merineth

[next basalt]

Is my lambda = coefficient which i got to be 8/3

multiplied with my planes normal vector

[queen crater]

What lambda are you talking about

[next basalt]

My teacher says lambda is the coefficient

[next basalt]

he says this is "lambda"

[queen crater]

Oh okay

[queen crater]

Then yes

[next basalt]

[queen crater]

But you projected onto the normal vector

not onto the plane

[next basalt]

Oh

So i did it wrong?

[queen crater]

Well, depends

Were you going to do something else after that?

[next basalt]

No?

I just took

[queen crater]

Then yeah you did it wrong

[next basalt]

$\frac38 (1,-,1,1)$

[glossy valve]

Merineth

[queen crater]

This is just a normal vector of the plane

It doesn't give you L_1

[next basalt]

But he doesn't want the lines equation

only the projection

[queen crater]

L_1 is a line

[next basalt]

Yea

[queen crater]

L_1 is also a projection, as in the projection of L onto pi

The question asks you to find L_1, right?

[next basalt]

I'm not sure tbh he is very unclear:
It just says:

"State the ortogonal projection L_1 by L on the plane pi"

• State the ortogonal projection L_1

Is what i see from it at least

[queen crater]

Ok well add in "the equation of"

State the equation of L_1 [...]

[next basalt]

He does say that on other questions

but not on this one

[queen crater]

It's just how you express a line

[next basalt]

Oh so he is just unclear then?

[queen crater]

Apart from drawing it

I mean it's not unclear to me, but it could be more clear I guess

[next basalt]

I can see that they have a point where L and L_1 cross.

But without the equation of L_1 i don't know how to figure that point out.

[queen crater]

You are thinking backward

You already have the point

[queen crater]

.

[next basalt]

But that isn't the point where L_1 and L cross each other?

That is when the plane and L cross

[queen crater]

Yeah

It's the same point

[next basalt]

The green point is what i found out

not the blue

[queen crater]

L_1 being the projection of L onto the plane, if you take that point on L and project it onto the plane, then it's just the same point

Oh okay you are not looking at the figure correctly

The green point is nothing

[next basalt]

OH

[queen crater]

The plane is behind the line there

[next basalt]

L is upwards away from the plane

[queen crater]

Yes

[next basalt]

Ahhh i see what you mean

[queen crater]

https://www.desmos.com/3d/1dc39e2e7b

[next basalt]

Doesn't that mean L_1 = (x,y,z) = (-3,3,2) +t(8/3, -8/3, 8/3)

[queen crater]

No, that's a line orthogonal to the plane

[queen crater]

It's the line that goes through the blue point here and goes in the direction of the normal vector

[next basalt]

It's getting a bit confusing

[queen crater]

You projected L onto the normal vector, so of course you get a line parallel to the normal vector

[next basalt]

So my problem is that i used the normal vector?

I should've used the plane?

[queen crater]

Kind of

[next basalt]

Oh

unless

So i'm trying to find a orthogonal vector

with the normal vector i just got?

[queen crater]

No there is an infinite number of those

[next basalt]

I see

[queen crater]

Ok here are three ways to do it

1. do the cross product between v and n to get v2, then do the cross product between v2 and n to get v1

2. pick another point on L, for example (2,1,3), project it onto the plane using the normal vector; that means set up an equation where (x,y,z) = (2,1,3) + (1,-1,1)*s and x-y+z=-4, solve for x,y,z; then you have 2 points to define L_1

3. directly project v onto the plane, but for that you need to first project it onto n (which you did) to get v3, and then subtract that: v1 = v - v3; L_1 is defined by (-3,3,2) + v1*t

[next basalt]

1. seems easiest afaik

But how would that look like graphically?

The first time i do the cross product between them, where would it point?

[queen crater]

3 is probably easiest, cross products can be a pain

[next basalt]

I have a really neat trick for cross product!

