help-33

[elfin berry]

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[stray hound]

the reason is bc between 1 and 4, the graph is below the x axis

and since you want to find area, you need to make sure all of the area is positive

so they split it between 0 and 1, and then between 1 and 4

all they did in the example is flipped the sign of the integral from 1 to 4 to make it positive instead of negative

[sonic pivot]

Why 1? Why not 0 and 2, then 2 and 4?

Is it because 1 is when it goes negative?

[stray hound]

because the graph interests the x axis at 1

yes

[sonic pivot]

Ohh ok I get it now

Thank you

[stray hound]

🫡

[sonic pivot]

🗿

.close

[marsh citrus]

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[marsh citrus]

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[jolly sigil]

I know the velocity would just be the first derivative if I have a function right. But here I'm just given coordinates. Am I supposed to construct a function for t?

[marsh citrus]

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[marsh citrus]

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[rustic hamlet]

Hey so I’m trying to do this trig sub but I’m getting stuck

[rustic hamlet]

Not sure what to do

How am I supposed to integrate the thing at the bottom also am i doing something wrong?

[outer palm]

what the question

[rustic hamlet]

^

[hardy slate]

should be 5^4 sec^4 (t), not 5sec^4 (t)

i think the main thing you're missing is that 1/sec^4 (t) = cos^4 (t)

[rustic hamlet]

Well how does that help the situation

[hardy slate]

??

ok so

whenever you have just tan and sec and csc and whatever

i like to just turn it all into sin and cos

it often helps

try it now

[rustic hamlet]

So true…

Lemme try that

[hardy slate]

also remember to make it 5^4

while you're at it

[novel juniper]

If you're just multiplying and dividing trig functions, it almost always helps

[rustic hamlet]

Is this correct?

[hardy slate]

nope

[rustic hamlet]

Hm

[hardy slate]

why are you multiplying by 1/cos^4 (t)

you were dividing by sec^4 (t)

so you should be dividing by 1/cos^4 (t)

so you should be multiplying by cos^4 (t)

[rustic hamlet]

Ohhhhhhhhh

Ok

Is that right cuz I’m pretty sure I can do that integral with u sub

[hardy slate]

almost right

it should be 5/5^4 i think

but yeah

now you can just do it with another u-sub

wait no

wait yeah

5/5^4

[rustic hamlet]

When I changed to multiplying by cos^4 I took the constant with me so it’s in the numerator

[hardy slate]

no

ok let's be clear on this

$\int \frac{\sqrt{x^2 - 5^2}}{x^4} dx$ \

$= \int \frac{\sqrt{5^2 \sec^2(\theta) - 5^2}}{5^4 \sec^4(\theta)} 5\sec(\theta)\tan(\theta) d\theta$ \

$= \int \frac{\sqrt{5^2 \tan^2 (\theta)}}{5^4 \sec^4(\theta)} 5\sec(\theta)\tan(\theta) d\theta$ \

$= \int \frac{5 \tan(\theta)}{5^4 \sec^4(\theta)} 5\sec(\theta)\tan(\theta) d\theta$ \

$= \frac{5^2}{5^4} \int \frac{\sin(\theta)}{\cos(\theta)} \cdot \cos^4(\theta) \cdot \frac{1}{\cos(\theta)} \cdot \frac{\sin(\theta)}{\cos(\theta)} d \theta$ \

$= \frac{1}{5^2} \int \sin^2(\theta)\cos(\theta) d \theta$ \

[rustic hamlet]

Should be 5^2

[hardy slate]

?

you see how the 5^4 stayed on the bottom right

[rustic hamlet]

The problem is sqrt x^2 - 25

[hardy slate]

oh sure

[elfin berry]

Kaisheng21

[hardy slate]

just an error in the first line, the rest was fine

but you see how the 5^4 stayed on the bottom right

[rustic hamlet]

Ohh I swear

See*

Yea makes sense

[hardy slate]

ok ok

it helps to just write it out all at once like this

instead of sorta... breaking it down and then maybe not putting it back together properly

[rustic hamlet]

Then now that I have the answer in terms of theta I use a triangle to get it in terms of x?

[hardy slate]

uhh

i mean i'd do another u-sub personally and then go all the way back

[rustic hamlet]

Yea after that

I have 1/3*5^2 sin^3theta

[hardy slate]

so we have x = 5sec(t) so we can say like t = arcsec(x/5) or whatever

and then simplify a whole bunch probably with trig identities

[rustic hamlet]

Ok I might have something

But I might be missing one thing @hardy slate

Sorry for bad handwriting

The answer says the numerator should be cubed but where does the cube come from

[hardy slate]

?

what's your final answer

[rustic hamlet]

What I got is what I took a pic of

The answer says it’s this

It’s (3/2)

The exponent

[marsh citrus]

@rustic hamlet Has your question been resolved?

[marsh citrus]

@rustic hamlet Has your question been resolved?

[marsh citrus]

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[marsh citrus]

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[quiet condor]

https://cdn.discordapp.com/attachments/363224154469826562/1209725248153587772/image.png?ex=65e7f76e&is=65d5826e&hm=375fd1d67b8b3b8036a377e2788e89bb23b13f762b4c25f4c89e06f98eb0aee5&

[quiet condor]

is blue the F^-1(x) of the red drawing?

excuse the bad drawing

[main idol]

,w plot arcsin(x)

[elfin berry]

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[quiet condor]

so the bottom one the f^1(x)

[main idol]

they're both the same graph

read the axes

[quiet condor]

so the F^-1(x) is the same with a slight difference im confused

?

