help-10

[timid silo]

awesome

is this a similar idea for integration?

[heavy thicket]

Why integrate?

[warm canopy]

eh not really, except for the fact you are substituting something

the idea behind subbing in integration is a bit different

but ig in general, subbing usually leads to some kind of helpful simplification that makes your problem easier, thats the link

[timid silo]

ah okay, tldr im taking a calc 1 placement so i learned everything in differentation this year in senior year

and i gotta teach myself integration but for now im just reviewing stuff i kinda forgot

[heavy thicket]

That’s a good first step

Because if you don’t know the fundamentals, you won’t be able to do the fun stuff

[timid silo]

tysm for the help!!

.close

[obtuse pebble]

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[alpine bison]

try to use P=id and D=A

[tardy epoch]

why would D=A? you know it's diagonal?

Use the definition of eigen vector and eigenvalue for A and your given two eigenvectors

have you ever diagonalized a 2x2 matrix?

i recommend watching
https://www.youtube.com/watch?v=2nLtBBatGt8

[obtuse pebble]

@slow timber Has your question been resolved?

[tardy epoch]

@slow timber did you watch the video

it basically gives you the whole answer

[pseudo swift]

because their software sucks

any order will give you a correct diagonalization

because you screwed up the determinant

your P^-1 was wrongly calculated that's my point

in the first attempt

,w {{1, 3}, {-2, -7}}

[warm shale]

Results provided by WolframAlpha

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

Requested by aPlatypus#3790

[pseudo swift]

oops

,w {{1, 3}, {-2, -7}}^-1

[warm shale]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by aPlatypus#3790

[pseudo swift]

you prolly forgot a minus somewhere while typing in wolfram

let's see it their software works

ah ok it is decent then

yeah I'm too quick on judging automated marking software lol

[obtuse pebble]

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[heavy thicket]

1 remains unanswered

[misty valve]

.close

[obtuse pebble]

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[fathom sedge]

How do i discover the interval of integration for a sphere out of origin?

[obtuse pebble]

@fathom sedge Has your question been resolved?

[rigid pine]

If you can agree the sphere is x^2 + y^2 + (z - 2)^2 = 4 then expanding gives us x^2 + y^2 + z^2 - 4z = 0.

Can you see how you can now use spherical coordinates? @fathom sedge

[fathom sedge]

not really, i mean, i see that i have a sphere and all, but idk how to go from that sphere equation to the interval i need with the right jacobian

[rigid pine]

You probably know a formula for rho in terms of x, y and z and also for z in rho and phi.

p^2 = x^2 + y^2 + z^2 and
z = p cos(phi).

[fathom sedge]

yep

[rigid pine]

What's the bound on p?

Hint. p^2 - 4pcos(phi) = 0.

[fathom sedge]

Okay, i think i got it

not enough to answer you, but i think i might be able to work some more
you just used the x=p*sin(phi)cos(theta), same for y and z in the equation and got it in terms of rho and phi
rho will go from 0 to 4cos(phi) i think?
that helped a lot to understand what to do

[rigid pine]

You could do that with x, y and z but you can get the result directly with a understanding of where spherical coordinates comes* from but you have the correct bounds assuming I've made no errors.

*Pythagoras.

[fathom sedge]

thats clever, i hadnt think about that tbh, but i can
i'm used to do x y and z cause its a recent subject for me, ty

[rigid pine]

Keep practicing you'll get it. Can you complete your problem now?

[fathom sedge]

i think i can, i'll try it, ty

.close

[obtuse pebble]

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[willow ravine]

Complex variables:

I got this, not sure if either is correct.

[obtuse pebble]

@willow ravine Has your question been resolved?

[willow ravine]

<@&286206848099549185>

[violet flame]

bro

u tried with euler rule?

cos + isin = e^itheta

ig it works with th

[willow ravine]

Yeah but e^z0

Not imaginary part

[violet flame]

assume z0 = costheta+isintheta

[willow ravine]

I know e^(x+iy)

[violet flame]

we dont have any relation for this na

[willow ravine]

??

That is equal to e^x(cosy + isiny)

In my problem I only have the e^x part

e^Z0

I need to use Rouche’s Theorem

[violet flame]

yh yh

one sec

[violet flame]

oh fine

i assumed z0 to be a complex number

[willow ravine]

No

Alpha is complex

[violet flame]

got it.

[obtuse pebble]

@willow ravine Has your question been resolved?

[slim cove]

you want your inequality for rouché's theorem to be the one that you already know to be true

so pick f and g such that f+g = e^z0 - az0, and f = e^z0

probably

I haven't actually tried this

[willow ravine]

So g is -az0

I’m trying to show that and e^z0

Have the same number of zeroes?

[slim cove]

yeah f(z) = e^z, g(z) = -az

[willow ravine]

I don’t see what good that does me to show these two functions have the same number of zeroes

[slim cove]

It will lead to a contradiction

if you take a moment to think about it

Remember the goal is to prove a contradiction

[willow ravine]

Let me think on it.

If I do this then I’m not sure how to simplify it. I get e^|z0| + a|z0| < e^|z0|

After using the triangle inequality

[slim cove]

oh

[willow ravine]

So I just use simple x + 1 < x

Is not possible?

[slim cove]

you don't need the triangle inequality

[willow ravine]

Hmmm

[slim cove]

for this problem

[willow ravine]

Do I use the fact that this assumption shows e^z -az and az have the same number of zeroes?

Then show somehow e^z - az does not have any zeroes?

But az does.

