help-27

[devout snow]

@thin brook Has your question been resolved?

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[devout snow]

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[stone silo]

Hi I have the set ${ ax+b | a,b \in R \ \text{with} \ a \neq 0 }$ that defines all polynomials with degree 1 and the relation ~ where $ax + b \sim cx + d$ if $\frac{a}{c} > 0$. I need to check whether its reflexive, symmetric, antisymmetric and transitive. Is my solution correct? \
(1) Its reflexive because given $ax + b$, since $\frac{a}{a} = 1 > 0$, $ax + b \sim ax + b$. \
(2) Its symmetric because if we have two polynomials $ax + b$ and $cx + d$, if $ax + b \sim cx + d$, there are two cases for $a$ and $c$, the first one being $a, c > 0 => \frac{c}{a} > 0$ and the second one being $a, c < 0 => \frac{c}{a} > 0$ which implies $cx + d \sim ax + b$ \
(3) Not antisymmetric because for example $5x + 8 \sim 6x + 4$ (as $\frac{5}{6} > 0$) and $6x + 4 \sim 5x + 8$ (as $\frac{6}{5} > 0$) even though $5x + 8 \neq 6x + 4$ which means it cant be asymmetric \
(4) It is transitive because if $ax + b \sim cx + d$ and $cx + d \sim ex + f$ then $a$ and $e$ always have the same sign meaning $\frac{a}{c} > 0$

[sullen island]

use $\sim$ for ~

[woven radish]

aPlatypus

[stone silo]

thx

[sullen island]

also if you want a line jump in the latex you need to add two new lines

[woven radish]

p1za

[stone silo]

sorry

[sullen island]

eh its fine we're not born latexers

not sure what you're doing in 2), how does that show symmetry ?

and a small typo in 4), you mean a/e right

@stone silo

[stone silo]

yes

[stone silo]

well we need to show that $\frac{c}{a} > 0$ given that $\frac{a}{c} > 0$. For the latter to be true, $a$ and $c$ need to have the same sign (and they cant be 0 anyways). If they both have the same sign then $\frac{c}{a} > 0$ is also true

[woven radish]

p1za

[stone silo]

and since $\frac{c}{a} > 0$ is true, $cx + d \sim ax + b$

[woven radish]

p1za

[sullen island]

yeah I guess that works too

alright

[devout snow]

@stone silo Has your question been resolved?

[stone silo]

so is everything fine? @sullen island

[sullen island]

yes

[stone silo]

alright thanks

.close

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[devout snow]

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[gusty flume]

Hi.

[gusty flume]

So I need some helps on an ODE. The question is $(x-y)dy=dx$ and I can't seems to find a correct answer.

[woven radish]

alstom.metropolis

[gusty flume]

Here's the original question.

[supple knot]

did you learn how to solve first order linear differential equation

[gusty flume]

Mhm.

[winter radish]

rewrite it to dy/dx = first then use substitution

[gusty flume]

Give me a second to get my scrap paper real quick.

since i already wrote it down before

found it

i'm checking the first part before concluding the ODE, but it doesn't have anything wrong

[winter radish]

think you need substitution to make it seperable

[gusty flume]

u-sub for the integration part is like the same though

since i already consider u=-y already at the tabular part

[devout snow]

@gusty flume Has your question been resolved?

[devout snow]

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[hybrid leaf]

How does 1/theta^2 + 1 become tan^-1 when integrated

I forgot how to do edy💀

[pseudo basin]

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

[woven radish]

ann.in.a.teacup

[hybrid leaf]

Is that just something I’m supposed to memorize?

[pseudo basin]

you can figure this out by differentiating both sides of x = tan(y) and doing some rearrangement shit -- but also this should be on your formula sheet anyway

[hybrid leaf]

Is there no explanation behind it?

[hybrid leaf]

Hmm

[pseudo basin]

there IS an explanation

[hybrid leaf]

Ohh ok ok then

Then I won’t bother highkey don’t wanna know💀🙏

Thanks for the help

.close

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[forest shuttle]

genuinely lost

[frozen aurora]

square both sides

[forest shuttle]

why can't i square every part and get (7-4x)+(3-2x) = 1

that ends up at 6x = 11 and it's not a valid answer

[frozen aurora]

(1+2)^2 = 3^2 = 9
1^2 + 2^2 = 1 + 4 = 5

$(a+b)^2 \neq a^2 + b^2$

[woven radish]

artemetra

[forest shuttle]

what

[iron kindle]

,tex .wrong square

[woven radish]

Bonk

[iron kindle]

$(a+b)^2=a^2+2ab+b^2$

[woven radish]

Bonk

[forest shuttle]

what are you guys doing

[winter radish]

$(a+b)^2 \neq a^2 + b^2$

[woven radish]

loopyloop

[forest shuttle]

why is everybody doing this

what's going on

[winter radish]

square both sides

[frozen aurora]

i showed you an example

you can't individually square it

it will be incorrect

[olive snow]

Cuz you typed that it was your idea

[olive snow]

.

[forest shuttle]

you're telling me to isolate one radical and square both sides that way

i am so tired

[winter radish]

Yes

[forest shuttle]

.close

[devout snow]

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[devout snow]

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[stone helm]

how to calculate dna share with my cousin?

[supple knot]

sir this is a math server

[lunar eagle]

LMFAO

[heavy current]

[lunar eagle]

this needs to be a sticker

pls

im begging

can i ping the mods

or i do i get ban for thta

[heavy current]

do not

[lunar eagle]

😔

[heavy current]

they will not add this as a sticker

[lunar eagle]

too good to be missed out

[heavy current]

I guess… I’ll close this then

.close

[devout snow]

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[stone helm]

....

HELP ME...

[iron kindle]

do you have a real quesiton?

[stone helm]

it is a question

it ends with a question mark

[iron kindle]

if you kiss your cousin you share 100% dna, happy?