[main idol]

,w plot y=sin(x), y=arcsin(x)

[elfin berry]

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Requested by rie.mann

[main idol]

look at the -pi/2 < x < pi/2 in the first plot and ignore the second plot

[quiet condor]

ah i see

the pi was throwing me off

.close

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[marsh citrus]

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[unborn prairie]

plz help

[unborn prairie]

find the lim of this

i get that sin(x-a)/x-a is 1

[fathom ridge]

so just sub x = a lol

[unborn prairie]

i dont get it

[unborn prairie]

what ya mean

<@&286206848099549185>

[fathom ridge]

you deleted your message

$\lim_{x \to \alpha} \frac{\sin (x + \alpha)}{x + \alpha}$

[elfin berry]

NEON

[fathom ridge]

This is where you got to right?

As x approaches alpha, do you see it to be any indeterminate form?

[unborn prairie]

i mean can we really split it?

[fathom ridge]

sure

[unborn prairie]

ok

sin 2a/2a

[fathom ridge]

yes

[unborn prairie]

thats the lim?

[fathom ridge]

yes

[marsh citrus]

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[cosmic raptor]

Can anyone help tell me why it went to sin(4k)?

[cunning basin]

$\sin{x}=\frac{e^{ix}-e^{-ix}}{2i}$

[elfin berry]

WhereWolf(ping if needed)

[cosmic raptor]

I haven’t seen this one before, what’s it called?

[cunning basin]

idk what it calls

but it's easy to prove using $e^{ix}=\cos{x}+i\sin{x}$

[elfin berry]

WhereWolf(ping if needed)

[cosmic raptor]

How would it go from this to that?

[cunning basin]

oh you need the insight that $e^{-ix}=\cos{(-x)}+i\sin{(-x)}=\cos{x}-i\sin{x}$

[elfin berry]

WhereWolf(ping if needed)

[cosmic raptor]

cool, got it now finally lol

thanks!

[marsh citrus]

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[prisma notch]

i need help to find the equation for this question in factored form.
(0, 1/3) is the hole, and there is no y-intercept

[cunning basin]

!status

[marsh citrus]

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 was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[prisma notch]

2 and 3

[cunning basin]

ok

can you show your work

[prisma notch]

i don’t really have anywork since i did on online

[cunning basin]

what's the answer you got

[cunning basin]

it should be x-3 instead of x-4

[prisma notch]

oh

damn

i can’t believe i spent an hour on this

over a simple mistake

😭

thanks dude for pointing it out

[cunning basin]

no problem

[prisma notch]

.close

[marsh citrus]

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[calm harbor]

How did you actually find that function 💀

[prisma notch]

what

[calm harbor]

Nvm

All good

[marsh citrus]

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[vocal cloud]

is this reasoning right to show its convergant?

<1/(3n^4+1)<1/3n^4 and 1/3n^4 is convergant

[sand fable]

no

can you guess why?

[strong sphinx]

it's infinite

is it not?

[sand fable]

are you saying the sum diverges or that there are infinitely many terms

[strong sphinx]

infinitely many terms

[sand fable]

that is true

[vocal cloud]

is it because my first equality is wrong

[sand fable]

this is true

[vocal cloud]

should be >1/(3n^4+1) right? and since its greater we dont learn anything

[sand fable]

yes

[vocal cloud]

got it

[sand fable]

so try to see what you can do with the cube root there (if it helps, you can think that a cube root is an exponent of 1/3)

[vocal cloud]

i think cube root of 3n^4 works

[gleaming pecan]

your series converges and you need to evaluate to the right, find a larger one that is definitely convergent, so reduce the denominator by 1 to keep convergent using the comparison test

[vocal cloud]

we can compare it with 1/n^4/3 right?

[gleaming pecan]

yes but i wud make it tighter, i wude write :

this temr is larger than yours

and is defintley covegeent using p = test

4/3 > 1 hence convegeent

and then yours as well

all )

[vocal cloud]

@gleaming pecan thankyou so much

got it

.close

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[grand horizon]

hey uh so

[grand horizon]

cant solve this question

[still temple]

if $\ds x^3+\41{x^3}= 8^3$ find $\ds x^3 -\41{x^3}$

[still temple]

did i note this down correctly?

[grand horizon]

no

its 83

lol

[still temple]

why not just directly solve for x and plug it in the right equation

i dont think its going to be rational anyways

[grand horizon]

the ans

is 756

[crystal wraith]

Isn't it x^2+1/x^2 = 83?

Or...

[grand horizon]

yess

no

no

its in the cube

oh wait

yes

sorry

square

shit

[crystal wraith]

👀

One last time.

Which is it?

[grand horizon]

i wrote 2

tho

he wrote 3

😭

[grand horizon]

this one

[crystal wraith]

He asked if it's correct.

[grand horizon]

thiss

[grand horizon]

oh

😭

[grand horizon]

its the square

um

[still temple]

you are going to get 4 solutions to that equation

so uh

[grand horizon]

one more quick question

can x be in the root form?