[queen crater]

I know about Sarrus's rule, it's still a pain sometimes

[next basalt]

Nope not sarrus rule afaik

2  x  3 =   3*2-4*1 = -1, 2, -1
3    4      1*3-2*2
-------
1    2
2    3```

for example

[queen crater]

Put all that in code:

 ` ` `
like this
 ` ` `

Damn it

without the spaces

That is Sarrus's rule

[next basalt]

Oh

[queen crater]

Looks a bit different but it's the same thing

So v is (5,-2,1), right?

And n is (1,-1,1)

[next basalt]

Well fuck

i just did n = (8/3, -8/3, 8/3) ?

[queen crater]

Works too, just inconvenient

[next basalt]

I always forget i can factor out

since the length doesn't matter

for the vector

:)

[queen crater]

How it looks graphically:
https://www.desmos.com/3d/5fb7823cb3
L in blue, n in red, v2 in gray, v1 (or L_1) in green

[next basalt]

How do i verify my answer?

[queen crater]

Graph it

If you take any point on L, you can get to it starting from (-3,3,2) by adding pv1 and qn, with some scalars p and q (p/q will be constant)

So (-3,3,2) + pv1 + qn = (2,1,3) + (5,-2,1)t, for any t, where p/q is always the same

That's a bit convoluted though

[next basalt]

I'll just trust my gut that it's right

(and cba to verify)

[queen crater]

It's just to say that the plane formed by n and v1 (and (-3,3,2)) contains L

[next basalt]

thanks nel

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[torn jolt]

Hey, how do I do this?

[torn jolt]

I need to do (b)

I'm presuming I need to find smallest $r$ possible such that $(x - 2)^2 + (y + 3)^2 + (z - 6)^2 = r^2$?

[glossy valve]

∞

[torn jolt]

For the sphere to be "touching" the yz-plane

[limpid junco]

Have you done a?

[torn jolt]

No, I haven't

I solely need to solve (b)

[limpid junco]

Well you know which axis is perpendicular to yz plane

[torn jolt]

Yeah, the x-axis

[limpid junco]

Yeah so imagine changing the sphere in x-axis

When will it touch the yz plane

[torn jolt]

What does this mean?

[limpid junco]

So imagine you have this $(x - n)^2 + (y + 3)^2 + (z - 6)^2 = r^2$

[glossy valve]

casework

[limpid junco]

What can you tell me how the sphere changes its place when you change n

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[torn jolt]

Its distance from the yz-plane changes

[limpid junco]

Yeah

Does that tell you what the answer is?

[torn jolt]

r = 2?

[limpid junco]

Yes

[torn jolt]

Ok, I understand the reasoning behind the answer... now, how would I be able to write it up?

I have this so far as my answer

[limpid junco]

yz plane is basically x = 0

And your sphere center is shifted from it for 2 units

[torn jolt]

Got you, thanks a lot

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[hasty igloo]

In a series a_n = (-1)^n +1/n
What would be the limit to this?
Could you have 2 for each signed value? Since 1/n converges to 0 and -1^n oscillates between +1 and -1, meaning that for each even n value, the series converges to 1 and -1 for odd n values, which is like having 2 curves or am i wrong in thinking this way? Also would epsilon have to extend to both -1 up to +1 or is it depending on the nth value?

[rapid rain]

if a sequence converges, then its limit is unique

One of the properties of convergence

so if by absurd a_n converged...

[hasty igloo]

So in this case it would have 2 unique limits, depending on n being odd or even?

[rapid rain]

do you know what "unique" means?

it means one

if l1 and l2 are both lim_{n-> infinity}(a_n), then l1 = l2

I do agree that $\lim_{n\to \infty}(a_{2n}) = 1$ and $\lim_{n\to \infty}(a_{2n+1}) = -1$

[glossy valve]

rafilou2003

[rapid rain]

but then what would $\lim_{n\to \infty}(a_{n})$ be?

[glossy valve]

rafilou2003

[hasty igloo]

Idk :c

[rapid rain]

that's the point

does it even exist?

