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[short parcel]

i dont know derivate this

[short parcel]

in exercise i have to derivate two times in x and two times in t
but i dont konw how i am do it this

r:

[tacit arch]

,tex .FTC2

[ocean seal]

rie.mann

[short parcel]

i dont know how make this in this case

hello?

[nimble fern]

why?

[nimble fern]

hello

[short parcel]

i know how to make but the process i dont know

[nimble fern]

did you mean the integral part or the sin² part?

[short parcel]

the sin² part

[nimble fern]

i see

d(sin²x)/dx = 2sin(x)cos(x)=sin(2x)

[short parcel]

This is the question mark

but I don't understand how my teacher arrived at this answer

[nimble fern]

for example:

Take $y=\frac12\sin²(x+2t)$, then

[ocean seal]

biscuityxd

[nimble fern]

$y_x=\frac122\sin(x+2t)\cos(x+2t)\cdot1$

[ocean seal]

biscuityxd

[nimble fern]

$=\frac12\sin(2(x+2t))$

[ocean seal]

biscuityxd

[nimble fern]

and arrived to ½ sin(2x+4t)

[lone heart]

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[elfin iron]

I have an assignment due for tomorrow, and Im struggling to pick between the 2nd and theid

Could you help me find out

.close

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[alpine sable]

gg

[lilac tendon]

<@&286206848099549185>

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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.

[sour dove]

also we can't help on quizzes or test

[lilac tendon]

oh

[sour dove]

I recommend deleting the question and typing .close with this warning. It's normally a bannable offense but since you're new we'll let it slide

[lilac tendon]

ok

.close

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[sour dove]

thanks! best of luck

[lilac tendon]

could i still ask people to explain a concpet that is related to a quiz question?

[sour dove]

if it's related to a quiz, no. Homework, yes

[lilac tendon]

ok

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[umbral acorn]

can someone confirm two of my answers for some questions?

[mental flame]

Send them so we can see

[mental flame]

Looks good 👍

[umbral acorn]

wb this?

@mental flame

[mental flame]

Looks good, but when finding the slope you put "y/x". Slope is defined as "change in y over change in x" not just y/x

[umbral acorn]

okay

ill change that

thanks

.close

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[mental flame]

Np 👍

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[opaque ocean]

how do i what even

[tardy stag]

oh that's a fun one

do you have access to a graphing calculator and/or desmos?

[opaque ocean]

desmos yeah

[tardy stag]

plot each of the options and see which one makes the most sense for a microphone's "field of view"

[opaque ocean]

how do i make it polar

the round graph

thing

[tardy stag]

uhhhh i remember having trouble finding that sec

[opaque ocean]

oh i was using x instead of theta oops

[opaque ocean]

so for the first question the 3rd option is a bit smaller and rounder and the 4th is a cardiod but a bit bigger

how do i know which is correc

they both point in the same direciton

[tardy stag]

yeah so clearly it's not the first or the second

we're micing a choir here, right? choirs typically are standing on risers like this

and you'd probably put the mic in front of them some distance away, so they all can sing into it

[opaque ocean]

so the cardiod

assuming the buttcheek part is facing the audience

cuz the other side conforms more closely to the shape of the stands

?

[tardy stag]

the buttcheek part
lmao

yes exactly

[opaque ocean]

lesgo

and then for 2nd question

r=3 makes the most sense ig

[tardy stag]

i don't see an obvious choice for that one tbh

[opaque ocean]

wait yeah

edgenuity bad

thanks for the help

.close

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[alpine sable]

I need help in understanding how they got to the right answer cuz I’m confused on this topic

End behavior refers to what happens to the graph as it approaches the bounds of negative infinity and positive infinity

What we're looking at here is a simple graph: y = -x^3

Right!

So take the limits of -x^3 as x goes to -∞ and +∞

Negative by negative equals positive

Therefore

[alpine sable]

Righht cuz 2 negs cancel put to become positive

$\lim_{x \to -\infty} -x^3 = \infty$

[ocean seal]

crimsondevil_rias

[alpine sable]

And

$\lim_{x \to +\infty} -x^3 = -\infty$

[ocean seal]

crimsondevil_rias

[alpine sable]

Oh so this is how they got the answer

I c

Yeah, this involves infinite limits/limits at infinty

Yes

[alpine sable]

I am just new to this concepts

[lament forge]

even if you don't recognise it as being specifically x^3 you can also just look at the graph in this case (which i think is probably what they intended)

[alpine sable]

Like this

[lament forge]

"x approaches negative infinity" means going left

and looking at the graph, as you go left it's going up
so as x goes left towards -infinity, f(x) is going up towards infinity

[alpine sable]

Right

@lament forge but the thing is

X is the middle line right

The thing is x is moving to the left side

So how’s that infinit

[lament forge]

not really
"x" is the horizontal position

so on that vertical line in the middle, x is 0

[alpine sable]

Yes

[lament forge]

to the left of that, x is negative, and as you go further left it becomes a bigger negative number

[alpine sable]

Rigt

Right is positive

[lament forge]

yep

[alpine sable]

Left is negative

[lament forge]

and then the question is asking about f(x), which is the vertical position

[alpine sable]

Right t

F(x) is horizontal

[lament forge]

...i'm not really sure what that means?

x is the horizontal position, and f(x) is the vertical position

[alpine sable]

Right

[alpine sable]

That’s what I said right maybe I might have said it wrongly

@lament forge I might need some help

In like few questions

Want to go into dms

[lament forge]

we can stay here
this channel might close if we stop using it but if it does you can just open another one

[alpine sable]

Okay

@lament forge there u go

Is this as x approaches infinity, f(x) approaches neg infinity

[lament forge]

yes

[alpine sable]

Mk

There you go

That's f(x) = -x^2

That graph approaches -∞ no matter if x goes to -∞ or ∞

So is this right

@alpine sable

[lament forge]

as x approaches negative infinity, f(x) approaches infinity
...how did you get that?