[crystal wraith]

Is that necessary though?

[still temple]

wdym

[crystal wraith]

Solving for x

Like you said

[still temple]

im just saying its a bit weird that there is a definitive, singular answer when x can be onr of four things

[crystal wraith]

Well yes.

There's two answers for x^3 - 1/x^3, hmm

[grand horizon]

👀

[still temple]

damn isnt it straight forward ?

[crystal wraith]

(x-1/x)^2 = x^2 + 1/x^2 - 2
And then difference of cubes?

[warm wyvern]

.reopen

[crystal wraith]

Yes it is.

But then x-1/x can really be one of two things.

But op just told there's exactly one answer.

[crystal wraith]

Too late mate.

[still temple]

oof wait i will send up somth

smth

do similarly for the second bracket

[crystal wraith]

and x-1/x doesn't have to be 9

[still temple]

complete the square

[still temple]

what

[crystal wraith]

What's (-9)^2

[still temple]

yeah

consider -9 :/

bruh ok

💀 you got me xd

[warm wyvern]

dam

[grand horizon]

so?

no sol for the questiom?

:(((

[crystal wraith]

Quite the opposite actually

[still temple]

bruh

[crystal wraith]

More than one.

[still temple]

isnt it

756 ?

[crystal wraith]

Yes?

[grand horizon]

but how??

[still temple]

For the second bracket

[grand horizon]

😭

[still temple]

Hope you understand
My bad handwriting

[crystal wraith]

I thought we both knew there's also -756?

[still temple]

Want it clean on a new page XD?

[grand horizon]

negative?

[grand horizon]

yes

please!

[still temple]

Dude I hate you

I don't even think there are two answers lol

[grand horizon]

its a grade 9th problem

lol

[crystal wraith]

You're not helping the op. Just handing out answers and not helping learn. There's clearly two answers for the problem the way it's stated. If you fail to acknowledge that then you are incorrect and aren't helping anyone.

[still temple]

Yeah I am sorry

[grand horizon]

😭

[still temple]

🙏

Sorry mate

[grand horizon]

so no ans?

[crystal wraith]

Two answers.

[still temple]

Yes ans

Yeah two

Agreed

[grand horizon]

if i write that in my exam

they gon kick me out

[still temple]

That full stop tho

[grand horizon]

👍

[still temple]

Exactly my point

[grand horizon]

😭

plus my teachers are hell

[still temple]

It's upto you

its correct obv as @crystal wraith stated

[warm wyvern]

\

[still temple]

@grand horizon if you wanna know how i thought of it and how i did it ask me

i just give away answers like the guy said

[grand horizon]

yes

i need help

i didnt get

no shit

[still temple]

lol okay

so

you know how to complete squares ?

[grand horizon]

uhh

no

[still temple]

like if i give you two numbers can you write them as a whole squared

[grand horizon]

yeah

yeaah

[still temple]

yes yes

so

x^2 +1/x^2

wait i will write it with my good handwriting and send

[grand horizon]

lol kay

@rabnoor aake apna doubt bhi solve krle bc

oh wait

[still temple]

indians everywhere huh

[grand horizon]

what

oh

WHAT

u indian?

[still temple]

ofc :/

even my name

[grand horizon]

😭

oh

hai naa

i didnt notice that before

welp

[still temple]

lol

[grand horizon]

soo?

[still temple]

writing oof

wait a min

[grand horizon]

okk

uhh so u joined dc today?

dam

[still temple]

nah

[grand horizon]

got hacked?

logged out?

[still temple]

add me will talk there

sending sol

[grand horizon]

👍

[grand horizon]

why is it -9?

[still temple]

nvm that

[grand horizon]

lol

k

[still temple]

+9 only

[grand horizon]

waait

where did

(x-1/x)^2 pop up from

wasnt it

positive?

😭

[still temple]

kaha pe ?

2nd page ya 1st ?

[still temple]

OH

I DIDNT THINK OF

THAT EQUATION

bro i struggled with that prblem forever

lol

[marsh citrus]

@grand horizon Has your question been resolved?

[marsh citrus]

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[tender oyster]

Let P be a convex polygon in the plane, and let P’ be an enlarged version of P, dilated by a scale factor of 2. Show that seven copies of P can completely cover P’.

[tender oyster]

don’t remember the source found this online

don’t know how to start rlly

[marsh citrus]

@tender oyster Has your question been resolved?

[marsh citrus]

@tender oyster Has your question been resolved?

[marsh citrus]

@tender oyster Has your question been resolved?

[pliant siren]

can always cover it?

how many edges does the polygon have?

[tender oyster]

I believe so

[tender oyster]

it’s an arbitrary convex polygon

[pliant siren]

so it can have 1000 edges?

oh and is the larger polygon warped?

[tender oyster]

wdym warped

[tender oyster]

It could

[pliant siren]

stretched more in one direction

[tender oyster]

no

[pliant siren]

hm ok

but if you enlarge a shape by a factor of 2

the area increases by a factor of 4, doesn’t it

so how are seven copies supposed to cover it

or can they overflow

and intersect

[marsh citrus]

@tender oyster Has your question been resolved?