[devout snow]

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[runic pumice]

can anyone double check my work?

[shadow brook]

Looks correct to me.

[runic pumice]

thank you!

can someone check this aswell?

[ionic coyote]

looks good!

[runic pumice]

did the calculations for this one my calculator because for some reason it wasnt letting me input fractions as the answer. Hoping thats just an issue with the question lol

[devout snow]

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@runic pumice Has your question been resolved?

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[prisma walrus]

what's the general formula to calculate t for any given number based on the pattern in this table

or at least how do i solve it

[prisma walrus]

<@&286206848099549185>

😭

[spare crypt]

need more context to be sure, but t=log2(T)

[prisma walrus]

alright i was told the answer was simmilar to y = $5 * 1.5^n (this was another equation from the same topic)

[prisma walrus]

I havent done much with log yet 😭

[spare crypt]

oh sure

so exponential stuff looks like y=a*b^n

[prisma walrus]

yeah

[spare crypt]

a is the number at n=0, and b is the multiplier change between n

so here it's a=1 and b=2

[prisma walrus]

so it's t=1*2^n

[prisma walrus]

okay I got it now but where did you get 1 and 2

[prisma walrus]

kk thanks

.close

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[devout snow]

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[still needle]

Am I going insane? If f(5) outputs 3 and f(3) = 5 how is this turning out incorrect?

[desert arch]

Teacher is a dumass mby

[still needle]

😭 😭

[tropic skiff]

Maybe f(5) doesnt exist?

[desert arch]

What grade is this?

[still needle]

its college pre calc im a freshman

i just started so theyre reviewing composition of functions

[desert arch]

Then it exists for sure

Its just 5 nothing more to say

Complete robbery

[restive river]

Try 4 for the meme

[still needle]

didnt work but thanks for the idea

[desert arch]

Real madrid typa robbery

[still needle]

i gotta email the teacher lmao

.close

[devout snow]

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[tropic skiff]

can someone check my work? not just the numbers, but preferably any minor issue that might get points taken off

[devout snow]

@tropic skiff Has your question been resolved?

[tropic skiff]

!helpers

[devout snow]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[tropic skiff]

<@&286206848099549185>

[eternal bobcat]

the first 3 sections look about right

[eternal bobcat]

i dont see anything particularily wrong lol

[tropic skiff]

Thanks!

Wait didnt you say youre taking this in the next unit

[eternal bobcat]

yeah we did this today lmao

[eternal bobcat]

^

[tropic skiff]

-# liar

(either way id prefer to wait for another person to verify)

[eternal bobcat]

makes sense lol

[devout snow]

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[devout snow]

@tropic skiff Has your question been resolved?

[devout snow]

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[midnight scroll]

okay someone PLEASE explain how i can find the height of this triangle

[acoustic leaf]

do you know the triangle area formula?

[midnight scroll]

a=1/2bh

[acoustic leaf]

out of those variables, which ones do you know? which ones don't you know?

[midnight scroll]

i know b = 5.9

[devout snow]

@midnight scroll Has your question been resolved?

[acoustic leaf]

any other variables that you know?

[devout snow]

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[vernal flame]

Could some one please help explain method of joints please?

[devout snow]

@vernal flame Has your question been resolved?

[devout snow]

@vernal flame Has your question been resolved?

[devout snow]

@vernal flame Has your question been resolved?

[vernal flame]

<@&286206848099549185>

[acoustic leaf]

https://engineeringstatics.org/method-of-joints.html

1. find external reaction forces
2. draw an FBD at each joint consisting only of the external forces and the internal forces of the ties connecting to that joint

[devout snow]

@vernal flame Has your question been resolved?

[restive river]

Method of joints

You basically first find global equilibrium, so do your support reactions and such

then, for each 'pin' just where two members meet, you want to find one where you have at least one known force, and then just solve for the other forces.

So, in this situation, you can start with joint A and find
FAB and FAF, then move on to the second joint with this information, id personally pick joint B, but you can also pick joint F. Keep doing this for all joints.

This smells like a Hibbler Statics problem

[devout snow]

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[devout snow]

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[bleak arrow]

how were these relationships derived?

[devout snow]

@bleak arrow Has your question been resolved?

[dry oxide]

is this from physics or vector calculus

looks like differential length elements in different coordinate systems

[bleak arrow]

engineering textbook

[bleak arrow]

yes, how did they derive them

[dry oxide]

i can explain the cartesian/rectangular one but i’m not too sure about the other two

[bleak arrow]

can we derive them through parametrization?

isn't dl = T dt?

= r'(t) dt

or is that for cartesian coordinates only

[dry oxide]

ye that works for cartesian coordinates since dL is just the derivative of the position vector

but for cylindrical and spherical, the basis vector change with position so it's not as simple as just differentiating r(t) like in cartesian

sorry but i barely have any knowledge about spherical and cylindrical

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[dry oxide]

why did you not react to this 😭

[bleak arrow]

it's like this

[dry oxide]

it wasn't like this before lol

[devout snow]

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[bleak arrow]

what do i do

[bleak arrow]

there you go

[dry oxide]

yea 👍

[bleak arrow]

do i post the question again

[dry oxide]

yea sure, i'm sure someone would help you

[bleak arrow]

https://cdn.discordapp.com/attachments/903486480075354182/1339661274425987173/image.png?ex=67af8849&is=67ae36c9&hm=9c7e99ac732e7531aec5d06a6e25cb997eeef94c148eb2b2f74485ff61fd8362&

how are these derived?

[devout snow]

@bleak arrow Has your question been resolved?

[acoustic leaf]

arc length formula $s=r\theta$

[woven radish]

cloud

[devout snow]

@bleak arrow Has your question been resolved?