[alpine sable]

Becuse it’s going downwards right

So theoretically it should be neg?

[lament forge]

...yes it is
so why would that mean f(x) is approaching infinity...?

[alpine sable]

Wiat it’s the last

[alpine sable]

It’s the last

[lament forge]

yep

[alpine sable]

I think

The answer is

The first

Correct

[lament forge]

correct

[alpine sable]

Because x^2 approaches ∞ on both ends

Alrt

[alpine sable]

Wait crimson

Is it not ll the last one tho

No

It's not the last one

Graph y = 1/2 x^2

Oh yes

[alpine sable]

I’m just so new to this lesson

That’s why I’m just kinda needing help

You're new to limits at infinity?

[alpine sable]

Kind of

Yes

[alpine sable]

I just started learning this

Like 2 weeks ago

Lmfao

All you have to know about limits at infinity is

Look at the graph and the x-axis

Yes

But some times there isn’t a graph yj

As you move left on the x-axis, you approach -∞

Yes

As u move to the right

It’s positive

+∞ yes

[alpine sable]

So if I’m right on this

It should be positive

Right

Graph 5/4 x^3

Right

X is positive

If x goes to +∞, f(x) should go to +∞ as well

Yep

So everything’s positive then

Wait

What about if x goes to -∞

Negative cubed is negative

Thus, f(x) → -∞ as x → -∞

Yes

[alpine sable]

The cubed is not negative tho

That’s the thing

In this case the cubed is positive

[alpine sable]

Right

Wait

[alpine sable]

So the answe is

[alpine sable]

That’s what I also got

When I used desmos

The graph I showed you is 5/4 x^3

[alpine sable]

Yep

Look at the direction the graph is heading

Rigt

It’s going toward the right side

That's going from -x to +x

But what happens if x approaches ∞?

If x = ∞, f(x) = f(∞) = 5/4 ∞^3 = ∞

Right

[alpine sable]

If x approaches neg

Then it’s positive tho

Right

Look again

Start at x = 0

Oh shoot mb it’s postifie

Then move left on the x-axis

Posit

You can clearly see that f(x) = 5/4 x^3 approaches -∞

Right

Cuz it’s going toward the left side

Ergo

So it’s this

Is this what the answer is

$\lim_{x \to -\infty} \frac{5}{4} x^3 = -\infty$

[ocean seal]

crimsondevil_rias

[alpine sable]

Correct

There you go

Yes

So wait my answer is right then

@alpine sable

It’s that or it’s

I mean C

Yeah

Cause the roots are where the graph crosses the x-axis

Wait it’s C right

It is C cause you're referring to the equation of the graph

Yep it’s a W

Yep

The answe is C

y = (x - 4)(x - 1)(x + 2)(x + 5)

But in actuality

[alpine sable]

?

Yeah, that equation is off

What is C? The highlighted answer?

Yes

No no

@alpine sable this one

the one which is highlighted nis w

Is C

Yes

That's the correct one

Yep

x = -4, x = -1, x = 2, x = 5

That's where the graph crosses the x-axis

But be careful of double or even triple roots

Is it the one which is highlighted

I think I’m getting a hang on this

Just a bit more practice and il be good to go

Yes, that's correct

But I have to warn you about double and triple roots

Sometimes the graph will touch y = 0 at a certain x

If that's the case, that root occurs twice

The question is in photo 1

Answer in photo 2

Correct

Because x + 4 = x - (-4)

And x = -4 is a root of the graph

You're getting the hang of it!

Ok

Is this right

7

Ohhh

This is gonna be tough

You have to look at the roots of each graph

And then work out their equations

Okay

I’ll do that

The first one on the top left has roots at x = -8, x = -3, and x = 2

[alpine sable]

Yew don’t worry about it

I think I got it

So the equation of that one is y = (x+8)(x+3)(x-2)

The same applies for this question

But you see how the graph in this question touches 0 then goes away at 2 points?

That's the "double root" I warned you about

Double roots are roots that occur twice in the equation of the graph

For example

(x-2)^2

Okay

[alpine sable]

Right

Oh yes I remember

This is the graph of (x-2)^2

Or in expanded form, x^2 - 4x + 4

Right

Alrt thx man

I got the practice right

I got 100% man

Thx a lot for ur help crimson

Now I’m ready for the test

I hope I can get a good mark on my test

Good luck man

@alpine sable thx 4 everything man

[alpine sable]

Thx bro

👍

Alrt I ama go have fun now

I’ll do the test late

I’ll let yk what I got for my test

In dms

If you need more tips, you can always DM me whenever I'm free

[alpine sable]

Which is

What time for u now

It's currently 5:15 PM for me

Oh Alrt

And no

Where u from

I do not live in Australia

Oh where do u life

I’m from India

No no

I came for summer

I’m here for summer vacation

Guam

[alpine sable]

I c

Alrt I’ll msg u in 2 hours then

Thx have fun bye

Shoot

.closw

.close

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[alpine sable]

.reppen

[lone heart]

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[civic flame]

I have 2 variable, namely a and b both variable can be negative or positive, it can also be fraction, but it is always real number. I want it so that when both a and b have value of between 0 to 1 the function will give 0 value, if any of the value in a or b is not in between 0 to 1, then the function will give 1 as value

[civic flame]

can someone think a function that can do that?