[marsh citrus]

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[vague stone]

A given sinusoidal function has an amplitude of 8, an axis at 𝑦 = 12, a period of 9𝜋, and a maximum at 𝑥 = 3𝜋. Determine an equation for the function, and find all intersections of the function and 𝑦 = 16 between 𝑥 = −5𝜋 and 𝑥 = 12𝜋.

so far i have f(x)=8cos(2/9(x-3pi))+12
supposedly from the question, graph should have an y=12, im stuck there

[static quarry]

that looks ok, what is your concern?

[vague stone]

its correct?

[static quarry]

yea, the +12 gives you an axis at y=12

[vague stone]

i assume its wrong since it show y=8 on desmos

[static quarry]

can you clarify, what shows y=8 on desmos?

[vague stone]

when x=0

graph is at axis (0,8) with the equation i have

[static quarry]

,w plot 8cos(2/9(x-3pi))+12

[elfin berry]

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[static quarry]

is that the graph you expect to see?

[vague stone]

thats what i have rn

[static quarry]

it satisfies all the requirements, doesn't it?

which one are you concerned about?

[vague stone]

"axis at 𝑦 = 12"

[static quarry]

yea, notice that the max is 20 and the min is 4

so the amplitude is (20-4)/2 = 8 as desired

and the axis is at the midpoint: (20+4)/2 = 12

[vague stone]

ur correct

if want to find specific intersections

do i simply equal those values to the equation?

[static quarry]

yea, if the specified numbers are y values

[vague stone]

yes there are y values

[static quarry]

yea, it's asking where does y=16

so just set your formula equal to 16

[empty badge]

How can I solve these equation: 3x + 1y = 155 and 2x + 3y = 220

[static quarry]

!occupied

[marsh citrus]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[empty badge]

My bad

[vague stone]

am i allowed to give another diff question

mb i just reread the rules

anyways thank you Bungo

real g

[tender oyster]

a common misconception

consider a regular 1000-gon (so basically a circle)

It’s easy to see that four copies won’t fit the bill

just as one counter example

[marsh citrus]

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[flint pebble]

I know what this does, it flips the matrix, but i'm not sure how to describe it geometrically

like it mirrors about y=x and and y=-x

but how should i describe that generally

[marsh citrus]

@flint pebble Has your question been resolved?

[flint pebble]

<@&286206848099549185> any input?

[still temple]

u reflected it about the y axis?

is how i'd put it

ig

[flint pebble]

well its not about the y axis. i think those two photos might be confusing. each one is a seperate graph with the orignal and the transformed vector

[still temple]

oh

u flipped it across y=x?

reflected it about y=x

idk, i'd use reflected about ___

[flint pebble]

yea but one is about y=-x and the other is about y=x

and i need to explain that in general terms for both vectors lol

[still temple]

ohh

yh im not sure how you'd phrase that tbh

[flint pebble]

yea lol

[flint pebble]

<@&286206848099549185> any ideas how i would describe it?

[vivid sierra]

Symmetry across y-axis, and 90° rotation clockwise ?

[limber gale]

Symmetry across y-axis, and 90° rotation clockwise ?

[marsh citrus]

@flint pebble Has your question been resolved?

[marsh citrus]

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[sleek crater]

d/dx of ln(x)/ln(a)

[sleek crater]

this is solved using quotient rule right

[cunning basin]

Ln a is just constant

[sleek crater]

but isn't ln still a function

we're dividing functions

so quotient rule?

[cunning basin]

a is a constant

So ln (a) is also a constant

being in a function doesn't mean it's a variable

[sleek crater]

i see

[iron meadow]

ln(a) doesn't vary with x

as x changes ln(a) doesn't change

[marsh citrus]

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[marsh citrus]

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[topaz ore]

I have simplified this to (I/I zero )+1 = exp(vp/kt)

but i dont know what to do next

[main idol]

$\ln(exp(a)) = a$

[elfin berry]

riemann

[topaz ore]

i see

thanks

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[thin osprey]

Is this a problem where an integral is necessary?

[thin osprey]

I know W = F required x distance traveled

So I have currently 1000 kg x 40 m

Am I missing any other necessary changes?

[cobalt sentinel]

imma say this and dip: use kinematics to find the initial velocity right before it starts to slow down. from there, u can use work energy thm

[cunning basin]

W = FD = maD

[wise jackal]

I believe u can avoid that

[wise jackal]

and just use this

[thin osprey]

40,000 is incorrect, it says

[cobalt sentinel]

oh yeah, oops

[wise jackal]

should be negative, as the work is acting opposite to the motion

[thin osprey]

Says it's wrong, again

[wise jackal]

wait

[wise jackal]

you forgot to factor in the acceleration

[thin osprey]

Would I then divide by the acceleration? Sorry, my first time encountering a problem like this.

[wise jackal]

no, the force is mass * acceleration right

[thin osprey]

Correct!

[wise jackal]

yeah so work = force * displacement

therefore work = mass * acceleration * displacement

[thin osprey]

so, 1000, times -6, times 40

?

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[wise jackal]

[marsh citrus]

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[dense igloo]

or does the fact that its a sub n matter

nth term stuff

like the a_n is just a variable right 😭

or function name

i mean

[nocturne glen]

since this is an explicit formula for the sequence yeah u would just do it

[dense igloo]

so i just ignore that its in that notation

[nocturne glen]

yeah ig

[dense igloo]

so the answer is just the lim as n approaches inf for sqrt(n^2+n)-n

[nocturne glen]

yes

[dense igloo]

okay

thank s

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[still temple]

so for every integer there is a rational no and for every rational no there is an integer but..for 0.000000001, there is no integer

[sand fable]

sounds more or less correct

do you have a question about that?

