help-33

[leaden monolith]

df should be a column vector

[marsh citrus]

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[earnest sundial]

Given a function $f: \mathbb{R}^3\to\mathbb{R}, f(x,y,z)=x^2+y^2-z^2$ and an $a\in\mathbb{R}\setminus{0}$
Show that $f^{-1}(a)$ is a $C^\infty$ submanifold of $\mathbb{R}^3$ with dimension 2.

[earnest sundial]

How would I even begin to prove something like this?

[elfin berry]

Bob Goldham

[earnest sundial]

I understand why it'd be 2dimensional I think? As soon as we fix 2 of {x,y,z} the third one is then determined by our value of a, so we really only have 2 "degrees of freedom"

How do I show that we're even dealing with a submanifold at all, let alone that it is C^infty

[marsh citrus]

@earnest sundial Has your question been resolved?

[earnest sundial]

inserting numbers from the definition we use gives me a definition along the lines of

[devout ridge]

Hm

[earnest sundial]

A subset $M\subseteq \mathbb{R}^3$ is called a 2dimensional $C^\infty$ submanifold if for any $p\in M$ there exists an open neighborhood $V\subseteq \mathbb{R}^3$, an open subset $U\subseteq\mathbb{R}^2$ and a $C^\infty$ immersion $F: U\to \mathbb{R}^3$, which homeomorphically maps U onto the open subset $F(U)=(V\cap M)$ of $M$.

How would I prove any of that

[elfin berry]

Bob Goldham

[earnest sundial]

this is basically a verbatim copy of how we defined a submanifold

with the 2, 3 and infinity plugged into the relevant places

... can we just apply the implicit function theorem here?

[earnest sundial]

I actually don't understand why it's necessarily 2d anymore

fix x and y

you then get z²= x²+y²-a

which for fixed x,y has 2 real solutions z

couldn't we be dealing with 2 parallel planes?

nvm we're dealing with dimensions of manifolds, it "behaves like a 2dimensional space around any point a" because we'd have to "jump" to get to the "second plane" even if it existed

okay I understand that part again

[marsh citrus]

@earnest sundial Has your question been resolved?

[slender folio]

@earnest sundial haven't u seen any result about the preimage of a function
being a submanifold?

[earnest sundial]

Someone suggested that the implicit function theorem combined with the fact that the gradient of f is never 0 for any such a should be sufficient to prove this, but I don't really see how.
Is that at all correct?

[slender folio]

If u want an abstract way of why f^-1(a) being a 2 dimensionnal submanifold makes sense, we could tell that in a given neighborhood of any point of M, the function behaves like it's total derivative, which is a linear map from R^3 -> R which has rank 1 everywhere excepted in (0,0,0). Therefore f^-1(a) is equivalent to the 'kernel' of this map which has dimension 2

[slender folio]

This is actually True but it's not that easy to show i think

[earnest sundial]

isn't that true for any function by definition of what a total derivative is?

[slender folio]

Intuitively yes , but u need to make assomptions on the regularity of the function in order to get cool results

U should actually take a look at the constant rank theorem

This is the key theorem to prove this result

With that, i feel like u should be able to prove it, but without this theorem, good luck ah ah

[earnest sundial]

isn't that just the implicit function theorem in more general terms?

I don't think we've covered the constant rank theorem, but according to Wikipedia "The inverse function theorem (and the implicit function theorem) can be seen as a special case of the constant rank theorem, which states that a smooth map with constant rank near a point can be put in a particular normal form near that point."

So I'd assume the IFT is just the special case we care about here?

[slender folio]

I don't remember exactly the proof given in my course but it seemed to me that we did use the constant rank theorem and not the IFT

Let me check

[earnest sundial]

also is the gradient being non-zero just required so that an inverse exists for purposes of the implicit function or is there some sort of deeper reasoning to it?

[earnest sundial]

well if you can prove it using IFT, then you can definitely also prove it using a generalization of that

[slender folio]

Oh okay my bad, i just realised that in my course, we did not use the same definition of a manyfold, we used a more restrictive one, which explains why we needed the constant rank theorem

Sorry

[slender folio]

No i would say this is the main reason to it's use given your definition

[earnest sundial]

so would it be correct to assume that "if the IFT applies we're always dealing with a C^infinity manifold"?

[marsh citrus]

@earnest sundial Has your question been resolved?

[marsh citrus]

@earnest sundial Has your question been resolved?

[marsh citrus]

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[wary hawk]

I need help with the fundamentals of TQF, I'm confused on how this equation under the root becomes positive

[vernal forge]

what does -4 * -4.9 equal to?