[spare crypt]

the complete way is called the Jacobian where you write all the coordinate conversions from x,y,z to angle coordinates, put all the derivatives in a matrix and take the determinant. the quick and dirty way is just drawing a tiny bit of space from a small change in radius and angle and approximating the side lengths of the "cube"

er I guess for dL specifically it's just like chain rule, you can convert with stuff like dr = (dr/dx dx + dr/dy dy + dr/dz dz)

[devout snow]

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[devout snow]

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[junior cradle]

The base of a solid is the region bounded by the graphs of y = root x and y = (1/8)x^2.
a. Find volume if the cross sections are semi circles on the x -axis
b. find volume if the cross sections are equilateral triangles on the x-axis

i am super confused on how to get this set up, but once it’s setup i know i can integrate it

[junior cradle]

for part a how do i find the radius

[tardy edge]

vertical distance between the curves

i think

given that it revolves around x axis right

[junior cradle]

i don’t think this is revolving

[tardy edge]

oh

but

since they're both

functions of x

then it's dx

therefore yeah

it is

vertical distance between the curves

would be your radius

mb diameter*

radius would be half of that

[junior cradle]

it’s not necessarily a solid of revolution

[tardy edge]

yeah

[junior cradle]

ik that area = 1/2 pi r ^2

but in my notes i just what what r =, and i have no idea how i got there

[tardy edge]

i still believe the radius is half the distance between each curves

[junior cradle]

so

(rootx - x^2/8)/2

wait i think im doing it

nvm

i got that completely wrong

ok

did algebra wrong that’s all

.close

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[north mountain]

pls help

[north mountain]

hmwk pls, let n be positive integer

prove that there is no solution for 2^{n+3} | 3^n +7 exists

[topaz crag]

You mean that no n exists such as 2^{n+3} | 3^n +7 ?

[north mountain]

yes dat

[topaz crag]

Thanks

[faint gorge]

induction?

[north mountain]

no weak proofs pls

[faint gorge]

what

[north mountain]

do it

[thin inlet]

modulo 3 implies n must be even

[north mountain]

ik

leads no where

tried playin with 9^k + 7 = (1+8)^k + 7 = sum binom(k,i) 8^i + 7 mod 2^{2k}

i go sleep, i wake up pls make proof & ping me, night night

[faint gorge]

maybe in your dreams that happens

[north mountain]

no weak proofs no induction or zesty goofy ahh proofs, wont bother readin

[faint gorge]

xddd ur tripping

[topaz crag]

Induction is weak ?

[north mountain]

ye

[topaz crag]

How high are you today 😄 ?

[north mountain]

use axiom of choice & p-adics

that will get job done night night goofy ahhs

.close

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[devout snow]

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[granite anvil]

is the first line not wrong here?

[signal ginkgo]

what a name in a math server

[granite anvil]

im getting $[sec^2(xy)]x\dv{y}{x}$

[woven radish]

gamer75431

[signal ginkgo]

nope

first line is right

[granite anvil]

how so

[signal ginkgo]

you remember product rule?

[granite anvil]

dont we take the inside argument after taking deriv of tan

[solar goblet]

you need to use product rule to differentiate xy

[granite anvil]

oh

flip

[signal ginkgo]

yea

its chain rule, then product rule

[granite anvil]

okok

thank you

.close

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[cerulean rock]

is anyone familiar with r studio?

[devout snow]

@cerulean rock Has your question been resolved?

[cerulean rock]

are u stupid

cant u see

that no one answered my question

fricking bot

[devout snow]

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[sick lantern]

Can anyone explain to me a clear definition of a gradient vector in MVC

[devout snow]

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[devout snow]

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[steel plank]

i know this is a question of physics but can anyone help me in this ?

[steel plank]

<@&286206848099549185>

[wispy oyster]

lenz' law

when the north pole comes closer to the ring

magnetic flux is increasing but the ring does not like that so it opposes that by generating its own magnetic flux in the opposite direction (causing an anticlockwise current)

try figuring out when the magnet is moving away from the ring

[devout snow]

@steel plank Has your question been resolved?

[steel plank]

then it should generate current in clockwise direction

[wispy oyster]

indeed

[steel plank]

but when magnet is passing through the coil after sometime the coil will interact with south pole and clockwise current will be generated so answer should be D

but ans is C

@wispy oyster help

[wispy oyster]

ah I see, the question is only asking what happens when the magnet is brought near to the ring

it is not asking what happens when the magnet is pushed away from the ring

[steel plank]

damn i noticed one thing

the loop is attached to the wall

[devout snow]

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[vernal ferry]

We draw five slips from a hat with each slip labelled 1-100. Each time we draw a slip, we can label it "largest," "second largest," "third largest," "fourth largest," or "fifth largest." Our goal is to maximize the chance that these five numbers we pulled are in exact order as labelled. Describe a strategy that we could do and provide estimates of the probability that that strategy would be successful.

[pseudo basin]

as in, we draw these slips one at a time and we have to give each one a label before we draw the next?

[vernal ferry]

yes i think so

[pseudo basin]

uhh

wait did you translate the problem or was it given to you in English

[vernal ferry]

it was given to me in english

[pseudo basin]

ok

well there is kind of a stupid strategy that seems to just be intuitively the best possible

for the first slip, assign it one of the 5 labels according to which of the intervals 1-20, 21-40, ..., 81-100 it falls into

[vernal ferry]

thats kind of what i was thinking too

but how would i estimate the probability this wwould be successful

is where i got stuck

[graceful cosmos]

A very rough estimate is just finding the probability of drawing 5 labels 1,2,3,4,5 in the right order

[vernal ferry]

that would just be 1/120 no?

[graceful cosmos]

Exactly. I imagine that's a low estimate

[vernal ferry]

yeah

ig thats the most simple estimate i cud give

[graceful cosmos]

I'm sure better can be done, I'll think on it

[vernal ferry]

haha yes thank u very much

[graceful cosmos]

Or no, wait, I didn't get my idea right. Instead, let's say we can draw from 1,2,3,4,5 with replacement. Given we do 5 draws, what's the probability we get one of each?