[tardy stag]

yes it's the function that takes two inputs a and b, both real numbers, and outputs 0 if both a and b have a value between 0 and 1, and outputs 1 otherwise

[civic flame]

you know such or similar function Hayley?

[tardy stag]

what i mean is - you've already defined such a function

[civic flame]

but I don't know how to write them...

[tardy stag]

do it piecewise

if you want notation

[civic flame]

how?

[tardy stag]

$f(a,b) = \begin{cases} 1 & \text{ if } 0 < a < 1 \text{ and } 0 < b < 1 \ 0 & \text{ otherwise } \end{cases}$

[ocean seal]

Hayley

[tardy stag]

oh it's the other way around

still

[civic flame]

maybe something that don't use if? I can't use if here

maybe something with multiplication, mod or something?

[tardy stag]

can you use floor?

[civic flame]

yes

I can

[tardy stag]

testing $\floor{x}$

[ocean seal]

Hayley

[tardy stag]

ok good

$f(a,b) = \floor{|a|} + \floor{|b|}$ correctly outputs 0 for 0 < a,b < 1 and outside that region outputs a positive number

[ocean seal]

Hayley

[tardy stag]

that's floor of absolute value

uhhh if we define g(x) like this:

[civic flame]

one without word please

I need it like uh... normal math formula?

[tardy stag]

$g(x) = \frac{\frac{\floor{2x-\frac12}}{|\floor{2x-\frac12}|} + 1}{2}$ maps 0 to 0 and all positive numbers to 1... i think

[ocean seal]

Hayley

[tardy stag]

this is a very normal math formula

[civic flame]

wait which one is a and b?

[tardy stag]

none, we're going to take the output of f(a,b) and put it into g

[civic flame]

oh!

wait let me try

[tardy stag]

it will be a mess

[civic flame]

this is the output of f(a,b) btw

[tardy stag]

did you include the absolute value

[civic flame]

yes

[tardy stag]

oh oops it outputs 0 for negative numbers too dang it

[civic flame]

I can also use clamp which force any value to be between 0 to 1, like -5 will output 0, and 4 will output 1.
I just can't use if

[tardy stag]

oh clamp seems helpful

if we clamp to (-1, 1)

[civic flame]

can we do it?

[tardy stag]

idk

[civic flame]

hmm...

[tardy stag]

what else can you use?

• • • / *
• sqrt, square
• abs
• clamp
• floor, ceil

oh it's just flip the order of the floor and abs i think

[civic flame]

mostly all other things, just not if

mod, sin cos, log

[tardy stag]

try $f(a,b) = |\floor{a}| + |\floor{b}|$

[ocean seal]

Hayley

[civic flame]

wow amazing

[tardy stag]

that's what i wanted

and then just clamp it to (0,1)

done

[civic flame]

thanks! ❤️

[tardy stag]

that was fun!

[tardy stag]

i'm glad you didn't have to implement this disaster

[civic flame]

lol

[lone heart]

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[mental flame]

When we arrange numbers in that format, the first number would be the lesser of the 2

[sly mantle]

.close

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[jagged plume]

How many solutions does the equation 2x+3(pi)sinx = 0 have? between 0 and 2pi

[lone heart]

@jagged plume Has your question been resolved?

[nimble fern]

this looks so hard
what have you tried
or even
what can you use?

[jagged plume]

i have no idea

ive never seen that before

Normally its like the 2 trigs

[nimble fern]

true

i kinda cheated and use Desmos to draw it out, looks like it has 3 roots in [0,2pi]

but I'm still thinking how to proof that

[lone heart]

@jagged plume Has your question been resolved?

[alpine sable]

$$2x + 3\pi \sin(x)=0$$
To figure out the number of roots, you must consider how many times this function crosses the $x$-axis over the interval $[0,2\pi]$. \
First, let's verify the most obvious solution which is $x=0$.
$$f(x) = 2x + 3\pi \sin(x)$$
$$f(0)=0+0=0$$
So, we already know there must be at least one root over this interval. Now, since we have implied this crosses the $x$-axis once, we need to consider if it turns back to cross it again. So we need to look for a stationary point.
$$f'(x)=0$$
$$f'(x)=2+3\pi\cos(x)$$
$$f'(x)=0=2+3\pi \cos(x)$$
$$\cos(x)=\frac{2}{3\pi}$$
Over the interval $[0, 2\pi]$, the above mentioned equation will have at least two roots. And we know they will definitely not be $0$ or $2\pi$. \
We know $f(2\pi)>0$, f(x) does cross the $x$-axis at least once and has two stationary points over our given interval. So, we can conclude it will have 3 roots.