[wary hawk]

19.6

sorry C is meant to be negative

@vernal forge

so you end up getting -2352

and then 8^2 + -2352 is -2288

[fickle compass]

wait waita iwat

are u solving a quadratic?

[vernal forge]

C?

[wary hawk]

im doing a physics problem involving TQF

[wary hawk]

yeah C

ax^2+bx+c

[vernal forge]

c as in ax^2 + bx + c

ok

[wary hawk]

yes

i know how to get the right answer, its just the essence of it confuses me

how does this equation end up becoming positive

i dont understand the logic

the answer under the square is 2416

[vernal forge]

show the equation

[wary hawk]

-4.9x^2+8x-120

[fickle compass]

is it equal to 0

[wary hawk]

yes

[fickle compass]

then u must make the coefficient of x positive

multiply by (-1) on both sides

[vernal forge]

it doesn't

[wary hawk]

but it does

[vernal forge]

it is -2288 under the square roor which means there are no solutions

[wary hawk]

it isnt

the answer to the equation under the root is 2416

[vernal forge]

then the resource you are following has a typo somewhere

[vernal forge]

that would be the case for +120, not -120

[wary hawk]

i think i see where i went wrong

@fickle compass was right

i made the equation so

0 = ax^2+bx+c

instead of ax^2+bx+c=0

i think that might've done it

[fickle compass]

not really

its just for easier calc

but the roots are imaginary

[wary hawk]

okay hold up

@fickle compass @vernal forge are you guys familiar with kinematikcs

kinematics*

[fickle compass]

kinda

[wary hawk]

i can just send the problem here so its easier for you to understand what im saying

[fickle compass]

yeah thats beter

[wary hawk]

im using TQF to solve for time

A helicopter is ascending vertically with a speed of 8.0 m/s; at a height of 120 m above the earth, a
package is dropped from a window. How much time does it take for the package to reach the
ground? [5.8s]

5.8 seconds is right and it is one of the answers i get when the equation under the root is equal to 2416

[fickle compass]

ok wait

[wary hawk]

i can give you all the values

v1=8m/s, d=120m, a=-9.8

[fickle compass]

what kinematical eq did u use

[wary hawk]

@fickle compass

in the way i did it; your quadratic equation ends up being -4.9t^2+8t-120

[fickle compass]

well according to ur answer

the equatio should ne

[wary hawk]

i even consulted chatgpt on ts and it got the same answer as my answer key

[fickle compass]

4.9t^2−8t−120

thats how u should get it

did u do some calc wrong

or signs

[wary hawk]

no, 4.9t^2 can't be positive because its negative acceleration

[fickle compass]

wouldnt vt be negative

[wary hawk]

vt?

v1t?

[fickle compass]

the vt term

v1

[wary hawk]

no, v1 is positive because the helicopter is initially moving upwards at 8m/s

if you use the equation i gave you

you'll get the same thing as me

[fickle compass]

oh wait

as u said

g would be negative

[wary hawk]

yeah

i know im missing something but i just dont know what

[fickle compass]

hmm yeah thats the eq ur getting

i think theres something wrong with the sign

[wary hawk]

do you wanna work it out and tell me what you get

if ur up for it

because 5.8 seconds is the answer

[fickle compass]

u said u got root2416 in one of the answers right>?

[wary hawk]

yes

[fickle compass]

then isnt that the answer

[wary hawk]

yes but im confused on how im getting that

from this

[fickle compass]

oh

[wary hawk]

this should be negative

but im missing something clearly because it isnt

like its just my understanding of TQF thats causing the issue

[fickle compass]

maybe the q is wrong?

[wary hawk]

q?

[fickle compass]

there might be some wrong sign conventions

cuz the roots cant be imaginary

s is obv positive

u is also positive

aka v1

and g is negative

[wary hawk]

s and u?

are you using different names for variables

[fickle compass]

uh s is d and u is v1

yeah

its in my terms

[wary hawk]

yea

so this is what im confused ab

where am i going wrong

if you organize it with that motion equation you'll get the same answer i did @fickle compass

[fickle compass]

i tried checking on internet

and its stupid

they put the d negative 💀

[wary hawk]

dawg what

what i did to get a positive answer

is i just did

64 - (-2416)

or sorry

-2352**

so 64 - (-2352)

because i was thinking that the -4 within TQF was still constant and needed to be applied to the equation

@fickle compass

[fickle compass]

how is a simple kinematics eq so tuff dang ;-;

[wary hawk]

its just the quadratic formula bro

worst thing ever made

like theoretically i could just plug 5.8 or whatever numbers i get out of tqf into another motion equation to see if it gives me the same value as a variable i already have to begin with

but im just so lost with how TQF works in this, because it shouldn't

[fickle compass]

u might wanna ping helpers ig

im just dumb

;-;

[wary hawk]

<@&286206848099549185>

if any of you guys are familiar with physics could you please help me?

im really stumped on this

[still temple]

with what ?