Basically Roache’s theorem

Wait

I can’t use az in roache’s

[tardy epoch]

🪳's theorem 🤣

[willow ravine]

Because z= r will always contain z = 0

[willow ravine]

Yeah, I’m degrading this for a reason

I hate this theorem and I know it’s going to be on the exam

[tardy epoch]

Why

Oh cuz it's hard for you?

[willow ravine]

For some reason it is so hard to understand the many different ways to use it, decide a function, show a contradiction.

I am not enjoying it at all.

[tardy epoch]

If every theorem was given a degrading name because one student had a hard time understanding it, everything would be roach's theorem

[willow ravine]

We should do this. I like this idea.

That way when I use something I don’t have to think of a hard name, everything can just be roach’s theorem.

[tardy epoch]

"hard name" lol

[willow ravine]

I am lost in the sauce

[willow ravine]

Would this imply that e^z -az has one zero then set it equal to zero and then you got e^z = az

[slim cove]

wait what are you choosing as f and g again?

[willow ravine]

F: e^z0
G: -az

I think I had it backwards

I should have then e^z -az has the same number of zeroes as e^z

[slim cove]

wait

state rouché's theorem again?

[willow ravine]

Wait f and g

Not f+g and f

So az and e^z have the same number of zeroes

Which is a contradiction since e^z is not zero?

[slim cove]

yes

[willow ravine]

This shows there exists at least one |z0| = r such that: (insert the problems statement again).

@slim cove

[slim cove]

yup

[willow ravine]

Okay! Thank you so much! I’m going to get some sleep now. Have a good one @slim cove

[slim cove]

good night!!

[willow ravine]

.close

[obtuse pebble]

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[woven mantle]

How to add and subtract polynomials

[proven zephyr]

need more context

what polynomial

[woven mantle]

Polynomials in general

That’s what it says I need to learn

[proven zephyr]

https://youtu.be/0aJBtUOXpsk like this?

[woven mantle]

Yeah

[obtuse pebble]

@woven mantle Has your question been resolved?

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[sonic cape]

hello! Could anyone kindly explain to me the underlined portion of the solution, thanks 🙂

[tardy epoch]

do you understand why six vectors are in the plane x + y + z + w + c = 0 for some constant c

[sonic cape]

hm not really :😓

I know that there must be dependence since 6 vectors cannot exist independently in R^4

[tardy epoch]

(1, 1, 1, 1) is a normal vector to the plane above

so when all the v vectors are dotted with that normal vector result in zero, that implies all 6 vectors are in that plane

a 2d version of that would be having vectors u1, u2, u3, u4 and when dotted with (1, 1) results in zero, then all of the u are on the line x + y + c = 0

or y=-x + c

[sonic cape]

Oh I see it now, each v . (1, 1, 1, 1) gives 0, hence all the V(s) lie in this plane?

plane with normal of (1,1,1,1)

[tardy epoch]

yes

[sonic cape]

wait but if it's a plane, means the basis of it is 2 independent vectors

but v1, v2 and v3 are independent?

[tardy epoch]

errr, i mean hyperplane* everywhere

[tardy epoch]

this is the equation of a hyperplane in 4d

[sonic cape]

ohhh okk i get it now!

thank you 😄

[tardy epoch]

[sonic cape]

.close

[obtuse pebble]

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[brisk grove]

Hi, I need help proving a limit of a sequence using the definition.
Showing that for all e > 0, there exists N which is a natural number, such that for all n > N, |a_n - L| < e

[brisk grove]

I understand the definition and the objective, but I am the type that likes everything to be perfect so I'm finding it hard to formulate the proof.

[tardy epoch]

what are you stuck on?

[brisk grove]

I am able to find the N that I need,

but I don't know how to write it formally

[tardy epoch]

just show the problem and your work

[royal basin]

what is the sequence and what is your N?

[brisk grove]

$a_n = \sqrt{\frac{4n+1}{n}}, L=2$

[warm shale]

RedstonePlayz09

[brisk grove]

So I took:
$\abs{a_n - L}$

[warm shale]

RedstonePlayz09

[brisk grove]

And found out that it is less than 1/n

So I can choose:
$N = \ceil{1/\epsilon}$

[warm shale]

RedstonePlayz09

[brisk grove]

But now I don't exactly know how to write the proof

[tardy epoch]

show your full work for this

[brisk grove]

Okay

[royal basin]

how did you establish that |a_n - L| < 1/n?

[brisk grove]

How do I type <= in latex?

[royal basin]

\leq

(less or equal)

[brisk grove]

how do I do less?

just <?

wait

I have a mistake

well I'll come back when I manage to get through this part

But one of my questions is

do I start with < epsilon

or just simplify |a_n - L|

[royal basin]

your goal is to present a value of N (in terms of epsilon) and verify that it works.

[brisk grove]

And the work to find that N should be in a draft right

not part of the proof

[royal basin]

yup

you've already presented N := ceil(1/epsilon)

we care not how you found this

but do verify that it works

[brisk grove]

it's wrong I think

[royal basin]

well whether it's wrong or not you still have yet to present any work

[brisk grove]

$N = \ceil{\frac{1}{\epsilon * (\epsilon + 4)}}$

[warm shale]

RedstonePlayz09

[brisk grove]

this should work

so I just

start with this? without showing how I got it?

[royal basin]

yes my previous point still stands

[brisk grove]

okay I'll try

[royal basin]

use \cdot if you want an explicit multiplication symbol btw

(or just write nothing at all)

[tardy epoch]

example of "work" for this proof

more examples here: https://dsury.com/content/images/104/104-lec4.pdf

[brisk grove]

alright I'll try doing that

[tardy epoch]

this could be right. you just haven't justified it with the level of details required like the examples.