[hasty igloo]

Thank you for breaking it down like this, i understand now. Is it like the irrational roots of polynomials?

[rapid rain]

uh no I don't see the connection

it's just that IF the limit existed, then it would either be 1 or -1

[hasty igloo]

Oh sorry, the irrational roots just dont make sense to me, so i thought it might have a connection. I can only find the real ones.

[rapid rain]

but if it were 1, then the odd terms don't approach the limit (impossible)

[hasty igloo]

Right! This clears it up. Thx a ton!

[rapid rain]

and if it were -1, vice versa with even terms

so the limit cannot exist at all

(a_n) diverges

[hasty igloo]

A great day to ya!

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[brittle nebula]

What does this mean?

Why is b not C(3,1), but d is 3?

[static bramble]

why do you think b is C(3,1)?

[brittle nebula]

Because it's not C(4,1)

7 - 3 = 4, but it's not C(4,1)

C(3,1) has 3s in it

[static bramble]

indeed it isn't C(4,1) because you have to choose two more elements

[brittle nebula]

?

[static bramble]

your subset has 5 elements, and 3 are fixed

that leaves 2 to be filled

[rapid rain]

In the notation C(n,k), n represents the list of stuff you can choose FROM

[static bramble]

if 2, 3, 5 must be included, how many more elements of S can you choose from?

[brittle nebula]

4?

[static bramble]

4

of those 4 elements, how many do you need to complete your subset?

[brittle nebula]

2?

[static bramble]

2

[brittle nebula]

2?

[rapid rain]

yes 2

[brittle nebula]

C(3,2)?

[rapid rain]

why 3?

you still have to choose 2 elements to complete your subset

how many elements can you choose from?

[brittle nebula]

4

[rapid rain]

yes

so how many choices do you have for choosing 2 elements amongst those?

[brittle nebula]

2

[rapid rain]

?

Remind me, you have to choose how many elements from a list of how many?

[brittle nebula]

7 elements

[rapid rain]

no

on this question

2 3 5 have already been chosen

[rapid rain]

answer both those in one sentence

"I still have to choose __ elements from a list of __ elements"

[brittle nebula]

2, 3

[rapid rain]

Why 3?

you still have to choose 2 elements to complete your subset
how many elements can you choose from?

(you answered this question earlier correctly)

[brittle nebula]

4

[rapid rain]

yes

[rapid rain]

I think you can attempt this again

[brittle nebula]

2, 4

[rapid rain]

correct!

So how many possibilities?

[brittle nebula]

16?

[rapid rain]

uh no

Remind me how many possibilities when having a list of n stuff, choosing k of them?

[brittle nebula]

6?

[rapid rain]

ok it's 6

C(4,2)

because you choose 2 elements from a list of 4

[brittle nebula]

ok

thx

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[gusty carbon]

the exponents are confusing me

[rain yacht]

when multiplying like terms, all you gotta do is add the exponents

[gusty carbon]

then what

it wants me to simplify

i got when i multuplied the square root of 9x^19 * y^6

[rain yacht]

ye, so you multiply the coefficients (3 in this case), multiply the variables, and then if the number is a perfect square, just square root the number.
as for the variables, I'll gibe you an example:
√(x⁵), if you expand it, it is √((x)(x)(x)(x)(x), you pair the variables, so it'll be x²√x

I'm not the best at explaining, but do you kind of get it..?

[gusty carbon]

yeah kinda

would it be 3x^9 * y^3 sqrt x?

[rain yacht]

yesss, you got it!

[gusty carbon]

yipppee

[rain yacht]

I'm learning that too, and it's confusing to me as well💀

[gusty carbon]

ok I think i just wasnt simplifying correctly or not at all

[rain yacht]

ye prob

[gusty carbon]

thank you :)

[rain yacht]

ofc!
good luck :))

[gusty carbon]

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[full forum]

@torn jolt Has your question been resolved?

[brisk obsidian]

That's basically what the instructions did.

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[grim oak]

how would u plot this on an argand diagram

[full forum]

@grim oak Has your question been resolved?