[still temple]

um so

[stark apex]

are you talking about the infinite hotel problem

[sand fable]

??? no

he's talking about the ideas often associated with the "infinite hotel problem"

[stark apex]

Ok

[still temple]

I read on google somewhere that rational numbers and integers have the same size of infinity and when i asked something related to this on discord, they told me about bijection so i related bijection to this and anyway so acc to this bijection, there should be an integer for that rational number but there is not. @sand fable

[sand fable]

what do you mean "there is not" ?

[still temple]

0.0000001

[sand fable]

right

[still temple]

oh

[sand fable]

so what this means is that there exists a "bijection" between the rational numbers and the integers

[amber birch]

1/10^n for any integer n will appear somewhere on this line

[still temple]

truee

[sand fable]

basically, every rational number can reasonably be "mapped" to an integer after applying some rule or transformation

there are numerous ways to prove this, one of which is posted above ^

[still temple]

ye

thank youuu

guys

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[leaden monolith]

@still temple sorry I was cooking dinner

[still temple]

nah nah its okay, ty, u helped a lot. My problem is solved : D

[leaden monolith]

When I said something like that

[still temple]

yea

[still temple]

continuee

[leaden monolith]

I meant that bijections are specific

A bijection is both surjective and injective

And both of those have very particular definitions

So I meant “something like that” as like a “that’s the general idea of surjectivity and injectivity at the same time”

[still temple]

yea i just googled itt

thank you for letting me know

[still temple]

ye

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[upper blade]

can someone help me understand a statement from a book

[upper blade]

the textbook says

If a and b do not coincide, then they are opposite sides of a parallelogram, and lie in the same plane

however i think this is inaccurate since they can be like this ( imma send it )

They do not coincide and they dont lie in the same planr

[marsh citrus]

@upper blade Has your question been resolved?

[marsh citrus]

@upper blade Has your question been resolved?

[marsh citrus]

@upper blade Has your question been resolved?

[calm harbor]

@upper blade has your question been resolved?

[calm harbor]

May I check out your textbook of the part describing this?

Ping helpers next time when nobody solves your question

[upper blade]

Nope

[upper blade]

I texted the paragraph that stated it but just wait a second imma send you the full part

[calm harbor]

Sure

[upper blade]

i dont know how to do that

[calm harbor]

@

[upper blade]

Ohh

[upper blade]

oh okay thank you. Didnt know i could tag ppl to help me

[calm harbor]

Your vector do coincide

You know you can move vector, right?

[upper blade]

yea but can we move it randomly?

[calm harbor]

yep

[upper blade]

ohh

okay so if it doesnt coincide in any way they’re parallel right

[calm harbor]

That’s why it says " be parallel"

Yes

[upper blade]

ohh okay

thanks

what about z

i just got to the section where i am seeing z btw

[calm harbor]

Nvm, I’m an idiot

[upper blade]

thats why im asking in case i see it in the future

[calm harbor]

They are applied in xyz space as well

[upper blade]

ohh okay lol its alr

[calm harbor]

Gl

[upper blade]

thank you

[calm harbor]

Type
.close
To close the channel

Cya

[upper blade]

alright

Cya

.close

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[lean imp]

does this look okay?

i figured that a intersect with the compliment of c and c intersected with the compliment of a means that b was included, so i added it in like that and it should be the same thing right?

because i figured that the m (a cap b cap c^c) is a subset of m(a cap b) right???

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[wet crystal]

any suggestion on solving this integral

[main idol]

did you try u sub

[wet crystal]

yes thats what I am trying rn

i wanna substitue e ^-x with u

[hardy slate]

what did you get

[marsh citrus]

@wet crystal Has your question been resolved?

[wet crystal]

stil doing

i am rn on integral sqrt(1+u^2) with bounds e^-1 and 1

[hardy slate]

ok cool

that looks right

[wet crystal]

thanks

[marsh citrus]

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[devout quartz]

Finding distance between point (10,100,1000) and plane [x,y,z]=[1,1,1]+s[5,-5,3]+t[-1,1,3].
How I am solving this problem:

• taking cross product of the vectors in plane equation,
• creating a line using the cross vector and the given point,
• finding intersection of the plane and the newly formed line
• finding distance between the point of intersection and the given point

I am stuck on the 3rd step, I am struggling to figure out how to find the intersection, any advice?

[devout quartz]

so far I have the line r=(10,100,1000)+t[-18,-18,0].

[marsh citrus]

@devout quartz Has your question been resolved?

[marsh citrus]

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[lavish owl]

second one is translated but it got wrong values

can someone help me?

for task a

I said that the model B(x) tells us that Roar's vehicle has a 60 liter tank of gas. The model also shows that for every mile driven, the vehicle uses 0.5 liters of petrol.

[twilit arrow]

yes

[lavish owl]

so that one is right?