[ocean seal]

brotherimusttalk1234

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[alpine sable]

Suppose that $X$ and $Y$ have the joint pdf shown below: \
$f(x,y) = \begin{cases}
cxy^2 & 0 \le y \le x \le 1 \
0 & otherwise
\end{cases}$\
a) Find $c$. (solved; c = 15)\
b) Find the following probabilities: \
i. $P(X < 1, Y < 0.5)$ (solved; $P(X < 1, Y < 0.5) = \frac{5}{8}$)\
ii. $P(X + Y < 1)$ \
c) Find the marginal pdfs for $x$ and $y$. (Solved; $f_X(x) = 5x^4, f_Y(y) = \frac{15}{2} (y^2 - y^4)$ \
d) Are $X$ and $Y$ independent? (Solved; No, since $f_{X,Y}(x,y) \ne f_X(x)f(Y_y) \forall x,y)$\
e) Find the expected value of $Y$ ($EY$). \
f) Find the conditional probability function of $Y$ given $X = x$ is defined ($f_{Y|X}(y|x)$).

[ocean seal]

crimsondevil_rias

[alpine sable]

I'm still stuck on part B, ii

This is what the region looks like for P(X + Y < 1)

Had to consult WolframAlpha to confirm, since Desmos cannot handle 2 chained inequalities

[foggy copper]

y can't go above 1/2

[alpine sable]

How is it supposed to look like then?

Consulted libretexts, this is what 0 <= y <= x <= 1 looks like
Honest to goodness truth

[lone heart]

@alpine sable Has your question been resolved?

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[chilly pike]

I need help understanding the second fundamental theorem of calculus

[chilly pike]

,rotate

[wispy ibex]

What are you not understanding ?

[chilly pike]

the whole thing

[wispy ibex]

Do u know what an integral is?

[chilly pike]

yeah ik what integrals and derivatives are

I just don't understand the theorem

[wispy ibex]

This theorem merely states that F'(x) = f (x)

[chilly pike]

I don't get the parts where f(t) is introduced

[wispy ibex]

You integrate f between a and x, this is F(x)

t is just a mute variable

[chilly pike]

ah I see

the integral of f(t) between x and a is F(x)

and the derivative of this results in f(x)

so the main point of the theorem is that the derivative of F(x) results in the original function

I understand now, thanks for the help

.close

[lone heart]

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[wispy ibex]

Yeah exactly and this will be the main way to compute an integral

[lone heart]

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[alpine sable]

I know that sin(2pi) = 0

so we can neglect the fraction inside hte logarithm

but what do I do next

[zealous lichen]

what is this monster

[worn fox]

Just start simplifying

[lone heart]

@alpine sable Has your question been resolved?

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[scenic wing]

so what is x here

[lone heart]

Please don't occupy multiple help channels.

[rancid trail]

Sorry

[scenic wing]

.close

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[keen furnace]

I don’t understand

[keen furnace]

So in integration

Dx means integrating in respect to x

But in integration by part

I can integrate stuff and differentiate in respect to that

While I can also differentiate stuff to integrate it

It’s so confusing

[zealous lichen]

are you confused about integration by parts?

integration by parts is really just the product rule reversed

[keen furnace]

no, I understand the concept

I think this can help me explain

What I don’t get is

Why can I integrate e^-x

And make the integral to be I respect of -e^-x

I understand we need it in this form for integration by part to work

But I want to know why we can do so

[zealous lichen]

ok

do you remember the product rule

when doing differentiation

[keen furnace]

Yes

So it’s just f(x)*g(x)=f’(x)g(x)+g’(x)f(x)

When differentiating

[zealous lichen]

what happens if we integrate both sides

[keen furnace]

It is reversed?

[zealous lichen]

just answer me

[keen furnace]

It becomes

I think my expression is incorrect

[zealous lichen]

yeah

[keen furnace]

Ah

I have to write is as

D/dx f(x)*g(x)

= d/dx f(x)*g(x) + d/dx g(x)*f(x)

[zealous lichen]

you forgot brackets

[keen furnace]

I think I kind of get it

Give me a second

Ah

thanks I got it

I think so

.close

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[zealous lichen]

wow ok

you didn't even need my explanation lol you are so smar

t

[keen furnace]

no lol

I was looking through my books

because

When u gave me the tips

I just kinda referred back to a chapter in my book

[lone heart]

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[alpine sable]

please explain me what is factor theorem

[alpine sable]

please

i am so confused so heres the definition i found, if in any polynomial x=a=0, then z-a is factor of that poly

[gray isle]

poor description / mix up with variables

[alpine sable]

but lets say x-4 be a polynomial and we know x=4=0 then x-4 shiuld be factor which is not

[mortal trellis]

x-4 is a factor of itself

[gray isle]

don't conflate =0 with is a zero

[alpine sable]

how

[gray isle]

x-4 = 1 * (x-4)

[alpine sable]

wut is this

[gray isle]

x-4 is 1 multiple of x-4

[vale wigeon]

x-4 is a factor of x-4 though

anything's a factor of itself

[alpine sable]

yeah but if we put x-4=x in the polynomial x-4 we dont get 0

[gray isle]

if x=a is a zero of a polynomial p(x), p(a) = 0
and (x-a) is a factor

[vale wigeon]

bad notation + you're confusing factors with roots

[alpine sable]

what a factor

then

isnt factor=value whjich we put in pooly to get 0

[vale wigeon]

no, those are roots.