[royal basin]

and for the purposes of the proof

[brisk grove]

I got it from |a_n - L| < epsilon

[royal basin]

really nobody gives a shit about justification

[brisk grove]

by isolating n

[royal basin]

redstoneplayz09, when you presented your value of N for the first time, i asked you to present your work for finding |a_n - L| < 1/n

but you backed out

why was that?

[brisk grove]

I sent it

but noticed a mistake

so I deleted it

its wrong

I wrote something + 2 < something

which is... false

[brisk grove]

this is something i got from another attempt I had before that

but wasn't sure how to continue

[royal basin]

so ok

can you send your rough work or something

(by the way, it is true that |a_n - L| < 1/n.)

[brisk grove]

some of these aren't obviously true

since I did these steps backwards at first, so I still have to justify a few things

also forgot to add abs at the end

[royal basin]

i suppose this might work, though it does feel a little inelegant

[brisk grove]

yeah this is why I wasn't sure about starting with < epsilon

[royal basin]

i can show you what i came up with

my scratch work, if you will

[brisk grove]

also u said that it is true that |a_n - L| < 1/n

if it is, I could choose N = ceil{1/epsilon} and it should work right

[royal basin]

indeed

[brisk grove]

so basically start with |a_n - L|, and show that it is less than some other expression

[royal basin]

\begin{align*}
\absv{\sqrt{\frac{4n+1}{n}} - 2} &= \absv{\frac{\sqrt{4n+1}-2\sqrt{n}}{\sqrt{n}}} \ & = \absv{\frac{4n+1-4n}{\sqrt{n}(\sqrt{4n+1}+2\sqrt{n})}} \ & = \frac{1}{\sqrt{n}(\sqrt{4n+1}+2\sqrt{n})} \ & = \frac{1}{\sqrt{4n^2+2n} + 2n} \ & < \frac{1}{2n}
\end{align*}

[brisk grove]

okay and u did that because 1/2n is easily to isolate n from

that's the point right?

[royal basin]

yes

[brisk grove]

the first time was good

[royal basin]

hold on

[warm shale]

Ann

[royal basin]

there

[brisk grove]

okay so

[royal basin]

in fact an even stronger statement is true

[brisk grove]

is this part of the proof or a draft

[royal basin]

but you do not really need to find "the best" N that works

[royal basin]

well this chain of inequalities can go either way tbh

if i were writing a proof i would include this almost as-is

[brisk grove]

so the proof will go like:
choose N = ceil{1/2epsilon}

then, since n > N

[royal basin]

ceil(1/(2ε)) if you insist

[brisk grove]

n > N > 1/2n > .... > |a_n-L|

[brisk grove]

yes I meant that

[royal basin]

n > N > 1/2n > .... > |a_n-L|
bad

[brisk grove]

okay so why is that bad?

[royal basin]

N > 1/(2n)?

are you sure this is what you want to say

[brisk grove]

=

?

Oh

epsilon

wait

[royal basin]

$|a_n - L| < \frac{1}{2n} < \frac{1}{2N} \leq \frac{1}{2 \cdot \frac{1}{2\ep}} = \ep$

[brisk grove]

im getting confused

[royal basin]

(even this is kind of pushing it)

[warm shale]

Ann

[brisk grove]

okay so I get this work but

im still confused..

give me a sec to think about it

I would do it like this
n > N = ceil{1/(2ε)} >= 1/(2ε) > ... > |a_n - L|

so I dont understand

[brisk grove]

how you thought of doing it this way

And also u didn't explain here why 1/(2n) is bigger than |a_n - L|, that would require the draft

[tardy epoch]

you should read this carefully

[brisk grove]

what do you mean

[tardy epoch]

did you read it carefully?

[brisk grove]

yes

thats the draft

it shows that |a_n - L| < 1/(2n)

right

[tardy epoch]

yes

[tardy epoch]

which is the first inequality here

[brisk grove]

yeah

ok wait give me a sec

[tardy epoch]

then why don't you understand this

[brisk grove]

I dont understand how they thought of doing 1/(2n) < 1/(2N) ....

[tardy epoch]

oh

[tardy epoch]

"for all n > N"

[brisk grove]

wait

I think I got it

n > N = ... > 1/(2ε)

therefore n > 1/(2ε)

ε > 1/(2n) > |a_n - L|

[brisk grove]

so my only problem now is showing that this is true in both directions

in different problems it might not

okay you know what, I think I understand enough to try it on my own now

@royal basin @tardy epoch thanks for the help!

[tardy epoch]

read the links

[brisk grove]

yes I will

right now

.close

[obtuse pebble]

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[loud egret]

I found the solution

but idk how to work it out manually

to find which one belongs to ker(A)

nvm

I thijnk

I figured it out

.closed

.close

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[ornate pollen]

How do I go about solving this?

Considering its a multiple choice question, I assume it needs to be done pretty quickly but I just can't figure out how to solve it.

[tardy epoch]

you can differentiate all your answers and see if they equal your integrand

[ornate pollen]

so that'll eliminate B and C

[tardy epoch]

lol okay maybe not

[ornate pollen]

since I can easily integrate them in my head

oh unless I've done something wrong 😂

[scarlet mango]

Use
$$ \tan^2 X = \sec^2 X -1$$

[warm shale]

D3VC

[haughty coyote]

You need to find which one has a derivative of tan^2

[ornate pollen]

ah okay that makes a lot more sense

I'm pretty washed up with my trig

ok thanks!

.close

[obtuse pebble]

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[next vigil]

hello 👋. i need a not very easy, but not hard definite integral that solves in 524. i’m not that good at integrals and can only think of simplest integrals, but maybe someone can think of an interesting integral which solution is 524?