[hollow wyvern]

Hmm first of all can you simplify arg(z²)?

[grim oak]

no

no

i can

😭😭

but there's a lower boundary and stuff

and that just makes it confusing

dunno what the boundaries r

[grim oak]

Upper

idk what im saying

[hollow wyvern]

Okay, in your class, what was the convention for arguments?

[grim oak]

wdym convention

[hollow wyvern]

Like a circle keeps going round and round you need to agree in where to start and stop

[grim oak]

-pi to pi

[hollow wyvern]

Ok

So you have θ>π/4, θ<π

[grim oak]

what

[hollow wyvern]

2θ>π/2 means...?

[grim oak]

that's what I did before

I think

And apparently it's wrong

[hollow wyvern]

Oh ok

What's the right answer?

[grim oak]

my tutor went through it

but I was like asleep

and I tried to take a photo

and this is all I got but he was considering all these other scenarios

[grim oak]

idk

[hollow wyvern]

Oh ok I see

Alright so does this diagram on the bottom of the screen make sense?

[grim oak]

No

[hollow wyvern]

Arg(z²) is given in the question to be >π/2

[grim oak]

Ah

Yea

i see

[hollow wyvern]

Hence the shaded region for z²

[grim oak]

i get that now

[hollow wyvern]

Now the shaded region can be referred to with a couple different names

The obvious way to call it is π<θ<π/2

Right?

[grim oak]

yes

[hollow wyvern]

But if we only consider that we miss out on the full answer

[grim oak]

are the grater than symbols other way around

[hollow wyvern]

Because squaring does weird stuff where it causes space to double over on itself

[hollow wyvern]

Yes I'm sleepy too

[grim oak]

yep

👍

[hollow wyvern]

So upper left quadrant is π/2<θ<π

But it can also be thought of as -3π/2 < θ < -π

Think about it

[grim oak]

ok

i remember my tutor saying that

[hollow wyvern]

Now θ here is arg(z²), ie twice the angle you want

[grim oak]

yea

[hollow wyvern]

So halve these inequalities/ranges to get...

[grim oak]

..

[hollow wyvern]

π/4 < θ/2 < π/2
Or
-3π/4 < θ/2 < -π/2

Where θ/2 is arg(z)

Both of these ranges work, right?

[grim oak]

yea

that makes sense

wait

[grim oak]

can u explain this

[hollow wyvern]

Ok so you know how π and -π basically refer to the same angle

[grim oak]

yes

[hollow wyvern]

Similarly, π/2 is the same as π/2+2π or π/2-2π or really π/2 + any integer multiple of 2π

Cuz going around in a circle any number of times doesn't change anything

[grim oak]

yea

[hollow wyvern]

So π/2 is the same as -3π/2

So there are two ways to write the shaded range

[grim oak]

oh

okk

that's so quirk

[hollow wyvern]

Notice we only considered two ways of writing angles here, that's because of the z²

If you were doing a problem with z³, you'd have to consider three different ways to write the range

[grim oak]

💀

what's the third way

[hollow wyvern]

So like -π = π = 3π

And so on

[grim oak]

ok

but than for like 4pi

[hollow wyvern]

And if you had z⁴... Well you get the picture

[grim oak]

would it b the same as 2pi

[hollow wyvern]

Yes, 2π=4π

Basically, any angles that differ by an integer multiple of 2π are equivalent

[grim oak]

i see

ok

problem solved

[hollow wyvern]

This really all boils down to how taking a square root gives you two answers

And you need to consider both

Like it's the same thing but at a more general level here

[grim oak]

yep

makes sense now

but I wouldn't think this up in a test

[hollow wyvern]

Yeah you need to have seen it before

[grim oak]

yea

Only one person in the class got it

in a class of over 150

and my tutor has been teaching since like 2010

[hollow wyvern]

Lucky you have practice problems so the test doesn't give any nasty surprises

[grim oak]

he said no one ever got it first try before

[grim oak]

perfect

ok thanks

.close

[hollow wyvern]

Aight cool

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[scenic stream]

For this question I have to find the arc length s of the vector function r(t) = 7cos^3(t)i + 7cos^3sin(t)j + k from t = 0 to t = pi/2.