[vernal ferry]

one of each as in like in not repeating

?

[graceful cosmos]

No need for getting them in order, just one of each

Yeah, no repeats

[vernal ferry]

thats like 90 percent

100times99tims98times97times96/100^5right?

[graceful cosmos]

I mean only from the labels 1,2,3,4,5

[vernal ferry]

ohhh

[graceful cosmos]

That one is 120/5⁵

I imagine that's like drawing from the intervals 1-20, 21-40, etc

[vernal ferry]

yeah so like 3 percent

OH

i see waht u mean

[graceful cosmos]

You have to draw from one of each interval

[vernal ferry]

what u mean

OHHH

[graceful cosmos]

I'm ignoring cases like drawing 19 and 20, so this is still a rough estimate

But yeah, I think it's a bit higher than 3%

[vernal ferry]

that makes a lot of sense

okok yeah but it gives a strong rough estimate

gotcha

thank u very much

[graceful cosmos]

Again I wonder if better can be done

[vernal ferry]

mhm

good point

i think

honestly ive thought abt this for almost an hr now so that much is good progress lol

[devout snow]

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[junior whale]

Hello, I'm trying to prove the following question:

You are given the two equations f, g : (0, π/2) -> R for which the following statements hold:
|g(x) - g(y)| <= (x - y)^2 for any x, y in (0, π/2]
g(x) = f(x) * sinx
f(π/4) = sqrt(2)

Prove that:
the equation of g(x) is constant for x in (0, π/2]

I've tried a lot of things but none of them seem to work. I need a push in the right direction, thank you for your time.

[devout snow]

@junior whale Has your question been resolved?

[junior whale]

oh wait

!show

[devout snow]

Show your work, and if possible, explain where you are stuck.

[junior whale]

ill do that rn

ill show something that i tried

i didnt arrive at a contradiction, so i couldnt prove that g(x) is in fact a constant function

and not a function where g(x) can be greater than or less than g(y), for some x in (0, π/2] and some y in (0, π/2]

[devout snow]

@junior whale Has your question been resolved?

[north mountain]

prove that g'(x) = 0

for any x in (0, pi/2)

$|\frac{g(y)-g(x)}{y-x}| \leq |y-x|$

[woven radish]

Goëtia

[north mountain]

then : $\lim_{y \rightarrow x} g'(y) = 0$

[woven radish]

Goëtia

[north mountain]

u can then deduce than g'(x) = 0 , thus g(x) = c => g(x) = f(x) sin(x), plug x = pi/4 , u got what u need @junior whale

[junior whale]

thank you

it does indeed come out that g'(x) = 0

ill close this then

tysm~

how do iclose

!close

.close

[devout snow]

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[junior whale]

.resolved

.resolve

why hasnt it gone back?

there arent many free channels left, i wouldnt wanna inconvenience anybody

[final dune]

.reopen?

[junior whale]

[north mountain]

not it will be free'd soon @junior whale

[junior whale]

it's resolved so...

oh okay

[devout snow]

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[severe prairie]

how do i do this? a is part 1

ik i forgot the rotation above origin at angle theta for a but idc ab that

also ignore the bottom issa diff q

<@&286206848099549185>

[silver crane]

please ping after 15 min

rotational transformation?

[devout snow]

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[silver crane]

and linear transformation?

not sure tho

[devout snow]

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[kindred mauve]

needs marking checking for corrections

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@kindred mauve Has your question been resolved?

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[hazy cradle]

whats the ques?

[pseudo basin]

to solve this system of inequalities, i would assume

OP is gone tho which is sad

[hazy cradle]

OP?

[pseudo basin]

original poster

[long drift]

Could someone solve this plesse?

[pseudo basin]

!noans

[devout snow]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[pseudo basin]

we won't solve this for you, but we can see what you've tried and give you pointers for how to proceed

if you are really completely absolutely stuck without even the smallest idea that's still ok (but tell us so)

[devout snow]

@long drift Has your question been resolved?

[pseudo basin]

wow ok hold on

ok i can see you are on the right track (and also i have made a mistake in my own solution) but this can probably be written a little better

so alright our goal here is obviously to solve each one of the inequalities to eventually take the intersection of their solution sets

[long drift]

Yes!

[pseudo basin]

for inequality II, you've got it broken down into sin(x)(2cos(x) + 1) ≥ 0

so what i would do here actually is plot its solution set on one circle rather than breaking it down into two

easier to show what i mean so let me sketch a pic

[long drift]

Alright.

Thank you very much!

[pseudo basin]

something like this

this is kind of an extension of the wavy curve method to a circle

the way i'm using it, inside the circle means negative and outside means positive

& i don't care much for it being to-scale radially, but all of the crossing points are explicitly marked here in green

0 and pi from the sin(x), 2pi/3 and 4pi/3 from the 2cos(x) + 1

so then i see for inequality I you decided to break it into two cases based on when the denominator was positive vs negative

wait actually

do you understand what im suggesting for II

i wanna make sure we're on the same page here before i move on to I

[long drift]

Yes yes!

Thanks!

The solution given is impossibile but jm not understanfing why!

[pseudo basin]

impossible huh

well let's just keep going with inequality I shall we?

so what i would do here is more or less follow you and first rewrite it as $\frac{1 - \sin(x) - \sqrt{3} \cos(x)}{\cos(x)} > 0$

[woven radish]

ann.in.a.teacup

[pseudo basin]

then also multiply both sides by $-1$ just to have a bit less minus signs to deal with, so $$\frac{\sin(x) + \sqrt{3}\cos(x) - 1}{\cos(x)} < 0$$

[woven radish]

ann.in.a.teacup

[pseudo basin]

and then also do the same thing you did with the R-formula (or whatever else you wanna call it) and reduce to $$\frac{2 \sin(x + \pi/3) - 1}{\cos(x)} < 0$$

[woven radish]

ann.in.a.teacup

[pseudo basin]

so far so good?