[wary hawk]

i appreciate the help tho

[wary hawk]

scroll up

everything is explained

[still temple]

this ?
Image

[fickle compass]

A helicopter is ascending vertically with a speed of 8.0 m/s; at a height of 120 m above the earth, a
package is dropped from a window. How much time does it take for the package to reach the
ground? [5.8s]

[still temple]

bruh i didnt send correctly

[fickle compass]

this q

[still temple]

ahh makes sence

wait this is really ez ?

[wary hawk]

its not

[still temple]

isnt this like o levels ?

[wary hawk]

because TQF works so weirdly with this

[still temple]

pardon ? whats the TQF

[wary hawk]

you need to use TQF to solve for time

[wary hawk]

quadratic formula

[still temple]

hmm in my country its diff

you mean -b + or minus ?

or the basic one

where you factorise ?

[still temple]

if you cant use the factorisation just use this tho ?

its not rly hard to use

[wary hawk]

@still temple

[still temple]

isnt that wrong ?

[wary hawk]

this is what your equation should look like under the root

[still temple]

ahh o

ok

[wary hawk]

no

i know the answer to the question

its 5.8 seconds

[still temple]

i thought it was the factorising

hmm ok

[wary hawk]

but i just dont understand because that equation under the root

[still temple]

lemme do it by myself

[wary hawk]

gives you a negative

[still temple]

yeah

but it always gives u two answers

the answer which is positive

is the right one ?

[wary hawk]

but i tried it another way where i did 64 - (-2352) and i got the correct answer

[wary hawk]

yeah

[still temple]

are they both negative

[wary hawk]

the number under the square should = 2416

[still temple]

so u js miscounted ?

[wary hawk]

no

[still temple]

ight lemme work from where u told me to go

[wary hawk]

if you can i would just work out the problem

i can send you the equation that i used

@still temple

[still temple]

is it this ? Where a = 4.9 , b = -8.0 , and c = -120 .

[wary hawk]

yes

b

shouldnt be -8

[still temple]

ight ill do it quite quickly then

[wary hawk]

because the velocity is upwards

b is positive

@still temple make sure you consider b to be positive

its v1

[still temple]

yeah i get it

mate

:/

i got the correct answer

there are 2 possibilities

5.83

[wary hawk]

yes

[still temple]

or -4.2

[wary hawk]

i know

[still temple]

the negative canceles out

so u stay with 5.83

[wary hawk]

the equation i got was

-4.9t^2+8t-120

[still temple]

so b was positive ?

[wary hawk]

yes

[still temple]

hmm ok

[wary hawk]

because v1t is = to 8(t)

[still temple]

let me work around it rq

[wary hawk]

1/2(a)(t)^2 is = -4.9

[still temple]

hol up

u used this right ?

[wary hawk]

because a is -9.8

[wary hawk]

yes

[still temple]

but

wouldnt

it be (-9.8)t to the power of 2 ?

that means the t would be positive

:/

[wary hawk]

that would mean t^2 is 96.04

you dont square -9.8

[still temple]

so the formulae u have at this point should be $ -120 = 8.0 t + \frac{1}{2} (-9.8) t^2$

oops made a gap

mb

$-120 = 8.0 t + \frac{1}{2} (-9.8) t^2$

[elfin berry]

Exornion

[wary hawk]

why is your d negative

[still temple]

but mate

[wary hawk]

the helicopter isnt underground

[still temple]

hol up yeah your right

[wary hawk]

d has to be positive on that side of the equation

yeah

[still temple]

NO

ITS THE DISPLACEMENT

;-;

[wary hawk]

displacement = height

same thing

[still temple]

yeah ik

hol up r u in a level or o level ?

[wary hawk]

what o level

im canadian

[still temple]

16 year old or 17 year old ?

[wary hawk]

16

[still temple]

there is 17-18 which is a level

[wary hawk]

gr11 physics

[true folio]

I have a 10th grade entrance exam and i cant find this question on google, can anyone help?

[still temple]

ahh makes sence

[wary hawk]

open up another forum

[true folio]

wha, how?