[scenic stream]

I have gotten to the point where I took the derivative and have sqrt((-21cos^2(t)sin(t))^2 + (21sin^2(t)cos(t))^2) but I keep getting stuck from there

[gritty rose]

Show your entire work

[scenic stream]

I erased it all I kept getting lost and the answer was wrong

[torn jolt]

how are we supposed to know what went wrong with what you did then?

Anyways, remember that the arc-length for a vector-valued function in an interval $[a,b]$ is defined as: [
L = \int_a^b \s{\p{\dv[x]t}^2 + \p{\dv[y]t}^2 + \p{\dv[z]t}^2}\dd t
]

[gritty rose]

Redo it then

[scenic stream]

Fair point I guess I don't know how to integrate the simplified magnitude. sqrt(441cos^4(t)sin^4(t) + 441sin^4(t)cos^4(t))

[scenic stream]

Redid it

[scenic stream]

Redid it

<@&286206848099549185>

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@scenic stream Has your question been resolved?

[full forum]

@scenic stream Has your question been resolved?

[minor vigil]

I got $\frac{21\pi \sqrt{2}}{16}$

[glossy valve]

Αρχιμήδης

[minor vigil]

I'll take a look at your working and find the mistake.

How did you go from line 5 to line 6?

$21\sqrt{2} \int cos^2(t) sin^2(t) dt$ = $2^{3/2} \int v^2$

[glossy valve]

Αρχιμήδης

[scenic stream]

I'm confused how to take the integral of that stuff

[minor vigil]

Have you learned integration by parts?

[scenic stream]

Yeah but it's been a long time since I learned it

[minor vigil]

Alright. Give me about 5 minutes and I'll write up a quick formula you might need.

[scenic stream]

Thank you!

[minor vigil]

Almost done

The last line in the proof is the most important. What I did with the original question was I took cos^2(x) sin^2(x) and rewrote it as:

sin^2 (x) cos^2 (x) = sin^2 (x) (1-sin^2 (x))

This is because cos^2 (x) + sin^2 (x) = 1, cos^2 (x) = 1- sin^2 (x),

Then I took
sin^2 (x)(1-sin^2 (x)) = sin^2(x) - sin^4(x),

Then I used the formula below to integrate the two. There might be a nicer way to do it, but this is the way I've always solved questions like this.

.

\begin{align*}
\int \sin^n(x),dx&=\int \sin^{n-1}(x)\sin(x),dx\\
\text{Using IBP:}\\
\begin{matrix}
u=\sin^{n-1}(x) & dv=\sin(x),dx\
du=\cos(n-1)\sin^{n-2}(x_,dx & v=-\cos(x)
\end{matrix}\\
\Rightarrow \int \sin^n(x),dx&=-\cos(x)sin^{n-1}(x)+(n-1)\int \cos^2(x)\sin^{n-2}(x),dx\\
&=-\cos(x)\sin^{n-1}(x)+(n-1)\int \sin^{n-2}(x)(1-\sin^2(x)),dx\\
&=-\cos(x)sin^{n-1}(x)+(n-1)\int \sin^{n-2}(x),dx - (n-1)\int \sin^n(x),dx\\
n \int \sin^n(x),dx&=-\cos(x)sin^{n-1}(x)+(n-1)\int \sin^{n-2}x,dx\\\
\int \sin^n,dx &= \frac{n-1}{n}\int \sin^{n-2}(x),dx-\frac{cos(x)sin^{x-1}(x)}{n}
\end{align*}

[glossy valve]

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

[minor vigil]

The formula for cos is the same, but you add the fraction on the right instead of minusing it.

I found the mistake in my proof. In the IBP substitution, it should be:

du = (n-1) [sin(x)]^(n-2) cos(x) dx

[scenic stream]

Thank you so much!