a factor is itself a polynomial by which your polynomial is divisible

[alpine sable]

huh

[gray isle]

the polynomial $\blue{x-4}$ being divided by $\red{x-4}$:
$$\frac{\blue{x-4}}{\red{x-4}} = 1$$

[ocean seal]

ℝamonov

[gray isle]

by factor theorem, since x-4 is a factor of the poylnomial,
x=4 is a zero of that polynomial,
when x=4, the polynomial x-4 will be 0
subbing x=4
gives 4 - 4 which is indeed 0

[alpine sable]

thanks @vale wigeon @gray isle i get it i was confusing factor with roota this all time

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[flat roost]

is 2(a) correct? ive checked multiple times but i dont think the answer can possibly be 0. sorry if im bad at understanding partial differentiation as im new to it, thankss

[flat roost]

^the question reads "Use chain rule to find ∂z/∂r and ∂z/∂θ for the following functions:"

[zealous lichen]

I think it is correct

to verify just find z(r,theta) and find dz/dr that way

[tawny tapir]

yes..it is correct

i just worked for it now

[flat roost]

ohh thanks guys

but how can the rate of change possibly be 0? im really confused sorry

[zealous lichen]

it means z doesn't change when changing r

nothing special

[flat roost]

oh

for total differential of f(x,y) must i write total differential = (∂f/∂x) dx + (∂f/∂y) dy?

or is it (∂f/∂x) ∂x + (∂f/∂y) ∂y?

[zealous lichen]

I think it's dx and dy

not sure though

yeah the wikipedia use dx

[flat roost]

ohh

thanks very much

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[warm bolt]

This is really weird what im going to ask

[near apex]

Weird is fine. Inappropriate is not.

[rancid trail]

Ask it, no problem if it is weird (and also if it is related to math)

[warm bolt]

So im on a grade 6 level right? i was wondering if you could give me some practice questions on Algebra. The stuff im doing is like: 2+X =5. Like that but a bit more complicated

[west girder]

Are you in 6th grade?

[warm bolt]

yes

[ornate ginkgo]

How old

Are you

[rancid trail]

You are comfortable with multiplication and division?

[warm bolt]

yea

[warm bolt]

umm... do i have to answer?

[ornate ginkgo]

Yes

[gray isle]

3x + 7 = 5x - 4

[rancid trail]

5 * x = 20

[west girder]

6+3=1

[warm bolt]

12

[west girder]

bye

[ornate ginkgo]

Sorry

[warm bolt]

4

[rancid trail]

Bro I struggle with equations with variables on both sides 💀 I'm going to attend University next year 💀

[warm bolt]

ya how do i do this

[rancid trail]

Sorry Lil bro you can't use discord yet, so you will probably be banned

[west girder]

Make the x on the same side

[alpine sable]

add/subtract same value on both sides to isolate a term with x

[west girder]

Yes subtract both sides by 3x

[warm bolt]

no im 12 turning 13 in october

[rancid trail]

According to discord tos, you have to be 13 before you use discord

[gray isle]

<@&268886789983436800> tos violation.
we can get into 🔥💧
they'll probs have a reasonable response

[warm bolt]

would the answer be 11?

[gray isle]

no

[warm bolt]

oh

well

[gray isle]

meanwhile give this a read, save it while you still can

,tex .algebra lesson

[warm bolt]

?

[gray isle]

wait for it

[ocean seal]

ℝamonov

[warm bolt]

oh ok

[warm bolt]

OK thank you so much

i have to go now

Bye

[warm bolt]

have a great day

[rancid trail]

Use .close

Just type .close

Before you go

[warm bolt]

ok

.close

[lone heart]

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[pale mason]

I have a quite large question, I encourage anyone who would like to help me solve it to read the whole question before responding, just to avoid clutter

[pale mason]

I have a fun but light differential geometry question for anyone interested in helping me, the first premise goes…
“You're standing on the surface of the Earth. You walk one mile south, one mile west, and one mile north. You end up exactly where you started.
Where are you?”
You are obviously at the North Pole. After further thought, you realize that this works anywhere on the planet. Any square that is formed on a sphere (definition of square being all 90* angles) has 3 sides. This introduces premise 2…
“On any spherical geometry, following the edges of a 3 sided square will always result in the return to the original point in 2 (90*) turns”
Because of this we can reasonably infer that this would work on any spherical geometry, including earth, at any scale. If we define a distance for each of the legs of our triangle as x and then take the limit as 1/x goes to infinity, we see smaller and smaller 3 sided triangles (which are navigable) are produced. With this logic, we can claim that standing anywhere on a sphere and walking x distance, turning 90* walking x distance, then turning 90* and walking x distance, will result in you standing in the same spot as you started. This should work for even the smallest possible distances as we proved earlier.
Yet in practice it doesn’t… this might seem jarring after we mathematically proved it. To anyone who doubts that this doesn’t, I challenge you to get into your car and repeat the process for any given distance, I guarantee you won’t end at your starting point. If you can’t drive currently, then go into an open room in your house and try it with distance x = 2 meters.
The point is it doesn’t work even after it mathematically should. Why is this? As a pure math guy, I immensely struggle to see why this doesn’t work!