[wary hawk]

okay but do you understand the issue that im getting at

[still temple]

not in this chat pls go to #help-1

[still temple]

wait lemme try to explain why the displacement is negative

[true folio]

thank you

[still temple]

its a lil complex

(it is dropped is 120 meters.)

[wary hawk]

oh my god

i understand

holy fuck

thank you

[still temple]

...

yeah

[wary hawk]

its dropped

[still temple]

np

[wary hawk]

dude

[still temple]

ik

(which will be -120 meters, since the package moves downward)

[wary hawk]

that is so annoying

wow

[still temple]

i was boutta paste this

ong

us physicists keep messing up our plusses and minuses

[wary hawk]

i hate when teachers dont give extra context like this

[still temple]

ong

[wary hawk]

its so bad

[still temple]

it was hidden

bcz it says it was "dropped" i noticed it bcz ive been doing this for so long

[wary hawk]

these questions are literally meant to trick you into giving a wrong answer

[still temple]

for an average student it would be rly hard

[wary hawk]

yea just memory

this is exam review

from our first unit

[still temple]

yeh

A helicopter is ascending vertically with a speed of 8.0 m/s; at a height of 120 m above the earth, a
package is dropped from a window. How much time does it take for the package to reach the
ground? [5.8s]

[wary hawk]

so i havent done any kinematics since like

janurary

[still temple]

tbh tho

this isnt the type of question

you get

in the exam

its alot simpler

idfk why you all got this

[wary hawk]

i already know that theres a question that involves TQF

[still temple]

yeah but not with hidden clauses

[wary hawk]

every time you have to find time in one of these problems you need TQF

[still temple]

its very VERY rare

[still temple]

mhm

[wary hawk]

to be honest id rather be overprepared than underprepared

this review is forcing me to read super carefully

[still temple]

i suppose so

[wary hawk]

theres so many really weird trick questions

[still temple]

js dont waste too much time on overdoing yourself

the examns require a lot of other stuff as well

[still temple]

thats true

[wary hawk]

yeah i have one unit left to review

which is dynamics

i hated dynamics

[still temple]

oh that ones easy

why ?

its quite easy

[wary hawk]

i didnt understand potential and kinetic energy well

[still temple]

i hated energy

[wary hawk]

did you ever get a roller coaster

[still temple]

oof

[wary hawk]

question

[still temple]

tbh it depends

but sorta like that

[wary hawk]

i had one like that on a test and it was just purely deriving one formula

and it was so confusing

[still temple]

;-;

[wary hawk]

thanks for the help @still temple

[still temple]

np

also

you can close the channel with .close

when you are finished

peace

[wary hawk]

yeah i know, ive had to come in here a lot lol

.close

[marsh citrus]

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Closed by @wary hawk

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[marsh citrus]

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[exotic marsh]

how do I approach this

[elfin berry]

TheRuleOfEngineering

[exotic marsh]

idk where to go from there

[latent coral]

do you know the specific points for triangle ABC, or are we assuming it's some generic (a, b, c)?

[exotic marsh]

generic

nothing else has been given

[latent coral]

surprisingly the normal vector is one thing you don't need

[exotic marsh]

oh

[latent coral]

hm actually

i thought i found a solution but i'm not sure if it works

[exotic marsh]

anything is appreciated

[latent coral]

the idea was to draw a line from the origin through the three vertices of the triangle to their respective intersections with the plane

something about scaling the vector (a,b,c) to get the points on the plane

so something like p * (a,b,c), then x = pa, y = pb, z = pc

[exotic marsh]

so if A = (a,b,c)
then we have the vector <a,b,c> going from the origin through A through the projection of A

i see

[latent coral]

then since x + 7z + z = 32, p(a + 7b + c) = 32 so p = 32 / (a + 7b + c)

but then you want (x,y,z) which you know is (pa, pb, pc)

and you just found an expression for p

in terms of only a, b, and c

but i think there might be something wrong with this approach, i don't feel too confident

[exotic marsh]

so 32a/(a+b+c) and so on

[exotic marsh]

well its something to start so ill look into it

[latent coral]

32a / (a + 7b + c), and so on

[exotic marsh]

yeah mb 7b

[latent coral]

hope that helps but i honestly haven't seen this kind of problem before

[exotic marsh]

me either

this couse was all about double and triple integrals

and somehow im here now

[latent coral]

and i think there might be an assumption there about the vertices being equidistant from the light and about the plane being oriented parallel to the triangle

[exotic marsh]

ok so your approach IS promising

we land at
$\bar A = (\frac{32a}{a+7b+c}, \frac{32b}{a+7b+c}, \frac{32c}{a+7b+c})$

[elfin berry]

TheRuleOfEngineering

[exotic marsh]

theres no "one correct answer" for this i think

because how

just random values of a,b,c that match a+7b+c=32

one such set could be 2,4,2

wait no

[elfin berry]

TheRuleOfEngineering

[exotic marsh]

so $t = \frac {32}{x_0+7y_0+z_0}$

[elfin berry]

TheRuleOfEngineering

TheRuleOfEngineering

[exotic marsh]

yeah idk

[marsh citrus]

@exotic marsh Has your question been resolved?