@long drift

[long drift]

Right!

[pseudo basin]

so when it comes to doing the circular wavy curve method, the denominator changes sign at 3pi/2 and pi/2 while the numerator changes sign at pi/2 and 11pi/6

which i believe you also worked out in your work

sin(x + pi/3) = 1/2 etc.

[long drift]

Yes!

[pseudo basin]

the upshot is that at pi/2, numerator and denominator both change sign, so the sign of the fraction does not change there

[long drift]

Okk

[pseudo basin]

so we get this kinda thing

i've got colored pens on me so i am using two different colors on one circle

[long drift]

Alright!

Thanks

[pseudo basin]

and i've also put hatch marks of the relevant color in the regions where each function is the right sign

if you dont have colors what you could do is draw a number line from 0 to 2pi that's like your circle unrolled into a straight line

and plot the intervals for each inequality that way

and then take their intersection

[long drift]

Yes yes i did that

[pseudo basin]

so anyway you can see there is no overlap here

that's why the solution set is empty

@hazy cradle what's with the skull react?

[long drift]

Oh right!

So what mistakes did i do?

Maybe with sin2x...

[pseudo basin]

no, II is perfectly fine as far as i can tell.

[hazy cradle]

nvr thought shit wud be that complex...

[pseudo basin]

i think your biggest mistake was trying to do this all symbolically rather than using unit circle diagrams and keeping yourself in check visually

with rational inequalities you can sorta get away with this, but with trigonometric ones it is much much harder

[long drift]

Alright!! Thankss!! Ill check maybe better thank you very much!

Yes this one is tricky!

[pseudo basin]

it's trigonometric inequalities, stuff is complicated with these

and this system in particular imo is quite beefy

[long drift]

Yess thank yiu very much!

[hazy cradle]

like i nvr solved it with such type of a graph

[pseudo basin]

do you know the wavy curve method

like for non-trigonometric ineqs

this thing i'm doing here is an extension of that to trig

which i think is pretty elegant

[hazy cradle]

damn

[pseudo basin]

its the same idea but just wrapped around a circle

anyway yeah ok

[hazy cradle]

sheesh, gotta look it up

might be useful for my studies

[pseudo basin]

@long drift do you have anything else to ask for now? or are we good

you can .close the channel if you're done

[long drift]

I ve also found this

[pseudo basin]

it looks the same in terms of the overall idea but there is a lot more table-bashing and unintuitive visuals for my taste

[long drift]

Altight

Ill keep it open for a second just to check something

[devout snow]

@long drift Has your question been resolved?

[devout snow]

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[pure mesa]

We are given a function ft(x)=tsin(tx) and are told that two neighboring zero points have tangents which touch at an angle of 28 degrees for a certain value of t

[pure mesa]

this is what I did, but I do not think that it is correct

can someone tell me why I am wrong and how to fix it?

(my approach btw is that if the angle of them together has to be 28 degrees and they have the same slope then the angle with the x axis has to be half of that at 14 degrees)

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[devout snow]

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[devout snow]

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[pure mesa]

We are given a function ft(x)=tsin(tx) and are told that two neighboring zero points have tangents which touch at an angle of 28 degrees for a certain value of t

[pure mesa]

this is what I did, but I do not think that it is correct
can someone tell me why I am wrong and how to fix it?
(my approach btw is that if the angle of them together has to be 28 degrees and they have the same slope then the angle with the x axis has to be half of that at 14 degrees)

<@&286206848099549185> #help-27 message

[devout snow]

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@pure mesa Has your question been resolved?

[supple knot]

the angle (1/2) * alpha being 14 degrees gives you the slope. so find the equation of the tangent line and solve for the slope by finding the angle beta.

[devout snow]

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[granite anvil]

is the inverse wrong here?

[granite anvil]

my teacher wrote y = loga(x)

[supple knot]

where exactly there's like 10 things written on that screenshot

[granite anvil]

she said "we know that the invrse of y = a^x is y =loga(x)

[supple knot]

this is correct

[granite anvil]

but then right beside that she wrote what i think is right?

yeah

shouldn't it be loga(y) = x

[supple knot]

this is the work that proves ^

[granite anvil]

how

[granite anvil]

isn't that not what she wrote here

i feellike this one is right

[supple knot]

x and y are just dummy variables

the inverse of y = 2x is y = (1/2)x

you can just think of it as convention to write the inverse function of $y=f(x)$ as $f^{(-1)}(x)$ as a function of the same variable $x$

[woven radish]

riemann

[granite anvil]

im still confused kinda

how can they both be right when they are saying different things?

[supple knot]

uh huh

they're not saying different things

[granite anvil]

isn't y = loga(x) the same as a^y = x

[supple knot]

again, the inverse function is, by convention, written in terms of the same variables y = f^(-1)(x)

do you agree that $f(x) = 2x$ and $f^{-1}(x) = (1/2)x$ are inverses of each other?

[woven radish]

riemann

[granite anvil]

y = 2x
x = 2y
y = x/2
so ye

[supple knot]

great. same logic applies to $y = a^x$ and $y = \log_a(x)$

[woven radish]

riemann

[granite anvil]

x = a^y

loga(x) = y

[supple knot]

it's right there

[granite anvil]

yeah

that one makes sense

i understnad ur saying its js convention for the other

but i dont get what you mean rly cuz dont variables have to serve a prupose

like how can x and y be interchangeable

when she writes y = loga(x)

[supple knot]

did you agree with this or not?

[granite anvil]

yes

[supple knot]

"ye" sounds like yes

[granite anvil]

i do

[supple knot]

you swapped x and y there

[granite anvil]

yes

oh

oh

okok

[supple knot]

so why are you asking this

[granite anvil]

so then this isn't the inverse?