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[eager holly]

can someone explain the highlighted part?

[eager holly]

Why does that set the index k to 0

and how did they determine C to be 1

[next sail]

well when k is 0, it's q^0 - 1 which is just 0

so it's like, sum from 1 to infinity is equal to sum from 0 to infinity

because you're just adding 0

[viral jasper]

$\sum_{k=1}^{\infty} (c\cdot q^k)=c\sum_{k+1}^{\infty} q^k$

[glossy valve]

SWR

[eager holly]

oh god thats still confusing but i appreciate the help

i know constants are allowed to be put in front of the sum

but that infinity rule is a bit too much

so its basically just saying it doesnt matter if k starts from 1 to infinity or k+1 to infinity? because its both going infinitely long?

[next sail]

no

SWR made a typo

$\sum_{k=1}^{\infty} (c\cdot q^k)=c\sum_{k=1}^{\infty} q^k$

[glossy valve]

Kaisheng21

[next sail]

it should be this

@eager holly

[eager holly]

oh that rule i know

thats pretty standard i guess

[next sail]

yeah

[eager holly]

was just wondering how they got k to be 0 by subtracting c

they do it quite often in this answer sheet if thats something i can just do i guess ill learn it? but it doesnt make sense to me

[next sail]

$\sum_{k=0}^{\infty} (q^k - 1) = (q^0 - 1) + \sum_{k=1}^{\infty} (q^k - 1) = 0 + \sum_{k=1}^{\infty} (q^k - 1) = \sum_{k=1}^{\infty} (q^k - 1)$

[glossy valve]

Kaisheng21

[next sail]

so it's just because the k = 0 term is 0

[eager holly]

oohh

so when k = 0 its 0 anyway so it doesnt affect the equation

?

[next sail]

yeah

[eager holly]

oh that makes more sense, ill keep that in mind thanks ^^

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[vague panther]

Please help : For how many pairs of positive integers (a, b) do we have minimum comun multiple (a, b) = 2024

[limpid junco]

!status

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[sly solstice]

how do u derive sin2x/cos2x

[sly solstice]

the 'worked' answer is very useful

[fierce chasm]

blud nawg

quotient rule

also u should memorise that d/dx tan(x) is sec^2(x)

[hot herald]

solution has a typo

[sly solstice]

where at

[hot herald]

the dy/dx

[sly solstice]

is it just 2/cos(x) not squared

[hot herald]

no

should be 2/cos^2(2x)

they forgot about the 2 in the argument

[sly solstice]

i see

but still how cuz i did quotent

\then

[hot herald]

show what you did

[sly solstice]

$(2(cos^2(2x) - sin^2(2x)))/cos^2(2x)$

[glossy valve]

™Vlad The Lad

[fierce chasm]

bomboclat

[sly solstice]

lol

i still dk how to use the thingy

shell be right

[hot herald]

you messed up some signs

[sly solstice]

cuz before innit
(2cos(2x)cos(2x) - 2sin(2x)sin(2x))/cos^2(2x) (straight form quotent)

[hot herald]

is incorrect

what's the derivative of cos(2x)

[sly solstice]

-2sin(2x)?

[hot herald]

yes

use that

[sly solstice]

ah f

i forgot to - * -

but ye then its just 2 on numerator

cool

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[dark shore]

$\lim_{{x \to 3}} \left( \left| x - 4 \right| \right)^{\frac{1}{\left| x - 3 \right|}}$

[glossy valve]

Alex

[dark shore]

Find the left and right limits.

im not sure how to handle this

does the limit changes depending the side? I assumed kt wiuld be the same because of absolute value

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[torn jolt]

for this since you know they intersect, just find normal and solve system right

i did it but my answer looks kinda wrong.

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[vague panther]

How do I find all pairs of numbers with minimum comun divisor=2024

[eager obsidian]

are you sure you meant to write "minimum common divisor" there

[vague panther]

Yes