[marsh citrus]

@exotic marsh Has your question been resolved?

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[mighty pagoda]

I was doing an exercise with radicans and ran into a curious scenario i think im just being silly but.

we have an exercise:
$5\sqrt{2}*\left( \frac{\sqrt{ 2 }}{2} \right)$

Now nothing special here really.

I solved the problem with 2 methods.
but the first method makes me quesiton my understanding even if i got the right answer.

method 1. (something fishy here)

$5\sqrt{2}*\left( \frac{\sqrt{ 2 }}{2} \right)$

Combine the terms:

$\frac{5\sqrt{ 2 }\sqrt{ 2 }}{2}$
Since $\sqrt{ a }\sqrt{ a }=a$
$\frac{52}{2}= 5$

**Here is my question **
if $5\sqrt 2$ can be written as
$(\sqrt{ 2 })(\sqrt{ 2 })(\sqrt{ 2 })(\sqrt{ 2 })(\sqrt{ 2 })*(\sqrt{ 2 })$

we have a curious case since this is equal to 8 not 10.

and thus the previous example should give answer = 4

I got a second more reliable method of calculating this but im just wondering is there some radical rule im missing here or is it simply fundamentally wrong?

[elfin berry]

HydroH

[mighty pagoda]

Sorry for the horrible syntax

[tired oxide]

$5\sqrt 2$ can’t be written as
$(\sqrt{ 2 })(\sqrt{ 2 })(\sqrt{ 2 })(\sqrt{ 2 })(\sqrt{ 2 })*(\sqrt{ 2 })$

[elfin berry]

smaysurable set

[mighty pagoda]

yea because the last sqrt 2 cant be interpeted as 6\sqrt 2 i think i just realised this...?

[tired oxide]

no not that either

what you wrote was $\sqrt 2 ^5$

[elfin berry]

smaysurable set

[mighty pagoda]

yea pardon the extra was related to the exercise

[tired oxide]

with the correct number of sqrt 2’s

like the exercise is just fine

the beginning part is indeed correct and has correct reasoning

what we can say about $5\sqrt2$ is that it’s $\sqrt2+\hdots\sqrt2$

[elfin berry]

smaysurable set

[tired oxide]

it won’t be multiplication of 5 sqrt 2’s together

[mighty pagoda]

oh

[tired oxide]

which can’t really be simplified as much

as you might have suspected

or maybe not, which is also fine

[midnight folio]

sqrt 2 / 2 can be simplified to 1 / sqrt 2

[tired oxide]

that as well

[mighty pagoda]

so just to be clear since $5\sqrt{ 2}=\sqrt{ 2}+\sqrt{ 2}+\sqrt{ 2}+\sqrt{ 2}+\sqrt{ 2}$

the radican rule $\sqrt{ a }\sqrt{ a }=a$ wont be applicable

[elfin berry]

HydroH

[main frost]

what the sigma 💀

[tired oxide]

it will after you distribute out the product over the addition

[main frost]

sorry for the brainrot

[mighty pagoda]

i dont understand

[tired oxide]

but you can just leave it as $5\sqrt 2$

[elfin berry]

smaysurable set

[tired oxide]

and trust that it works out if you wanted

[main frost]

let sqrt(2) be equal to x
now 5*sqrt(2) = 5x
you can also write 5x as x+x+x+x+x AND NOT AS 5*5*5*5*5

its not your fault tho i too do silly stuff like this when im brain fatigued

[midnight folio]

Bro wrote it before me

Couldve told me to not waste time

[main frost]

idk i just saw a silly porblem and wrote a silly answer

i guess he got confused due to the root sign so i simplified it into a variable

[midnight folio]

I love you (platonically)

[main frost]

i love you too bro (platonically)

[midnight folio]

5 x 6 = 6 + 6 + 6 + 6 + 6

Sinilarly

5 x sqrt 2 = sqrt 2 + sqrt 2 + sqrt 2 + sqrt 2 + sqrt 2

[tired oxide]

the computation to show that it work either way is to EITHER: do it the way you did in your image (which was perfectly acceptable) or write out
$$
(\sqrt2+\hdots+\sqrt2)(\frac\sqrt22)= \frac{\sqrt2\sqrt2}2+\hdots+\frac{\sqrt2\sqrt2}2= 1+\hdots +1$$