[supple knot]

wut

what

can you use complete sentences

[granite anvil]

how can log_a(y) = x and y = loga(x) both be inverses of y = a^x

i understand how log_a(y) = x is the invrese

but if that one is the inverse of y = a^x, then how can the other also

[supple knot]

you wrote this right?

[granite anvil]

yes

[supple knot]

x = 2y

is like log_a(y) = x

[granite anvil]

yeah

what about the other one

thats js writing it in logarithmic form isn't it?

[supple knot]

what

[granite anvil]

from which she js swaps the x and y

[granite anvil]

in the screenshot of log_a(y) = x, is this just the expression of y=a^x as a logarithm while the line log_a(x) = y is the inverse of y=a^x

[supple knot]

$y = a^x$ implies $\log_a(y) = \log_a(a^x) = x$

[woven radish]

riemann

[supple knot]

no inverses happening ^

$y = \log_a(x)$ is telling you that the inverse of $y = f(x) =a^x$

is $y = f^{-1}(x) = \log_a(x)$

[woven radish]

riemann

riemann

[granite anvil]

yeah ok that makes sense

thank you

.close

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[devout snow]

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[runic loom]

Hey I'm working on solving DEs by substitution and i've been given a bernoulli equation. I'll show the steps i've taken so far up to where i've gotten stuck

\begin{align}
\dv{y}{x}&=y(xy^4-1)\\
\dv{y}{x}&=xy^5-y\\
\dv{y}{x}+y&=xy^5\\
n&=5\\
u&=y^{1-n}=y^{1-5}=y^{-4}\\
y&=u^{-\frac{1}{4}}\\
\dv{y}{x}&=-\frac{1}{4}u^{-\frac{5}{4}}\dv{u}{x}\\
-\frac{1}{4}u^{-\frac{5}{4}}\dv{u}{x}+u^{-\frac{1}{4}}&=xu^{-\frac{5}{4}}\\
-\frac{1}{4}\dv{u}{x}+u&=x\\
-\frac{1}{4}\dv{u}{x}&=x-u\\
\dv{u}{x}&=4(u-x)\\
\end{align}

[woven radish]

lILi

[runic loom]

steps 1-3 are manipulating it into proper form
steps 4-7 are setting up the substitution
step 8 is applying the substitution
steps 9-12 are manipulation

and by proper form, I mean
$$\dv{y}{x}+P(x)y=Q(x)y^n$$

[woven radish]

lILi

[runic loom]

anyways, i'm not sure where to go after step 12

[thin inlet]

linear ode

your goal with bernoulli odes is typically turning it from bernoulli ode into a linear ode

[thin inlet]

which is what you've done here

then you'll have u(x), and you can replace u with y

[runic loom]

So, if i've understood correctly, I do this and then work things out?

\begin{align}
u&=y^{-4}\\
\dv{u}{x}&=-4y^{-5}\dv{y}{x}\\
\dv{u}{x}=4(u-x)&\Rightarrow-4y^{-5}\dv{y}{x}=4(y^{-4}-x)
\end{align}

[woven radish]

lILi

[thin inlet]

errr

my point was more solve the linear ode

this was the whole point of getting it down to a linear ode

so solve the linear ode in terms of u, i.e. get u(x)

then using the fact u=y^-4, you get u(x)=(y(x))^(-4)

https://tutorial.math.lamar.edu/classes/de/bernoulli.aspx

[runic loom]

ah, my bad

[runic loom]

I realise how I got confused about your explanation. Anyways, the issue i'm having is just, algebraic manipulation, trying to solve for u

[thin inlet]

its a linear ode

https://tutorial.math.lamar.edu/classes/de/Linear.aspx

[runic loom]

solved it

.close

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[devout yarrow]

quick questiion, why is integral calc chaging the 2-3x? I'm guessing its the absolute value but im not sure why

[distant helm]

you're correct

its the absolute value

|A| = |-A|

[devout yarrow]

so abs value of 2-3x would also be correct?

[distant helm]

yea

[devout yarrow]

ok thanks

.close

[devout snow]

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[devout snow]

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[remote elbow]

how to do 49

[solar goblet]

it is helpful to first do a similar problem, but in 2d instead

[pseudo basin]

well, that or you can just bash it.

[winter patrol]

did you do 48?

[pseudo basin]

48 is harder, no?

[remote elbow]

no

[scarlet jacinth]

if annyone can help me with this pls do

[winter patrol]

the question involves distance between points
so consider applying distance formula

[winter patrol]

please claim your own channel as this is occupied #❓how-to-get-help

[devout snow]

@remote elbow Has your question been resolved?

[remote elbow]

can u write it out bro

how to apply it

to find the set

[winter patrol]

have you applied the distance formula before?

[devout snow]

@remote elbow Has your question been resolved?

[remote elbow]

yes

im not at home now

i can’t do it rn

[devout snow]

@remote elbow Has your question been resolved?

[devout snow]

@remote elbow Has your question been resolved?

[silk phoenix]

I think you need to find the plane equation between 2 points

[remote elbow]

yea

so i got midpoint

now how to get plane

[silk phoenix]

You just need vector of those 2 points though

And midpoint

[devout snow]

@remote elbow Has your question been resolved?

[devout snow]

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[broken wagon]

I'll send my work in a second

[broken wagon]

Just ping me when possible to let me know what I did wrong, I appreciate it

[low sentinel]

looks like you typed it wrong

🙂

[broken wagon]

Hi

Oh I did, that's not good but thanks for letting me know

[low sentinel]

you mean still not the right answer?