[kind hawk]

integral from sqrt(5) to 23 of 1/2 x

I mean how interesting are we talking

you could use interpolation to find some polynomial with p(0)=a and p(1)=b with b-a=524. then differentiate that polynomial to get f. then int from 0 to 1 of f dx gives 524

[next vigil]

it seems to solve into 131

[kind hawk]

ah sorry I meant integral from sqrt(5) to 23 of 2x

[next vigil]

yep, that fits! now i got at least one suitable integral, thank you very much. can you think of something a bit harder than this one? i’m not sure if we’re able to use trig functions since the answer should be integer.

[next vigil]

i’ll note this to play with it myself later, thanks aswell.

[kind hawk]

well in general the best way to tackle this is probably to find some function with f(x)-f(y)=524 for some y<x

and well you could define all kinds of functions with that

but most of them probably seem very constructed

maybe write 524=67.5^2-63.5^2 and then construct f(x)=x^2cos(a*x) so that f(67.5)-f(63.5)=524

and if you want integer bounds maybe shift the input by 1/2 to one side

[obtuse pebble]

@next vigil Has your question been resolved?

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[next vigil]

thank you!

[obtuse pebble]

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[jaunty lake]

Ive been trying to find out what values m will intersect and not intersect of the two lines below and I have no idea on how to find out, could someone please help?

[timid silo]

do you know how to solve linear systems using matrices?

[jaunty lake]

nope

what are matricies

[timid silo]

dont worry about that for now

you can use the fact that two lines (in 2d space) do not intersect if they are parallel (ie have the same slope) and aren't the same

[jaunty lake]

so how do i solve the equation with that

[timid silo]

find the slope of both lines, then see when they are equal

[jaunty lake]

first one is 1 and the other would be 1/2 right?

[timid silo]

no

the first one is right, the second isn't

write the second in the form y=ax+b

then a will be the slope

[jaunty lake]

m/2?

[timid silo]

yup

now see when m/2 is equal to 1

[jaunty lake]

when m = 2

[timid silo]

yup

[jaunty lake]

thank you bro

[timid silo]

yw

[jaunty lake]

.close

[obtuse pebble]

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[obtuse pebble]

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[timid silo]

What does it mean for tanx and tan2x must be defined? and How should I solve this question?

[warm shale]

Boytjie

Boytjie

[tardy stratus]

Let me know if this observation explains things or you need more help.

[timid silo]

use compound angle formula

$tan(2x)=\frac{2tan(x)}{1-tan^2(x)}$

[warm shale]

∆y/∆x=sin(x) + cos(y)

[timid silo]

so u can rewrite the equation as

$\frac{2 tan^2(x)}{1-tan^2(x)}=1$

[warm shale]

∆y/∆x=sin(x) + cos(y)

[timid silo]

@timid silo u there?

yeah, I never heard of the formula tho so I kinda got confused

[lusty cedar]

if you've heard of tan(x+y), then just put x=y in that

might help

[timid silo]

oh

wait

what grade are you

high school?

yeah grade 10

well I think they expect a different solution

coz u dont learn compound angle in grade 10

or have u?

the topic should involves anything between advanced functions to precalculus but harder than normal curriculums

[timid silo]

have u learned $sin(2x), cos(2x)$ formulas either?

[warm shale]

∆y/∆x=sin(x) + cos(y)

[timid silo]

ig compound angle is in pre calculus

tho I am not sure I didn't took pre calculus

[timid silo]

not really but I can look it up

[timid silo]

so like $\frac{2tan^2(x)}{1-tan^2(x)}=1=sin2x$

[warm shale]

Pikachuexeallen

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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[distant moth]

is (f+g)(x) always f(x)+g(x)?, and is it the same in reverse?

[distant moth]

does it work like (sin+cos)(x)= sin(x)+ cos(x)?

[fierce lagoon]

I uh

I mean, yes and no??

Who the hell would use (sin+cos)(x)

I mean if f = sin(x) and g = cos(x), sure, yeah I guess but like

No one writes like that

[distant moth]

ive just encountered this notation in book, so im curious what is going on

[fierce lagoon]

Yeah I mean stick to just single-letter functions

That are just dummy names like f and g

[timid silo]

Its prolly useful for question makers

[distant moth]

so f+g means f(x)+g(x) aways?

[fierce lagoon]

Yeah

[timid silo]

I have seen functional equation questions like it

[distant moth]

okay thanks that clears my doubt, and its the same for the other operators like f.g just means f(x).g(x)? and f/g means f(x)/g(x)?

[distant moth]

wait a min, how do u know xcalibur🤣

[timid silo]

Thats incredibly rare and impossible

[distant moth]

💀

[timid silo]

1 in millions

how do you know xcalibur

[distant moth]

some server named debate lounge

[fierce lagoon]

Tbh those are rarely used to the point i wouldnt stress them and just go with gut instinct

f • g is multiply
f/g probably means to divide

[distant moth]

alright then, hope i dont run into any problems

thanks for ur help

.close

[obtuse pebble]

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[spring trail]

For real values of k prove that 2kx^2+5x-k=0 always has two distinct real roots.

[spring trail]

So this question reduces to:

8k^2+25>0

And i can just write some explanation to prove it

but mathematically solving the inequality gets me unreal values

<@&286206848099549185>

[proven zephyr]

b^2 - 4ac > 0

so you already have the discriminant right?

[spring trail]

yes i do

in terms of k

[spring trail]

.

[proven zephyr]

ok

[spring trail]

...