[twilit arrow]

looks good to me

[lavish owl]

for task b i wrote

We know that the validity range must at least be within 60 to 0 liters. Roar's vehicle could have had room for more, but we don't know that with the given information from the task. X will vary according to how many liters are in Roar's vehicle. But basically, x's range of validity is from 0-120 miles if the vehicle has between 0-60 liters of petrol.

the range of validity for distance: 0-120 miles.
The range of validity for liters of petrol: 0-60 litres.

all that

@twilit arrow

[marsh citrus]

@lavish owl Has your question been resolved?

[lavish owl]

<@&286206848099549185>

[lavish owl]

.close

[marsh citrus]

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[south goblet]

how do I know this

[south goblet]

when this is the question

I know the formula for the differential equation of the second order linear equations

but how did he decided on the Ce^x

[twilit arrow]

from theorem 9.4.2

[south goblet]

what does that one say

[twilit arrow]

i have no idea, but they are saying that's where that came from

you should have access to the theorem

[south goblet]

is this the theorem?

[south goblet]

also how did he come up with f(0)^2 = 100

nvm I see it

[south goblet]

but I still don't understand why he chose f(x) = Ce^x

[twilit arrow]

f(0) = C

and you ahve the diffeq f'(x) - f(x) = 0

so the solution by the theorem is f(x) = Ce^(1x), since k = 1

[south goblet]

oohh

[south goblet]

so when f'(x) = k f(x) and f(0) = C u can use this theorem right?

[twilit arrow]

yes

[south goblet]

are there more theorems like this one

I feel like these ones are more of a remember thing

[twilit arrow]

they should also be intuitive

what function is its own derivative

i.e., satisfies f'(x) = f(x)

[south goblet]

oh

e^x

damn

[twilit arrow]

Ce^x

yes

[twilit arrow]

and yes

differential equations are a big part of mathematics

[south goblet]

that's if there is a constant behind e^Cx right?

[twilit arrow]

a lot of them dont have nice solutions but a lot of the common ones do

[twilit arrow]

can you rephrase this question

[south goblet]

I do know the 3 ones for second order lienar equations

[south goblet]

well f'(x) = Ce^x only if f(x) = e^Cx right?

[twilit arrow]

no

there's two things going on here

if f(x) = Ce^x, then f'(x) = Ce^x

[south goblet]

oh

[twilit arrow]

remember that constants don't really affect the derivation

[south goblet]

right

ye

[twilit arrow]

the other part is that

if (f(x) = Ce^{kx},) then
[f'(x) = Cke^{kx} \implies f'(x) - kf(x) = Cke^{kx} - k(Ce^x) = 0]

[elfin berry]

maximo

[twilit arrow]

so f(x) = Ce^{kx} is a solution to
f'(x) = kf(x)

since taking the derivative brings down that k

[south goblet]

ye

[south goblet]

so I know the second order differentaial equaqtions

3 of them

but the one we discussed is the first order differential equation?

[twilit arrow]

yes

first order linear ODE

[south goblet]

ODE?

[twilit arrow]

ordinary differential equation

[south goblet]

so there are more?

[twilit arrow]

many many more

[south goblet]

can u name only the important ones

[twilit arrow]

but you don't need to learn everything about them

[south goblet]

that IG otta remember

oh

[twilit arrow]

not really since i dont know what you need to know

just focus on what you're being taught

and if you want to learn more then look around and find some other interesting equations

but don't try to learn everything, no one can

just focus on what you're taught for now

[south goblet]

I was also being taught the seperable case

but I didn't find that intuively

[south goblet]

can u also explain about that one

[twilit arrow]

well the explanation usually goes like

consider the differential equation
[\dv{y}{x} = \frac{f(x)}{g(y)}]
then
[g(y)\dd y = f(x)\dd x \implies \int g(y),\dd y = \int f(x),\dd x + C]

by "multiplying" by dx

[elfin berry]

maximo

[twilit arrow]

this is the intuition you should have imo, no reason to understand the mechanics since this is a fine explanation at the level you're working on

[south goblet]

ye I solved a couple question of these

[south goblet]

so basically the reason why u seperate them is just to integrate them to get rid of the dy/dx right?

[twilit arrow]

yes, so as to have an equation with no differentials

just "regular" functions of x and y

[south goblet]

I see

[twilit arrow]

and in the best case scenario you get y = f(x)

though sometimes the solutions cant be a single function

[south goblet]

I don't think I have gotten it a single time

and there is usually x involved

[twilit arrow]

well the simplest subsets of separable equations you see in calc 1/2/3

[\dv{y}{x} = f'(x)]

[elfin berry]

maximo

[twilit arrow]

this is just an integration problem

and you get y = f(x) + c

[twilit arrow]

so you've seen them, just in disguise

[south goblet]

I always saw this as a definition

i'ts not?

[twilit arrow]

well y and f are different things

the definition is

or

"definition"

is

[\dv{f}{x}\ (x) = f'(x)]

[south goblet]

ohh

[elfin berry]

maximo

[south goblet]

when I did the questions they just called y instead of f(x)

I was doing excercise 5

for example

did I got the best case scenario when this was the answer?

because I just call the right side f(x) and that's fine I think

[south goblet]

how can the solution not be a single function actually I am bit confused about that

[twilit arrow]

consider [\dv{y}{x} = -\frac{x}{y}]
this is a separable ODE with solution
[x^2 + y^2 = r^2]

[elfin berry]

maximo

[twilit arrow]

for an arbitrary real value r

note that this is a circle of radius r, which cannot be written as y=f(x)

[south goblet]

oh

[twilit arrow]

if you solve for y, you’ll end up with an expression that is either not a function, or not the whole solution

[south goblet]

ye u can split the circle in up part and down part right?