[broken wagon]

No as in it's not good I typed it in wrong

[low sentinel]

ah ok

[broken wagon]

Unfortunatly

[low sentinel]

try 54.7164

[broken wagon]

I must be tired, thanks man

It worked

[low sentinel]

k

!done

[devout snow]

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[broken wagon]

.close

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[radiant ore]

for systems of equations how do you choose whether to eliminate y or x

[junior flax]

Personal preference

Or

[wind mason]

Depends on the system as well.

[junior flax]

Guass!

[wind mason]

Let's say you have a system like:
100x+y=0
-99x+y=3
You're obviously eliminating the y here.

[junior flax]

For the real ones (linear fellows)

[wind mason]

Yeah, both of these involve prime numbers. Just multiply out the ones that seem smaller.

[wind mason]

[wind mason]

I would personally eliminate the y first

[radiant ore]

just relised i did the second part wrong lol

[crude wasp]

tbh i think for like school it's probably easier just to do lots of practice rather than trying to optimise eliminating x or y, i don't think it makes too much of a difference here

[junior flax]

Go the unkindly path, eliminate only in the most algebraically confusing and computation intensive ways! (For the real ones)

[radiant ore]

i know i thought it made a difference

it doesnt

.close

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[crude wasp]

obviously in some examples like ^ you probably wanna eliminate y instead

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[pallid stream]

there are 15 points as in the picture. what's the probability of 3 randomly chosen points NOT being colinear?

[pallid stream]

now here's the thing

I got 1223/1365 while the correct answer is 1221/1365

so that means I'm off by 2, which means I can already be proud of myself lmaoooo

idk if I did it right (well obviously not since I'm mistaken) but I started by writing down all the ways you could choose the points that are colinear (so that I can subtract later in order to get the ones that aren't colinear)

and idk if there's a faster (and more importantly, correct) way of doing it

[restive river]

yeah the technique is good

probably you just missed some

possibilities

[pallid stream]

yeah

oh wait wait wait I got one more way

but that still leaves me with 1 way I don't have

well no

[restive river]

lol

[pallid stream]

okay so now I have 1222.5/1365

I need 1.5 way less

[restive river]

how do you get 0.5 in odds

[pallid stream]

well because I divided it by 2

in reality there are 2445

[restive river]

oh right

my bad

[west summit]

there are 2 with slope 1/2

[pallid stream]

yeah I got these

the extra one I just found is vertical

actually I confused myself bc of the fractions lol I originally needed 4 more ways and now I need 3

[west summit]

you got the horizontal ones I suppose?

[pallid stream]

I can show my solution

1. horizontal lowest layer
   5*4*3
2. horizontal 4th layer
   4*3*2
3. horizontal 3rd layer
   3*2*1
4. diagonal with 5
   2*5*4*3
5. diagonal with 4
   2*4*3*2
6. diagonal with 3
   2*3*2*1
7. slope 1/2
   2
8. vertical
   1

and all the ways are 15*14*13

[pallid stream]

and calling it "with slope 1/2" is very smart lol in my notebook I called it diagonal double jump lmfao

[west summit]

why are you multiplying diagonal with 3 by 4?

[pallid stream]

bc there are 4 ways you can get them

[west summit]

thank you lol

[west summit]

I only see 2

[pallid stream]

I'll draw it

[west summit]

hrizontal and verical

[pallid stream]

oh wait

you're right bruh

I also included the ones that are already in diagonal with 4 bruh

[west summit]

ohh, yeah

so what is the actual sum?

[pallid stream]

I'll fix it

[west summit]

do you double count the vertical?

[pallid stream]

uhhh bad news

now there's a lot more than it should be

[west summit]

...

[pallid stream]

1228.5/1365

I was closer when I started lmfao

[pallid stream]

no, why?

[west summit]

I guessed that you divided the whole thing by 2 as to not double count right-left and left-right.

and you cant do that with vertical/horizontal

[pallid stream]

I'm confused

well when you pointed out the diagonal with 3, I just swapped the 4 at the beginning for a 2

and that's it

[west summit]

my bad. Thought you counted in a different way

[pallid stream]

and 2*3*2*1 is indeed 12 ways that I "lost"

[west summit]

wait... 3 diagonal should be 1 if we don't care about order...

[pallid stream]

2457 instead of 2445

[west summit]

do we care about order?

[pallid stream]

this is 3 diagonal

there are 2 ways

[west summit]

yeah... here is 4x3x2x1?

[pallid stream]

well if I scrap these too then there's gonna be a whole lot more I missed lmao

[pallid stream]

but I changed it for 2*3*2*1 like I told you

I'll edit the message maybe

[west summit]

yeah, I just looked back, my bad

[pallid stream]

no worries

[west summit]

but 3x2x1 cares about the ORDER in which we choose the points

and that's goes for all the other ones aswell. Is that what the problem stated?

[pallid stream]

it doesn't really state anything

just "3 randomly chosen points. what's the probability of them not being colinear"

[west summit]

It doesn't matter(I think). You just have to d the same for the calculation of all points. And you do that

[pallid stream]

but for example if you chose 2 ones from the left diagonal "line" and 1 from the top of the right diagonal "line" then they wouldn't be part of the same line, so I'm choosing to count them separately as 2*3*2*1

idk if that's about order

well for some reason my solution is wrong and I'm now missing 12 ways

[rustic jetty]

you should get a fraction over 455, not 2730

because order doesnt matter

[pallid stream]

huh?

[rustic jetty]

the number of ways to choose three points out of the fifteen is 15c3

not 15p3

[pallid stream]

what's c and p

[west summit]

that's what I thought. But both should cancel since we're talking about probability

[rustic jetty]

ah okay sure

so just make sure everything is ordered then

[pallid stream]

also 455 is basically 1365 in the fraction

bc 1365 is like the expanded fraction

the correct answer is 407/455, but since I got 1223/1365, I multiplied it by 3 to figure out how wrong I was lmao

and 407/455 is 1221/1365

[pallid stream]

I still don't know what these stand for

[west summit]

I don't know n but c is for choose

[rustic jetty]

those shouldnt matter for the premise of this question

[pallid stream]

aight

[west summit]

where does 2730 come from?