Lol

[west girder]

It’s a very good question though

[zealous lichen]

It's literally wrong lmao

[west girder]

Yes but why?

[pale mason]

It’s wrong, but I can’t figure out why

Granted I’m a math major Lmaooo

[worn fox]

I don't think the premise is correct, start at the equator and go north, 2 90 degree turns aren't you going to get you back where you started

[zealous lichen]

it's simple generally if you walk south 1 mile, the west 1 mile, then north 1 mile it doesn't form a triangle/square or whatever you want to call it

[west girder]

You are

[pale mason]

Yes

That’s the point

[worn fox]

I don't know what you're asking then

[pale mason]

I am asking why there are discrepancies between the pure math and the real world

[zealous lichen]

your math is wrong I told you

you proved nothing

end of story

[pale mason]

Where is my math wrong

[tardy stag]

what does it mean to walk east on a sphere

[pale mason]

Turn 90* from when you originally walk south

[worn fox]

You've taken the fact that a specific set of moves from the north Pole gets you back to where you started and jumped to the conclusion that 2 90 degree turns from anywhere and equal length walks will get you back with no justification

[pale mason]

There is no east or west pole

I claimed only that triangles on spheres are 3 sided

Assuming that they are navigable

Where should I go to solve this then lol?

[tardy stag]

imagine starting like 10 feet south of the north pole and walking east

[mortal trellis]

you dont have to be here. go away if you are annoyed by math

[pale mason]

Wtf has this chat become

Discord is ass sometimes lol

[tardy stag]

<@&268886789983436800> someone named Hitler just posted and deleted something? not sure didn't get a look at it

[real gazelle]

Oh boy

[pale mason]

This was it

[tardy stag]

sorry

oh i probably had the reply open already

[mortal trellis]

(already banned in other thread)

[tardy stag]

anyway @pale mason think about what it means to walk east along a line of latitude

there's something strange about "lines of latitude"

[real gazelle]

By the way stop advertising your channel in other people's help channels.

If you're not helping, then don't post in this channel

[pale mason]

Anyway, here’s the og problem

Please read it before bsing lol

I have a fun but light differential geometry question for anyone interested in helping me, the first premise goes…
“You're standing on the surface of the Earth. You walk one mile south, one mile west, and one mile north. You end up exactly where you started.
Where are you?”
You are obviously at the North Pole. After further thought, you realize that this works anywhere on the planet. Any square that is formed on a sphere (definition of square being all 90* angles) has 3 sides. This introduces premise 2…
“On any spherical geometry, following the edges of a 3 sided square will always result in the return to the original point in 2 (90*) turns”
Because of this we can reasonably infer that this would work on any spherical geometry, including earth, at any scale. If we define a distance for each of the legs of our triangle as x and then take the limit as 1/x goes to infinity, we see smaller and smaller 3 sided triangles (which are navigable) are produced. With this logic, we can claim that standing anywhere on a sphere and walking x distance, turning 90* walking x distance, then turning 90* and walking x distance, will result in you standing in the same spot as you started. This should work for even the smallest possible distances as we proved earlier.
Yet in practice it doesn’t… this might seem jarring after we mathematically proved it. To anyone who doubts that this doesn’t, I challenge you to get into your car and repeat the process for any given distance, I guarantee you won’t end at your starting point. If you can’t drive currently, then go into an open room in your house and try it with distance x = 2 meters.
The point is it doesn’t work even after it mathematically should. Why is this? As a pure math guy, I immensely struggle to see why this doesn’t work.

Lmaooo

have a fun but light differential geometry question for anyone interested in helping me, the first premise goes…
“You're standing on the surface of the Earth. You walk one mile south, one mile west, and one mile north. You end up exactly where you started.
Where are you?”
You are obviously at the North Pole. After further thought, you realize that this works anywhere on the planet. Any square that is formed on a sphere (definition of square being all 90* angles) has 3 sides. This introduces premise 2…
“On any spherical geometry, following the edges of a 3 sided square will always result in the return to the original point in 2 (90*) turns”
Because of this we can reasonably infer that this would work on any spherical geometry, including earth, at any scale. If we define a distance for each of the legs of our triangle as x and then take the limit as 1/x goes to infinity, we see smaller and smaller 3 sided triangles (which are navigable) are produced. With this logic, we can claim that standing anywhere on a sphere and walking x distance, turning 90* walking x distance, then turning 90* and walking x distance, will result in you standing in the same spot as you started. This should work for even the smallest possible distances as we proved earlier.
Yet in practice it doesn’t… this might seem jarring after we mathematically proved it. To anyone who doubts that this doesn’t, I challenge you to get into your car and repeat the process for any given distance, I guarantee you won’t end at your starting point. If you can’t drive currently, then go into an open room in your house and try it with distance x = 2 meters.
The point is it doesn’t work even after it mathematically should. Why is this? As a pure math guy, I immensely struggle to see why this doesn’t work.

[tardy stag]

@stray turret shut UP

[pale mason]

Bro plz stop

[real gazelle]

@stray turret I told you to stop posting here

(that was roketto lol, not gonna take the credit)

[pale mason]

That was epic

Anyway, this has been dog shit lol… everyone have a good night!

.close

[lone heart]

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[tardy stag]

think about the question i sent you...