[midnight folio]

If we factor out a sqrt2

We get

[elfin berry]

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

[midnight folio]

Sqrt2 (1+1+1+1+1)

Which is 5, mashallah

[mighty pagoda]

yea i think i got it. Ty for everyone with the help

.close

[marsh citrus]

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[tough monolith]

Im having problems understanding whats being done in this table

[tough monolith]

i can send full question if needed, its using euclidean algorithm to work out gcd

[midnight folio]

Its finding the GCD

Hope this helped

[tough monolith]

i know it is lol

thats not what im confused about

ij ust dont understand whats actually going on in that table

[vital oracle]

do you have a screenshot of the sequence of remainders in regular order

[tough monolith]

i htought it was just, i take the 1 sub it in, 2nd line is 6 and i just keep feeding in each number and getting new result

but then i scrolled down adn thats not how you do it

so im kinda confused

[vital oracle]

dont know what you mean by that and I dont see a 6 anywhere

youre trying to look for two amounts of m and n where adding 324m + 149n is 1

in other words youre trying to see how many 324s and 149s you add/subtract to get 1

[tough monolith]

ohhh wait

so the last number is just whatever u multiply the other number by so it equals 1

i think i get it now okay

thank you

[vital oracle]

np

[marsh citrus]

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[still temple]

hi! Can someone walk me through this quesiton please

[still temple]

tan2x + tanx = 0 for 0<=x<=180

[marsh citrus]

@still temple Has your question been resolved?

[unkempt herald]

hmmm..

[ivory turtle]

Use tan=sin/cos and double angle identity?

[unkempt herald]

maybe try considering a right angle triangle and draw a angle bisector

[still temple]

uhhh

ok

[still temple]

the question shouldnt need that

[still temple]

sorry you want the (2tanX)/(1-tan^2X)

identity/

?

i did that andit gets me nowhere

[tired knoll]

I would analyze the function tan(x) + tan(2x) in each of those 4 intervals :

0° <= x < 45°
45° < x < 90°
and so on
(assuming 180 means 180 degrees above)

And you should be able to deal with each parts, for example, between 0 and 45 degrees, the function is strictly increasing

[still temple]

what do you mean by analyse

[tired knoll]

Is it increasing? decreasing?

[still temple]

in the first part its positive

[still temple]

like you mean using a graphing calculator?

[tired knoll]

positive? negative?

no, using tan properties

[still temple]

oh by logic ok

[tired knoll]

you already know how does the function tan behave, right?

[still temple]

bit foggy but yeah

0 and 45 its increasing and +ve

45 to 90 its increasing gbut -ve

[tired knoll]

What is the value of tan(x) + tan(2x) when x approaches 90° from below? for example, when x = 89.99°

[still temple]

infinity?

[tired knoll]

Yes

[tired knoll]

So this is wronf

wrong*

[still temple]

right

lmso

[tired knoll]

And what about when x is just above 45°?

[marsh citrus]

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[still temple]

.reopen

[marsh citrus]

✅

[still temple]

45 to 90 is increasing but +ve
0 to 45 is decreasing but +ve

[still temple]

negative

[tired knoll]

No, it is wrong

Between 0 and 45 degrees, the function is also increasing

[still temple]

oh

[tired knoll]

Correct

[still temple]

wait sorry how does this play into

the original question

[tired knoll]

For example, between 0 and 45 degrees

tan(0) + tan(2*0) = 0

and f(x) = tan(x) + tan(2x) is strictly increasing

meaning there is only one solution in the interval [0, 45 degrees[

Now in the interval ]45 degrees, 90 degrees[, it is strictly increasing, and it approaches -infinity close of 45, and approaches +infinity close of 90

Meaning there must be one (and only one) solution

This is called the intermediate value theorem, it is quite an intuitive result

[midnight folio]

This should work

[rocky kettle]

for equations involving trigonometric functions, you should always look for a possible substitution to make your life easier

[still temple]

i guess ill deal with this later thanks guys

.close

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[still temple]

I need help with this question, are the value wrong or am I doing it wrong?

[still temple]

blue and red lines are the tangents
those two points are where they are supposed to touch the parabola
no such parabola can ever exist?