[rustic jetty]

15x14x13

you ordered everything except the lines with slope 1/2 and the vertical line

so slope 1/2 should be 2 * 3 * 2 * 1

and vertical should be 3 * 2 * 1

which gives the the given answer

[west summit]

ohh, ok

So, we have:

horizontal lowest layer
5x4x3
horizontal 4th layer
4x3x2
horizontal 3rd layer
3x2x1
diagonal with 5
2x5x4x3
diagonal with 4
2x4x3x2
diagonal with 3
2x3x2x1
slope 1/2
2x3x2x1
vertical
3x2x1
and all the ways are 15x14x13

@rustic jetty

[pallid stream]

it isn't tho

[rustic jetty]

read after

[pallid stream]

OH BRUH I'm a total dumbass

idk why I didn't count them like this even tho I did it with everything else bruh

thank you so much

[west summit]

.close :)

[rustic jetty]

np lol

[pallid stream]

.close

[devout snow]

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[west summit]

It was fun while it lasted @pallid stream @rustic jetty

[devout snow]

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[granite anvil]

what is the question asking?

idk what f^6(0) is referring to

the 6th deriv when it is equal to 0?

[coral heron]

yes

[wind mason]

Yep

[granite anvil]

damn

ok

thanks

.close

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[uneven talon]

This is my explanation of why its true that the columns of an n x n matrix A are linearly independent when A is invertible. Is this a good explanation?

[solar goblet]

what's the clown emoji for

[uneven talon]

Mey doesn't like my explanation at all

[kind tiger]

i would say theres a few problems, firstly what is a pivot ? is this just some term i havent come across and has a defined meaning or did you make it up

if you made it up then you need to define what it is

[uneven talon]

its a term for the class

i thought pivot was a generally well-known term

[stone stump]

it is

[uneven talon]

https://en.wikipedia.org/wiki/Pivot_element

[kind tiger]

ah nws then

i would also like to see some more reasoning as to why the rows being linearly independant means the columns must also be

can be as simple as just saying det(A) = det(A^T) tbh, but i think it needs to be in there

[uneven talon]

i never mentioned the rows being linearly independent though

[stone stump]

they said nothing about the rows being lin indep

and its not needed

[kind tiger]

oh wow yeah i cant really

[uneven talon]

the identity matrix pretty obviously has pivots in every column imo

[kind tiger]

really dont know how i read that so wrong ahah

[uneven talon]

its ok

[stone stump]

yes the explanation is fine

[uneven talon]

even the last statement?

[kind tiger]

yeah sorry again for misreading so badly LOL

[uneven talon]

im worried that i made a jump from rref(A) invertible -> A invertible

[stone stump]

well presumably you are allowed to use that row reducing doesnt change linear (in)dependence between columns

[uneven talon]

Not sure i was explicitly told that in class

Not that I don't trust it's true

More that I think its up to me to justify that

[stone stump]

the whole point of row reducing is basically that fact

[uneven talon]

Yeah and it makes sense given the idea of elementary matrices

[stone stump]

or phrase it as solving the linear system

same solutions <=> same dependencies between columns

[uneven talon]

Oh right

I can easily prove that Ax = 0 implies x = 0

Which would mean the columns of A are linearly independent

[stone stump]

yes

[uneven talon]

Thank you! I appreciate it

[stone stump]

yw

[devout snow]

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[viscid wagon]

,rotate

[woven radish]

Couldn't find an attached image in the last 10 messages.

Feb 16, 2025, 2:31 AM

[viscid wagon]

,rotate

Question 2

,rotate

[viscid wagon]

I'll probably only need help with part 1

[distant helm]

Hint: $\frac{x^2}{9}+\frac{16}{x^2}=\left( \frac{x}{3} \right)^2+\left( \frac{4}{x} \right)^2=\left( \frac{x}{3} -\frac{4}{x}\right)^2+\frac{8}{3}$

gotta admit though, that one may be pretty hard to figure out

@viscid wagon

[woven radish]

TargetVN

[distant helm]

sry, it's this instead

[devout snow]

@viscid wagon Has your question been resolved?

[viscid wagon]

So the minimum value of A is the minimum value of x/3 - 4/x +8x/(x2+12)?

[distant helm]

you can do that, or just do the very obvious substitution u = x/3 - 4/x

do note that with this substitution, 1 value of u will output 2 values of x

[viscid wagon]

Oh bruh I'm stupid

[west summit]

but we can't gaurantee that those are real, can we...

nor can we gaurantee that they are double roots

[distant helm]

examine u(x) = x/3 - 4/x through u'(x)

[viscid wagon]

Is it just A = 16/3

Wait ni

2*sqrt(8/3)

[west summit]

ohh, smart

[distant helm]

,w 3A^2-32=0

[woven radish]

Results provided by WolframAlpha

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Requested by return.h3x.is.targetvn

[distant helm]

seems like it

[viscid wagon]

Huh how'd you get that quadratic

[distant helm]

u^2 + 8/3 = Au

which is just a quadratic

[viscid wagon]

Oh

[distant helm]

since 1 u gives 2 x, the quadratic must have 1 root only

[viscid wagon]

For part ii is it just checking if the AM-GM inequality holds

[distant helm]

i mean since we have found what is A, we can just find u

then find x

no need some fancy stuff here

also the word "minimum" is kinda deceptive lol, theres only 1 value of A satisfying the problem after all

[viscid wagon]

How are you able to see all this

[distant helm]

it all started from this

that one is obvious for me but not for anyone else

then just kinda work from that point and the solution pops up over time

[viscid wagon]

Wait but

A = u + 4/3u >= 4*sqrt(2/3)

But the AM-GM inequality doesn't hold