[proven zephyr]

hmm

im also getting sqrt of negative number

[spring trail]

bruh

ofc you'll too

I'm asking what that means

[proven zephyr]

why do you need to find k btw

[spring trail]

@tardy stratus would u mind helping?

[spring trail]

I want to find the range of k. Its not required.

[proven zephyr]

k can be any real values tho

[timid silo]

Square is always positive

[proven zephyr]

no

square of i is not positive

[spring trail]

i get that but

[timid silo]

We're assuming k is real

[spring trail]

.

[spring trail]

the question specifies k as real

[timid silo]

K can be any real value

[proven zephyr]

ye ik

[spring trail]

so the question is already answeres

d

but the inequality itself. and k's range is what i wanna know

[proven zephyr]

the range is -infinity to +infinity no?

it can be any number right?

[spring trail]

yep but solving inequalities doesnt rlly show that

[timid silo]

U can't use wavy curve

Cuz I think that's what I suppose ur using

Right?

The thing is it's never 0

[spring trail]

u mean the parabola?

[timid silo]

So u can't use factorization and solve it

[timid silo]

Wavy curve method of finding solution to inequality

[spring trail]

yes coz ksqr is always positive

[worn ocean]

U can use wavy curve on any rational solutions

[timid silo]

Yes but there's no solution ie 0 here

[spring trail]

idk what that is. i just factorize then check individual factors as inequalities. rather tedious. mind teaching me this wavy curve?

[timid silo]

Yes thus is essentially the gist of wavy curve but it doesn't work here

Cuz factors are imaginary

And imaginary numbers cannot have any greater/lesser relation

So u just use the fact that x^2>0 (x belongs to R)

[spring trail]

u sure abt that?

[timid silo]

Yep

[worn ocean]

The factors arent imaginary here tho

[timid silo]

5i isn't greater than 2i for example

[spring trail]

what r they?

[worn ocean]

They should be real numbers

[spring trail]

i'd say it is

[worn ocean]

Maybe they're irrational

[tardy stratus]

Taking a look now

[worn ocean]

Irrational =/= imaginary

[spring trail]

is it also true for complex numbers?

[tardy stratus]

Yes, okay I can see why you're getting imaginary solutions

The thing is, something squared is always >= 0, right?

so k^2 >=0

[worn ocean]

K^2 is always positive

[spring trail]

[worn ocean]

So its automatically true

[tardy stratus]

and then certainly 8k^2 >= 0

yes!

So when you try to solve, you're asking when 8k^2 + 25 = 0 as part of that

and that's.... that doesn't work out

so you get complex values

Does that make sense?

[spring trail]

y does the inequality not show its true colours and tell me its critical points?

[tardy stratus]

So can you run me through how you're solving it, and we can try understanding

[spring trail]

or are the unreal solutions i got equal to positive and negative infinity?

[tardy stratus]

Ah, no

[worn ocean]

What's the problem with this?

Axioms are made for a reason

[tardy stratus]

Quill, I don't think this is a conversation about axioms really, this is about a simple inequality :)

[spring trail]

i wanna mathematically get a range for k

[worn ocean]

Oof

[spring trail]

okay then imma give up and complete hw

[spring trail]

cool

[tardy stratus]

Right Shah, so unfortunately you just have to think about this case

[worn ocean]

I thought it was about proving how is 8K^2 + 25 is always greater than 0 ;;;

[tardy stratus]

bc sometimes it's just always true

[worn ocean]

[spring trail]

umm i specific at the start i already answered the question itself... anyways bye

.close

[obtuse pebble]

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[tardy stratus]

Good luck Shah

[obtuse pebble]

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[glass gorge]

how would i go about finding the linear relation

[fierce lagoon]

Find the slope

Find the y intercept

Use slope-intercept form

[glass gorge]

is it -4

ohh wait

is it

y=-4x+4

wait no

hmmmm

OH

@fierce lagoon I found it

y=4x-4

.close

[obtuse pebble]

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[obtuse pebble]

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[sour peak]

what would be the best method to approach this problem

[mighty comet]

When you have velocity as a function of time, integrate it with respect of time to obtain the position function as a function of time

That is,

[sour peak]

velocity is d?/dt right

because you separate the variables and integrate both sides with respect to t?

[warm shale]

calicals

[mighty comet]

yes

[sour peak]

so i can just write this as x(t) = integral and my function

[mighty comet]

Yep!

[sour peak]

okay that makes sense

i would use integration by parts for this right

also, would i assume that x(0) = 0 for this practice problem

[mighty comet]

Just asking, but why do you think that?

[sour peak]

this is practice and in a similar version of this problem they did ibp

ive been doing multiple practice probs for this type of diff eq

to prep for my final next week

[finite lance]

y = 4x-4

[mighty comet]

I think you can, this is a rocket launch anyway. At t=0, it's still on the ground 😄

[sour peak]

okay that makes sense, because for the first part we had to find the velocity of the rocket at any time

and we assumed the intitial velocity was zero

that part is sooo much easier lol

yeah i guess obvi x(0)=0 so its inferred then

[mighty comet]

Ahahahah, yepp

[sour peak]

would it be easier to use the property of logs of lnM-lnN = LnM/N

to do ivp this way

[mighty comet]

I mean, you can

[sour peak]

or would it not really help in this situation

[mighty comet]

Both are ok imo

[sour peak]

I think this is right

[mighty comet]

I don't think anything is wrong with that, congratulations!