[pale mason]

The original east west question was a rhetorical one to lead the reader into thinking about 3 sided squares

[grand oasis]

You should post this on the Math Stack Exchange rather than here. The question is quite long. And Discord is designed to be more quick.

[pale mason]

True

[lone heart]

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[silver quartz]

how to find volume?

[grand oasis]

What have you tried?

[pale mason]

Alright

[silver quartz]

this my first time doing this so not much

[pale mason]

Think about this 2 dimensionally first

[silver quartz]

ok

[pale mason]

We have a rectangle on top of a triangle

If we solve for the area of the rectangle and triangle then multiply it by 9 we get our volume

Does that help so far?

[silver quartz]

kinda

[pale mason]

Where are u stuck

[silver quartz]

volume

[pale mason]

Area is 2d and volume is 3d

For example, if I have a square inch of something, I can place that on a price of paper

Yet if I have a cubic inch, it takes up physical space

So for the house, we see the rectangle is 8m by 7m by 9m, what is the volume of the box?

[silver quartz]

508

[pale mason]

Very Close

Try that one more time

[silver quartz]

504

[pale mason]

Yep!

Now for the triangle

Find the area of the triangle and multiply it by 9

[lone heart]

@silver quartz Has your question been resolved?

[silver quartz]

can you solve it for reference

[pale mason]

5m by 8m by 9m

Then divide by 2 Bc triangle

[silver quartz]

done

that’s it?

[lone heart]

@silver quartz Has your question been resolved?

[lone heart]

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[ripe comet]

Hello! Ive been doing proofs but have been struggling. This is the problem on a clean pdf, I have already done the Understand the Problem and Make a Plan, however, for the Solve the Problem (Statements and Reasoning) I think my answer is incorrect. Please let me know how to fix it :))

[ripe comet]

This is what I have for my proof so far.

[lone heart]

@ripe comet Has your question been resolved?

[ripe comet]

<@&286206848099549185>

[ripe comet]

<@&286206848099549185>

[ivory fern]

hi

[ripe comet]

hello, do you know how to solve?

[lone heart]

@ripe comet Has your question been resolved?

[ripe comet]

no but I’ll keep trying myself

[lone heart]

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[ivory olive]

What did I do wrong?

I am basing my answer on this example that my teacher did in class

[lone heart]

@ivory olive Has your question been resolved?

[ivory olive]

<@&286206848099549185>

Never mind, I think I fixed it!

.close

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[chilly quiver]

i need help

[chilly quiver]

Can someone explain to me how part 2 simplified to part 3

[outer lark]

common denominator of the numerator and then (a/b)/c = (a/bc)

[chilly quiver]

huh

wait

ioooooh

wait

i dont get it nvm

[outer lark]

ok look at the numerator first

1/(x+h) - 1/x

slap that into one fraction by finding a common denominator

[chilly quiver]

how do you find the common denominator? (x+h) x = x (x+h) right?

[outer lark]

https://www.youtube.com/watch?v=N-Y0Kvcnw8g , I would suggest knowing how to do that before looking at first principles of derivatives

[chilly quiver]

i do

i probably just forgot

but let me watch that video asap

okay i remember

let me try it

okay i got x-(x+h)/x(x+h)

[outer lark]

nice

now that is all over h

$\frac{\frac{x-(x+h)}{x(x+h)}}{h}$

[ocean seal]

clarkie.

[chilly quiver]

yeah

do you have to make it multiplaction?

[outer lark]

what do you mean

[chilly quiver]

by rotating the top fraction?

[outer lark]

no, (a/b)/c = a/(bc)

multiplication is associative and commutative which means if we were to take x = (a/b)/c , then that implies xc = a/b or (xc)b=a, which can then be written as x(bc)=a or just x=a/(bc)

[chilly quiver]

hmm okay

[chilly quiver]

how does this become the 3rd part?

[outer lark]

by simplifying with what I just wrote

your a is x-(x+h) , b would be x(x+h), and c would obviously just be h

[chilly quiver]

okay let me take a look

oh

so what you basically did is you multiplied b with the numerator and denominator?

[outer lark]

sure

[chilly quiver]

okay lets continue to the 4th part

[outer lark]

thats just distribution and simplification

[chilly quiver]

yeah i see it now lmao

let me just do it then if i get stuck ill come back here

okay im good

but i have one last question

[outer lark]

[chilly quiver]

when do i know when to use the first principle of dervatives?

like if i told you to differantiate f(x) = 1/x

you won't use the first principle

[zealous lichen]

it's used for proofs

like proofing the product rules and chain rules

or trying to prove that d/dx x^n = nx^n-1

or prove that d/dx sinx = cosx

[chilly quiver]

ahh okay

thank you very much guys

this was helpful

.close

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[outer lark]

[lone heart]

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Channel closed

Closed due to the original message being deleted

[zealous lichen]

umm don't delete original message

the bot get confused

[dapper lynx]

Sorry bout that

[lone heart]

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[grand falcon]

how can you identify a 30 60 90 triangle without any angle values?

[zealous lichen]

calculate the angles using trigonometry

[plain flame]

sidelengths have a specific ratio

[long quarry]

https://youtu.be/LDQV2vUocyw

[zealous lichen]

?

[mental flame]

Don't clog help channels with yt links

[grand falcon]

ok, heres the question for refrence

im supposed to solve for a

which, according to the 30 60 90 rule, is 4.5 X sqrt3

unfortunately none of the answer choices come up

<@&286206848099549185>

[heady egret]

What is the slope of the ramp?

degree or..?