[south smelt]

It doesn't want you to find lines tangent to that parabola, it only tells you that the slope of said parabola at 3,2 and 2,3 are 31 and 14. So you know the derivative there is equal to those

that'll help you find the constants

[still temple]

Thank u

But but when you eliminate the constants
the parabola you find doesnt even satisfy the two points
(3,2) and (2,3)?
Idk if the question is wrong

[marsh citrus]

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@still temple Has your question been resolved?

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[worthy obsidian]

iss thiss all good or im missing something?

[ivory turtle]

,rotate

[marsh citrus]

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[marsh citrus]

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[rigid kelp]

.close

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[cinder lion]

My answer is not close to the mark scheme, they have different measurements on the lengths of the triangles. Have I done something stupid?

[cinder lion]

Sorry it’s all on its side

,rotate

[marsh citrus]

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[marsh citrus]

@cinder lion Has your question been resolved?

[marsh citrus]

@cinder lion Has your question been resolved?

[tacit marsh]

Bruh just use quadratic equation

[mild knoll]

I already did xd

[marsh citrus]

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[tranquil sparrow]

https://gyazo.com/242191f13b41bb8fc5fad8b860c5567c how is csc not negative as well also how is tan and cot wrong?

[tranquil sparrow]

https://gyazo.com/48ff282762620ad35645ce19e71cb5fc

[winged pier]

Csc is 1/sin(x) right

[tranquil sparrow]

yes

[winged pier]

The way you visualize the angle is wrong

Remember that theta is negative

So you’re doing a negative rotation

[tranquil sparrow]

wouldn't it be 2 full rotations

then negative 3pi/4?

[winged pier]

-13pi/4 radians is the same as -5pi/4 radians right

nuh uh uh

[tranquil sparrow]

omg

[winged pier]

2 pi is 8pi/4 lol

[tranquil sparrow]

im so stupid

SORRY lmao

its 3 full rotations and pi/4

[winged pier]

not your fault

wait 3 rotations?

It’s 3 half rotations

NEGATIVE rotations

And a negative pi/4 rotation

[tranquil sparrow]

oops

I'm tripping

thank you

[winged pier]

np

[marsh citrus]

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[marsh citrus]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[torpid marlin]

tangent

toa

no info on hypotenuse

so

tan 18 = x/12

x is on top so we multiple

so x = 12 tan 18

but I get a negative

[trim quest]

Make sure your calculator is in degrees

[torpid marlin]

how do I switch to degrees

[trim quest]

on that calculator Idk

[torpid marlin]

I tried doing this

on my phone calculator

but

I do 12

the tan

and it calculates tan 12

doesn't let me finish the euation on phone

[trim quest]

so type 18 first, then tan

to get tan 18

[torpid marlin]

ok

[trim quest]

and then multiply by 12

[torpid marlin]

3.9

[trim quest]

Looks good 👍 Maybe 3.90 since it says two decimal places

[torpid marlin]

ok

[trim quest]

but yeah that's right

[torpid marlin]

thnaks

thanks

[trim quest]

np 👍

[torpid marlin]

.close

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[feral scaffold]

Hey guys, this is a question for people familiar with NCERT books and isn't exactly a question about a specific math problem, rather about understanding what the textbook is trying to say. I am currently doing class 9 chapter 2.4 after finishing exercise 2.2, the topic before 2.4 was 2.3, finding zeros of a polynomial, which i understood with no issues, now it starts referring to some example 10 above which hasnt come up yet and also starts talking remainder theorem, which also havent been mentioned yet, and now i am really confused as to what's happening as its just going on about a topic that it expects me to know but it never mentioned up until now. Is this an error on their end or am I just tripping, what can I do about this, thanks. You can see the chapter here: https://ncert.nic.in/textbook.php?iemh1=2-15

[marsh citrus]

@feral scaffold Has your question been resolved?

[shadow ibex]

Looks like they messed up.
This looks like a preview page, so maybe they mixed pages from more than one edition

[marsh citrus]

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[feral scaffold]

what do you mean? that is the full chapter from the official website

[feral scaffold]

it isn't a preview

[marsh citrus]

@feral scaffold Has your question been resolved?

[shadow ibex]

B.c. of the watermark

Ok, but it still looks like an error. I couldn't find a reference to example 10 or a q(t) that has a factor 2t+1

Unless it's referring to the previous section...which it should have said if that was the case

[feral scaffold]

I can confirm it is indeed an error on their end

pretty silly from a government organization

but it is what it is

[feral scaffold]

that's there for the online copies, it's still the full textbook though

[shadow ibex]

Gotta make that textbook $$$

[feral scaffold]

😂

it's ok

I will ask my teacher for help

[shadow ibex]

Also

The remainder theorem

iirc, it just says that a linear expression divides a polynomial if the remainder is 0

If only there was a quick way for me to be sure...