[sour peak]

okay thats good

for this, would my u be ln(M0-at/mo)

[mighty comet]

You could simplify it first

To become $$1-\frac{at}{m_0}$$

[warm shale]

calicals

[mighty comet]

Use the fact that ln 1 = 0

[sour peak]

okay i can try that

if i factor out the -b and put it on the x(t) side, would that still be my u, and then my v would be -gt

[mighty comet]

Why don't you try to separate the integral

Since it's just gt, you can solve that easily and put it in the LHS

[sour peak]

so i can rewrite it as

$$x(t) = -gt^2/2 int(-bln(M_0-at)/M_0dt$$

[warm shale]

elleyes

[sour peak]

lol ive never used that before

[mighty comet]

Let me help

[sour peak]

i appreciate your help im just struggling hard so thank you

[mighty comet]

$$x(t) = \frac{-gt^2}{2} \int(-b\frac{\ln{(M_0-at)}}{M_0}\ \mathrm{d}t$$

[warm shale]

calicals

[sour peak]

yes

thank you

[mighty comet]

No problem!

[sour peak]

i can also factor out - 1/b so i can just focus on the mess of a fraction i have

and just put it on the left side

[mighty comet]

Yep

[sour peak]

that might make it more manageable because my professor has a video where he does that integration by parts but the -gt was added to this problem

since i separated the gt^2/2, once the IBP is done, can i just add it to the end of the problem

[mighty comet]

Of course, that's the purpose of evaluating it first 🙂

[sour peak]

because this is what the solution to the problem is without -gt

okay, well if thats all i gotta do now thank you! separating that made it so much easier

i appreciate your help.

[mighty comet]

Yep, good luck!

[sour peak]

thank you !

.close

[obtuse pebble]

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[little shuttle]

so

[little shuttle]

I want to figure out how not to fail trig

since this is a general question I wanted to ask main chat but they were too busy being dumb

I scored 79 and 72 on my last two tests

I am trying so hard but its just so fucking quick

I am taking a half semester class

so its a full course in half the time

its collegiate

im just so fucked rn

I cant fail or retake this class either

I need it as a prereq

[nocturne minnow]

Study

[solar trellis]

how much do u need to pass

[nocturne minnow]

Also, not really the channel to ask this in since it's not really a question you need math help on

[little shuttle]

very badly

oh you mean a score

a C

but a B is better

[solar trellis]

how much out of 100?

[nocturne minnow]

Even though a B is better. You're sitting at a C range so you're doing well enough to pass

[cedar lichen]

There's always Khan Academy

[little shuttle]

70

[little shuttle]

khan doesnt really help for this trig

[cedar lichen]

Why not?

[little shuttle]

its like advanced proof honors type trig

[cedar lichen]

Ah

[little shuttle]

like circumcenter diagonilization proof whatever

like he actually proves each identity we use in class

its kinda cool but theres so much to learn

[obtuse pebble]

@little shuttle Has your question been resolved?

[obtuse pebble]

@little shuttle Has your question been resolved?

[obtuse pebble]

@little shuttle Has your question been resolved?

[little shuttle]

.close

[obtuse pebble]

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[stable shale]

After multiple attempts (21), and emailing my teacher I still cannot figure out the answer to the last 2 parts of the question. It is probably simple and I am just tired of math, but I have seemed to have tried everything and havent gotten there. If it is any help, I am pretty sure you have to use Ce^rt on this problem at some step, but I have not figured it out yet. Any help would be greatly appreciated.

[distant wasp]

You have solved part a right?

[stable shale]

The first part, A., is correct

[distant wasp]

Then you need to solve the differential equation

[stable shale]

With the initial balance of 3500 though

[distant wasp]

I will be in the voice chat to solve this

[obtuse pebble]

@stable shale Has your question been resolved?

[stable shale]

It is still ongoing

[obtuse pebble]

@stable shale Has your question been resolved?

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[dusty steppe]

Help

[dusty steppe]

#6 please help
I don't know how to find the domain or range

[void pelican]

<@&268886789983436800> test

[dusty steppe]

?

Oh

Yes

[void pelican]

it literally says test on top

[dusty steppe]

Test

[void pelican]

we can't help with thay

literally cheating

[dusty steppe]

Welp it’s homework

Ah

Ok mate

[void pelican]

but why would it say test

[dusty steppe]

Idk it’s the Sunday and doing it for tomorrow morning

.close

[obtuse pebble]

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[main abyss]

I understand the FTC1, however what confuses me is with the first area where the bounds of x are [0,2)

hello sir this is occupied

[woeful jolt]

sorry

[main abyss]

so for calculating the integral from [0,2)

do I do essentially the integral of 1-2t from 0-x + F(2)?

nvm got it fuck sake

.close

[obtuse pebble]

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[timid silo]

Distinction between Variable and Constants,
When I studied Integral Calculus I saw that Constants can be represented by symbolically to. Before that I thought variables are unknowns but constants here can also be unknown.
I'd like to know what exactly is a variable.

[nocturne minnow]

Do you have an example of a constant being a variable?

[timid silo]

wdym

ig y=2x+C

[nocturne minnow]

Do you have an example of a constant being a variable

C is a variable still

[timid silo]

lets say we integrate 2

[nocturne minnow]

C is a placeholder variable because you don't know the exact value of that constant

[timid silo]

alright could you explain why we refer to it as a constant in integration

[nocturne minnow]

Because a constant has a fixed value. When we integrate and add that plus C, it denotes to the person that there's suppose to be some fixed value but we don't know what that exact value is. Because the derivative of constants, is zero

[timid silo]

could we say that a constant is a value for any pre-image has a fixed image,
ex f(x)=5?

[nocturne minnow]

Yes

[timid silo]

maybe im over complicating this

[nocturne minnow]

So if you integrate 2, you get f(x) = 2x + C. As I mentioned, C is a placeholder because we don't know what the exact value is

[timid silo]

2x+1=3,
x=1
Is x a constant in this case?