[twilit arrow]

yes

[south goblet]

how do u get an expression of the circle without it being a function?

[twilit arrow]

well (x^2 + y^2 = r^2) is already not a function

[elfin berry]

maximo

[twilit arrow]

but you could also have

[south goblet]

but solved for y

[twilit arrow]

(y = \pm\sqrt{r^2-x^2})

[elfin berry]

maximo

[twilit arrow]

which is also not a function

[south goblet]

oh

like that

btw

this one

[twilit arrow]

right, so that solution at the end

[south goblet]

is this not a function either then?

[twilit arrow]

is not a function

[south goblet]

ahh

[twilit arrow]

the solution is an implicit equation/implicit curve

[south goblet]

but how does that work will the solution of y be at the positive or negative part

[twilit arrow]

both

the problem is that if you wrote it as

(y = \sqrt{r^2 - x^2})

[elfin berry]

maximo

[twilit arrow]

you only get the positive part

and if you put a -

you only get the negative

[south goblet]

but I meant

say u got x = 5

how do u know

if that fits in the positive

or negative part

[twilit arrow]

they both work

that's why we use +-

[south goblet]

oh so u get 2 solutions?

[twilit arrow]

the solution

curve

has 2 y values for some x values

and so is not a function

[south goblet]

ahh

[twilit arrow]

if you recall the vertical line test

[south goblet]

ye

[south goblet]

oaky I think I got it pretty good in my head now thanks

[twilit arrow]

[marsh citrus]

@south goblet Has your question been resolved?

[marsh citrus]

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[foggy whale]

If a question says to find all values of n for:
||x|-n|-n=0 ,
would the answer be n=+-1/2x , or would it be n=all real numbers? (or are both fine?)

[marsh citrus]

@foggy whale Has your question been resolved?

[slow holly]

This can be rewritten as ||x|-n| = n, so note that n can only be positive or 0

[marsh citrus]

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[obsidian bobcat]

What is going on in the line labelled "divide by highest denominator power"? I understand that we must divide by the biggest term in the numerator and the denominator, I think. The numerator's biggest term is x and the denominator is x^2.
Only part I'm iffy about is why the x outside the square root in the denominator became 1. Presumably that would only happen if we did x/x, rather than x/x^2. Is this what's happening?: since there is a square root, the terms inside the square root are divided by the full x^2, but the x outside the square root is actually being divided by the square root of x^2, which is x, thus resulting in x/x which equals 1?

[sand fable]

they divided the numerator and denominator both by x

[obsidian bobcat]

but then why do we have 4/x^2?

[sand fable]

1/x becomes 1/x^2 inside a square root

[obsidian bobcat]

im confused haha. I see that the term 4 became 4/x^2. So we divided that one by x^2 right?

similarly, our x^2 became 1

and 6x became 6/x

all of those make sense to me with dividing by x^2

my only question is about the +1 outside the square root

it seems like that one was divided by x rather than x^2

[sand fable]

what i mean to say is:

if any part of this image is confusing i will explain

[obsidian bobcat]

hmm okay i see. so we first determine that the biggest power in the denominator is actually x and not x^2, because the x^2 is inside a square root?

so the division is by x

[sand fable]

yes

that's more or less how it works

[obsidian bobcat]

would my method of viewing the terms inside the square root separately from the term outside be "wrong"?

[sand fable]

just depends on which is higher

[obsidian bobcat]

i just viewed it as "inside the square root, x^2 is the biggest term, so lets divide everything inside by that."

outside the square root, the biggest term is x

so lets divide the stuff outside by that

but idk if thats wrong

[sand fable]

this approach neglects to consider that the x^2 is being "lessened" in some sense by the square root

as sqrt(x^2 + 6x - 4) grows really large, it mimics an expression of degree 1, not 2

you can think of the square root as an exponent of (1/2) and that it should cancel out the x^2 inside of it

[obsidian bobcat]

hmm got you. as far as the part where you brough the 1/x into the square root and turned it into 1/x^2, what "rule" is being applied there?

we can bring a multiplier into the square root

[sand fable]

do you agree that the square root of (1/(x^2)) is 1/x ?

[obsidian bobcat]

yes

[sand fable]

yeah that's pretty much what's going on

[obsidian bobcat]

hmm okay, ill try and find more problems like this one and see if i can handle them. if not i'll write again somewhere here haha, appreciate it

[sand fable]

sure

but i will note... as far as solving limits is concerned, that last step isn't really that necessary

it suffices to rewrite sqrt(x^2 + 6x - 4) as sqrt((x + 3)^2 - 13), and then approximate that as (x + 3)

[obsidian bobcat]

hmm okay. this is the rest, for clarity:

[marsh citrus]

@obsidian bobcat Has your question been resolved?

[obsidian bobcat]

yes, .close

.close

[marsh citrus]

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[unreal bridge]

The answer is C), but i can only get B). How do I solve this?

[marsh citrus]

@unreal bridge Has your question been resolved?

[spark otter]

i would advise you show your work

[spark otter]

how?