[grand falcon]

9/25 is the slope

[heady egret]

what is 9/25

degree?

[grand falcon]

no its the slope

[heady egret]

it has 36% slope

then the version without ramp 0% slope

36% slope means 36% slipped down

[grand falcon]

so what would be A then?

[heady egret]

It was 90 degrees, if it slides down 36%, then we have 90-32,4 degrees?

[grand falcon]

oh shoot

so how would we be able to solve for the side lengths of the triangle?

[heady egret]

lol

b = 57,6° and a = 32,4°?

sin(32,4) = 4,5/x

x = hypotenus

wow it is

it got complicated

..

did we make a mistake

[grand falcon]

what does the commas stand for

[heady egret]

I don't know how you use it there but

its like

5,5 = 5.5

[grand falcon]

ohhhh

[heady egret]

do you understand

[grand falcon]

buttttt, the answer choices dont make sense

[heady egret]

yes

[grand falcon]

so are you wrong or is this sheet wrong???

HMMMMM

[heady egret]

dont know

[grand falcon]

oh nvm solved it

since the rise in the slope is 9

and the value of B, aka the rise of the ramp is4.5

multiply by 2

just devide 25 by 2

and you get 12.5

alr

[lone heart]

Channel closed

Closed by @grand falcon

Use .reopen if this was a mistake.

[heady egret]

i dont understand

what did you do

[grand falcon]

.reopen

[lone heart]

✅

[grand falcon]

umm

you seeeee

[heady egret]

no i dont

4,5ft height

[grand falcon]

yup

the sloe

[heady egret]

9/25 slope

[grand falcon]

slope

the slope is merely a ratio

if the true height is 4.5

and the rise factor of the slope is 9

as in 9/25

the slope can also be 4.5/12.5

[heady egret]

yes

[grand falcon]

because 4.5/12.5 as a fraction is equal to 9/25

[heady egret]

but how did you reconcile the slope with the length

[grand falcon]

so A, which is the x value of the slope

is12.5

as you can see in the diagram

the triangle is the slope

[heady egret]

yes

[grand falcon]

the length of b, which we know as 4.5, is the y value of the slope

and the y value of the slope is 9, in 9/25

[heady egret]

what is the y value

[grand falcon]

the y value is the numerator

while the x value is the denominator

that is slope

[heady egret]

How do you know

[grand falcon]

according to the picture i sent

[heady egret]

the length of b is the y value

[grand falcon]

because B is the vertical line of this right triangle

A is the horizontal line

[heady egret]

How would you respond if someone said to you that "the hypotenuse is the slope i think"?

yes

[grand falcon]

thehypotenuse is the slope

wait

[heady egret]

lol

[grand falcon]

hold up

[heady egret]

okay

[grand falcon]

b is the y value

we know that

and a is the x value

so apparently the pythagoremtheorem

[heady egret]

okay but b/a is not hypotenus

[grand falcon]

we know that

however

we know that the slope is 9/25

9 bneing the y value

[heady egret]

okay

[grand falcon]

screw the hypotenuse

we are solving for a, the horizontal line

so since 9 is the y value of the slope

[heady egret]

ok but the slope is 9/25

[grand falcon]

and 25 is the x factor

[heady egret]

not b/a

[grand falcon]

but the triangle itself is acting as the slope

[heady egret]

How, is the bottom base of the triangle slope?

[grand falcon]

the bottom base is the x value of the slope

in this case 25, as in 9/25

[heady egret]

now i draw you any triangle

Can we say that the short right side divided by the long right side is equal to the hypotenuse?

[grand falcon]

alright

but the thing is the hypotenuse does not always equal the slope

[heady egret]

yes

i think the length of the hypotenuse cannot be the slope

[grand falcon]

so it doesent matter

[heady egret]

because we can expand the same shape

and we can get different lengths with the same angles, right?

I think we should focus on the angles, I don't know if I'm right..

[grand falcon]

itsn ot about the angles

its about the side lengths

in this contexxt

the side lengths are all we get

we dont get any angles besides one, and that one is a right angle

we know the vertical side of the right triangle is equal to 4.5 feet

which in the slope is equal to 9, as in 9/25

since the slope is a ratio, we can assume that 9/25 is also equal to 4.5/12.5

which according to the side length of B, the vertical side, is congruent

so since we know that the vertical side length is 4.5

the horizontal side length we are solving for, a, is 12.5

SIDE LENGTH SLOPE
vertical=4.5 Vertical=9
Horizontal=12.5 Horizontal=25

slope, once again, is a ratio

notice how, if you devide the vertical of the slope by 2, you can acheive the side length

consequently, wecan do the same for the horizontal

@heady egret

[heady egret]

ie 12.5 for a and 4,5ft for b

?

[grand falcon]

yup

[heady egret]

what is hypotenus?

[grand falcon]

176.5

wait

ya

pythagorem theorem

[heady egret]

okay but what is 9/25

[grand falcon]

ther ratio

[heady egret]

so I don't understand how the short side divided by the long side gives the slope

[grand falcon]

it can be the opposite

[heady egret]

He didn't tell us anything like that, he just said the slope 9/25, i dont know..

whatever