[nocturne minnow]

C could be 1 and that makes the derivative still 2, because that constant becomes 0

[timid silo]

I got that.

[nocturne minnow]

When you plug in a number, that becomes a constant. But 2x + 1 = 3, x is a variable

When you say x = 4, and plug in 4, it because a constant at that instance

[timid silo]

oh so for the value of 3, x is a constant?

so we are limiting x's domain to 1?

[nocturne minnow]

Here is a google definition between variable and constant

A constant does not change its value over time. A variable, on the other hand, changes its value dependent on the equation.

So if we make the variable a set value at that time, it's a constant

[timid silo]

Let's say for that the C(co-domain).
n(C)=1,
That's means it is constant?

[nocturne minnow]

That, I'm not too sure about. I haven't dealt much with that kinda of logic. The only difference I know between variable and constant is what I stated above

[timid silo]

So, x is a constant at 2x+1=3?

[nocturne minnow]

x is still being treated as a variable because it's an unknown value that you are trying to determine

[timid silo]

After finding the value does x become a constant?

[nocturne minnow]

I would say so because now you declared a value for x, but that occurs when you don't change that value of x

If you say x = 9 and don't change it then it's constant. If you say x = 9, then x = 3, you're changing the value of x meaning it's a variable

[timid silo]

Ok, I got it, if the possible values for x is a set and the cardinality of the set is 1 then it is a constant

[nocturne minnow]

Uh.... sure

[timid silo]

Thanks man

I've had this question for months but never asked anyone because I thought it was trivial.

.close

[obtuse pebble]

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[eternal dome]

i have a quick question

[eternal dome]

what is the slope of a slide

fuck

wait

worded that wrong

how do i find the slope of a right triangle

this is the q

[cedar lichen]

What's the equation for slope?

[eternal dome]

y=mx+b

[cedar lichen]

That's the equation of a line

Not slope

[eternal dome]

of slope is rise/run

[cedar lichen]

Yep

What's the rise of the slide?

[eternal dome]

11

so i would just do 11/7

[cedar lichen]

You sure it's 11?

What does the rise represent?

[eternal dome]

oh up

so 7/11

[cedar lichen]

Yep

[eternal dome]

thats it

thank you

.close

[obtuse pebble]

Channel closed

Closed by @eternal dome

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[uneven palm]

Wouldn't it be around 7 / 8.5?

[uneven palm]

I drew a diagram and it seems like pythagorean theorem is involved to find the slope

unless I'm horribly mistaken

although that is still greater than 0.55 so I suppose it doesn't matter

[cedar lichen]

I interpreted "how long" to mean the horizontal distance, but if it was referring to the hypotenuse, then you'd be right

In fact, you're probably right

[uneven palm]

yeahh because it's a slide

anyway that's fine as it doesn't affect the result

but you should probably keep that in mind for future problems @eternal dome

[eternal dome]

thanks

[uneven palm]

.close

[obtuse pebble]

Channel closed

Closed by @uneven palm

Use .reopen if this was a mistake.

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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>.
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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.

[eternal dome]

<@&286206848099549185>

[oblique sage]

!15min

[obtuse pebble]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[oblique sage]

Please read the rules and be a good person please

[eternal dome]

i was in another help thing first

and then i deleted the first question i had to send the second

but i have been in the help thing for a sum of 30 mins

<@&286206848099549185>

[mortal mica]

What's up

[eternal dome]

i need help with my question

i dont understand it at all

[kindred oasis]

Don't ping helpers before 15 minutes passed after you posted your question, and only ping once

[eternal dome]

ok

well i have been sitting here for too long

and i just want to learn

teacher is asleep and cant find anything on google

[kindred oasis]

What's the part of the question you don't understand? It's a pretty long question

[eternal dome]

idk what the diff of inside and outside diameter is

[eternal dome]

oh

so its like pizza

[kindred oasis]

You pool is something like this

[eternal dome]

and the crust is the outsidew

and the cheese and shit is the inside

[kindred oasis]

No there's an hole in it

[eternal dome]

oh

but just for knowing the outside and inside

[kindred oasis]

It's a pool, so...

[eternal dome]

yeah i get that

[kindred oasis]

r would be the outside radius

[cedar lichen]

It's like a pizza but it's all eaten and all that's left is the crust but you put the pieces back in a circle

[eternal dome]

yeah ok

[kindred oasis]

The radius up to the outside of the wall is the outside radius, the radius up to the inside of the wall is the inside radius

Do you get what I mean or I need to explain it better?

[eternal dome]

no i get it

[kindred oasis]

Ok

Let's divide the first request in two parts, let's find the volume of the walls first, then the volume of the floor ok?

[eternal dome]

ok

[kindred oasis]

I've never used the imperial system because I'm not from the US (and because it sucks) so I have no idea how to convert from cubic feet to cubic yards btw, but I can help you with the rest

[eternal dome]

im not from the us either

[kindred oasis]

I mean, wherever you use the imperial system

[eternal dome]

idk

[kindred oasis]

The volume of a cylinder is:
$$V=\pi r^2\times h$$

[cedar lichen]

3 feet in a yard. Cubic feet to cubic yards follows from there

[warm shale]

Andrea276

[kindred oasis]

Ok got it

[kindred oasis]

I'd use this formula to find the volume with the bigger radius (basically of the whole cylinder if it was full) and the I'd subtract the volume of the smaller circle (which is basically the hole in the cylinder).

Do you get what I'm trying to say here?