[unreal bridge]

its a symmertrix matrix

transpose of B = B

[spark otter]

uh huh

but why -2B then

if it's -B^T + B

[unreal bridge]

ah ok i get it

stupid mistake

thanks

.close

[marsh citrus]

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[marsh citrus]

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[latent coral]

i have this problem but it feels very vague and i don't know if i can provide an actual matrix or set of matrices as an answer. i know for the m = 1 case it must be the matrix with entry 1, already for m = 2 the matrices get more complicated. i know that A cannot be the zero matrix since it has to be rank m, and that A can always be the identity matrix. there still seems to be plenty more A that satisfy this. maybe they want me to characterize A more generally? the previous questions in this homework talk about projection matrices, so maybe something about A being a type of projection matrix (though I don't see the condition that A^T = A)

[slow holly]

First of all: what is the determinant of the product of two matrices?

Remember that if A^2 = A, det(A^2) = det(A); this will restrict the matrices you can use significantly

Additionally, the dot products of each row and column must be equal to the entry in both (which is simply by row-column multiplication)

[latent coral]

so from that i get that A has to have det(A) = 1

because then det(A^2) = (det(A))(det(A)) = det(A), and i know det(A) != 0 since A is full rank, so det(A) has to be 1

[slow holly]

So you're already restricted to the unitary matrices

[latent coral]

is there some special significance to matrices with determinant 1?

[slow holly]

They're known as "unitary matrices", they have some special properties but you probably don't need them as far as I know

I'm just illustrating that this question is more restrictive than it may seem; A^2 = A implies many things about matrix A

[latent coral]

does A have to be an orthogonal projection matrix?

we haven't covered determinants yet in class so i don't know if me characterizing A as the set of mxm matrices satisfying det(A) = 1 is the answer he's looking for

chatgpt thinks it's something like "A is a projection matrix onto its own column space"

[slow holly]

well each column vector must map to itself

under the transformation

[latent coral]

perfectly to itself? like a column vector (1 1 1) should map exactly to (1 1 1)?

[slow holly]

yes, because multiplying AB is like constructing a matric C such that column c_n is column a_n times B

[latent coral]

if i wanted to answer it as "A is a projection matrix onto Col(A)", does that make sense?

i'm trying to think about what the significance of that is compared to any other projection matrix

[slow holly]

A is a projection matrix onto Col(A) by definition

[latent coral]

could you explain what you mean by that being by definition?

i thought a projection matrix just had to satisfy A^T = A and A^2 = A, and i don't see the transpose condition here

[slow holly]

wait a second, I misunderstood "projection matrix" for a moment

A is not a projection matrix onto Col(A) by definition

[latent coral]

i know that A being full rank means that it will map vectors from R^m to R^m, and the A^2 = A thing means there's some kind of redundancy

i was going to say that meant that A had to be some sort of projection since applying it again to a vector doesn't do anything, but then that would imply that any vector in R^m would be getting projected onto itself, so A would have to be the identity

but A can be other matrices too i think

[slow holly]

By definition, this condition can be summed up as all matrices A for which the dot product of column row i and column j is a_ij

I think the implications of that definition, with the corollary that it must map all its own column vectors to themselves, should be what to consider

Also, i'm fairly certain it can be a singular matrix too by this question...

[latent coral]

sorry what does singular mean again? noninvertible?

[slow holly]

yes

det(A)=0

[latent coral]

oh shoot just because A is full rank doesn't mean it can't have 0 determinant?

i thought det(A) being 0 implied that A was not full rank

by invertible matrix theorem and stuff

[slow holly]

hmmm

oh true

where does it ask for a full rank matrix though

[latent coral]

mxm matrix A has rank m

[slow holly]

oh

[latent coral]

yeah so A has to be invertible

[slow holly]

ok

[latent coral]

wait by that reasoning, why can't i take A^2 = A and multiply both sides on the left by A^-1 since it must exist?

[slow holly]

Thus A = I

[latent coral]

then A^-1 * A * A = A^-1 * A, so IA = I and A = I

[slow holly]

Which is, I am fairly certain, the answer, since I is the only matrix that maps all vectors to themselves

Which is logically a criterion here

[latent coral]

oh perfect i thought there would be more and that I was one of them but i guess it's the only one

is it valid to multiply both sides on the left by A^-1 and assume the equality holds afterwards?

[slow holly]

should be

[latent coral]

ik matrix multiplication gets messy sometimes with equality based on if the null space is trivial or not

[slow holly]

Well the null space is trivial if the matrix is full rank

Since the nullity is 0

[latent coral]

oh yeah in this case then A would have to be one-to-one so inverting would maintain equality

perfect

[slow holly]

So it's definitiely a legal move then

[latent coral]

thanks for your help!

[slow holly]

anytime!

[latent coral]

.close

[marsh citrus]

Channel closed

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[marsh citrus]

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

[austere bramble]

How do I make this fibonacci sequence return an array... yet keep it recursive. I want it to only be a recursive fibonacci sequence ```ull fibonacci_r(int n, ull *ops)
{
ull *row = malloc((sizeof(ull)) * (n + 1));
//set base numbers
if(n >= 0)
{
row[0] = 0;
}
if(n >= 1)
{
row[1] = 1;
(*ops)++;
return fibonacci_r(n - 1, ops) + fibonacci_r(n - 2, ops);
}

}```