[eternal dome]

no

[kindred oasis]

We have something like in the figure above; to find its volume I think it's better to:
-first find the volume of the cylinder as if it didn't have a hole in it (diameter=13ft);
-then subtract the volume of the hole to remain with just the volume of the walls of the cylinder (diameter=12ft).

[eternal dome]

the radius is just the diameter x 2

right

[kindred oasis]

No, d=r×2

r=d/2

[eternal dome]

oh ok

[kindred oasis]

So try to find the two radiuses

[eternal dome]

so the radiuses are 6.5 and 6

[kindred oasis]

Correct

[eternal dome]

and then the volume for the outer is 199

[kindred oasis]

Wait, what's the height you used?

[eternal dome]

1.5

and for the inner it would be 169

[kindred oasis]

That's the inside height, as the problem says, but you also have a 0.5ft floor

[eternal dome]

so its 2 instead of 1.5

[kindred oasis]

To find the total height you need to sum the inside height (the height from the top of the pool to the floor) with the height of the floor

[kindred oasis]

Yes

[eternal dome]

ok so outer is 265

and inner is 226

[warm shale]

Result:

226.19467105847

[kindred oasis]

Yes

Subtract to get the volume of the walls

[eternal dome]

subtract both numbers

[kindred oasis]

Subtract the outer from the inner volume

[kindred oasis]

.

Since the inner volume it's just the volume of the hole, you want to subtract it from the volume of the whole cylinder

[eternal dome]

oh right

so then the outer is 39

typo

[kindred oasis]

Yes

That's the volume of the walls of the pool

Now you need to add the volume of the floor to find the total volume right?

[kindred oasis]

.

[eternal dome]

yes

so then for the floor do i use the 1/2 and 1.5

[kindred oasis]

How thick is the floor?

[eternal dome]

2

ft

[kindred oasis]

Just the floor of the pool, read the problem

[eternal dome]

oh

1/2 a ft

so 6 inches

[kindred oasis]

That's the height of the whole pool

[kindred oasis]

Yes, don't convert to another unit

[eternal dome]

oh ok

[kindred oasis]

0.5ft

[eternal dome]

so 1/2 ft

[kindred oasis]

Yes

[eternal dome]

ye

[kindred oasis]

You can think of the floor as an almost flat cylinder, with height 0.5 and radius [?]

[eternal dome]

radius of 12

[kindred oasis]

Yes

[eternal dome]

no

yes

[kindred oasis]

It's inside the walls, not below them

[kindred oasis]

.

[eternal dome]

but we have the volume already dont we

226

[kindred oasis]

But you have to use the height of the floor, that's what we're trying to find

[eternal dome]

so what do i swap

im kinda lost

[kindred oasis]

You found the volume of the walls

[eternal dome]

yes

[kindred oasis]

Now you're trying to find the volume of the floor, which is basically a disk right?

[eternal dome]

oh ok

[kindred oasis]

The thickness of the disk, or height, is 0.5ft

And it's radius is 12ft, since it's inside the walls

[eternal dome]

so V=pi 12^2 x 0.5

[kindred oasis]

Yes

,calc 6^20.5pi

[eternal dome]

i got 226

[kindred oasis]

Sorry, the radius is half the diameter

So it's not 12 but 6

I forgot to divide by 2 the diameter

It's the same radius as before

[eternal dome]

so 56

[warm shale]

Result:

56.548667764616

[kindred oasis]

Yes

I'd keep 56.5

Now you can find the volume of everything

[eternal dome]

ok

but to answer a

is that just the outer wall plus the floor

[kindred oasis]

Yes

Then you need to convert from ft³ to yard³ since it asks for cubic yards

[eternal dome]

how do i do that

[kindred oasis]

First calculate the volume in feet

[eternal dome]

i got 95.5

[kindred oasis]

$$3ft=1yard$$
Now, since it asks for cubic yards, and you have your volume in cubic feet, you can raise everything to the third power to get from ft to ft³:
$$(3ft)^3=(1yard)^3$$
$$27ft^3=1yard^3$$
So:
$$1ft^3=\frac{1}{27}yard^3$$
To convert from cubic feet to cubic yards you can just divide your answer by 27

[warm shale]

Andrea276

[eternal dome]

wow

[kindred oasis]

That took a long time lol

[eternal dome]

yeah i could tell

[kindred oasis]

But it looks good, so it's worth it

[eternal dome]

so i need to divide 95.5 by 27

because it is in a different unit

[kindred oasis]

Yes

[eternal dome]

ok

wait but do i still need to multiply it by the 3rd power

or no

[kindred oasis]

No, that's just the unit

[eternal dome]

i can just do 95.5

[kindred oasis]

yard³

[eternal dome]

ok

[kindred oasis]

Or ft³

It just means it's a volume and not an area or the length of something

[eternal dome]

thank you so much.

😘

[kindred oasis]

Yw

[eternal dome]

do you get paid to do this?

bc if not you are like a saint or some shit

[kindred oasis]

No, every helpers it's a volunteer

[eternal dome]

wow

that is crazy

[kindred oasis]

Lol thanks

It's great to help people when you can

Do you need help with the second part of the problem too?

[eternal dome]

idk

let me read it quickly

[kindred oasis]

It's pretty similar, you just need to find another height

[eternal dome]

so basically what it is saying is that the water is filled 3/4 to the pool

and it wants to know what 4/4 is

[kindred oasis]

No, it's filled up to 1/4ft, or 0.25ft, from the top of the pool

It's in feet, it's not a fraction of the height of the pool

[eternal dome]

oh

and the pool is 1.5 ft tall

[kindred oasis]

Yes

[eternal dome]

so it needs 6x more water