[feral scaffold]

the chapter doesn't mention anything abt dividing polynomials either

[shadow ibex]

No

[feral scaffold]

they've just removed some major parts from the textbook

[shadow ibex]

It's even more general

[feral scaffold]

I still don't understand 😕

the textbook is just missing pretty important information the topic before this was perfectly fine, it appears they removed some information between these 2 topics without patching it up

but thanks for helping, my peanut brain just can't comprehend this stuff 😂

hopefully this gets resolved

[shadow ibex]

Under division by a binomial (specifically, and more usefully, an expression of the form x-a), a polynomial p(x) has a remainder of p(a)

Here's an example of what that means

[feral scaffold]

so how would you divide a polynomial by a binomial?

[shadow ibex]

Say you have the polynomial

[feral scaffold]

alright

[shadow ibex]

p(x) = x^3 - 4x^2 - 7x + 10

[feral scaffold]

alright

[shadow ibex]

Evaluate this at 2

[feral scaffold]

meaning?

[shadow ibex]

p(2) = -12

[feral scaffold]

ohh

[shadow ibex]

Meaning plug in 2

[feral scaffold]

alright alright

I understand

[shadow ibex]

We got -12

Ok

[feral scaffold]

yep

[shadow ibex]

Now, if we had just done

p(x)/(x-2)

[feral scaffold]

so p(2)/(2-2)?

[shadow ibex]

We should still get a (constant part) remainder of -12

[shadow ibex]

No

[feral scaffold]

oh wait wait

I get it

[feral scaffold]

this entire polynomial

divided by (x/2)?

is that right?

[shadow ibex]

No, divided by x-2

I'll do it hang on

[feral scaffold]

Mb I meant x-2

[shadow ibex]

I'm going to use synthetic division b.c. it is fast

[feral scaffold]

alright

[shadow ibex]

My pen was running out of ink and I messed up a couple of times 👀

[feral scaffold]

that's fine

I think I understand now

[shadow ibex]

If (x-2) had been a factor, the last number would have been 0, but it was -12, so we write it over the divisor x-2

[feral scaffold]

ohh

I get it I get it

[shadow ibex]

That's good

[feral scaffold]

alright thanks for your patience

and help

[shadow ibex]

np

[feral scaffold]

I will close it now

.close

[marsh citrus]

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[reef trout]

is r/q equal to sin y°?

[warm goblet]

sin y degrees will be perpendicular/ hypotenuse...
which in this case is r/q
hence, yes r/q is equal to sin y°

[marsh citrus]

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[deft nova]

Idk how to approach this

I tried using the definition

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[novel juniper]

There is someone in the freshman class who doesn’t have a room-mate.
The negation would be everyone in the freshman class has a room mate

[novel juniper]

is this right?

[void elm]

yes.

[novel juniper]

thanks

wanted advice

[void elm]

hm?

[novel juniper]

how do you negate statements like this , wwhere multiple intereprtations exist ?

Like I could say there's noone in the freshman class who does have a room mate

[void elm]

oh. I hate those questions

[void elm]

yeah see, it's better to put things in quantifers then negate for this reason

"there is someone" => there exists
"everyone" => for all

[novel juniper]

I see

yeah, makes sense

[void elm]

so the original statement is

∃x such that P

and the negation is

∀x we have not P

where x is a classmate and P is not having a roommate

[novel juniper]

got it

TYSM!

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[void elm]

[leaden monolith]

@novel juniper this is generally how negations get "pushed through" quantifiers

the for all's turn into existence, the exist's turn into for all

and the statement gets negated

[novel juniper]

got it

thanks!

[void elm]

yes, I should've said that

my mistake

but you should convert English sentences to logical statements before negation still

[novel juniper]

got it

[void elm]

it's just less ambiguous that way

[marsh citrus]

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[dark crypt]

Hello, I got a question, so in this question do i need to find f(t) and then solve f(2)?

[brave marsh]

Considering they want you to estimate the amount of solid, it's likely that they want you to compute the derivative at t=2 and use the tangent line there as an approximation for t=3

[dark crypt]

Alrighty, so f'(2) would be equal to
-24 right?

[brave marsh]

Yeah that looks good

[dark crypt]

ok so now do i plug in t=3?

[brave marsh]

Well get the equation of the tangent line first

[dark crypt]

So this means y-y_0=m(x-x_0) right?

[brave marsh]

Yeah

[dark crypt]

so now i'm a bit confused,
how would i find the equation of the tangent line then?

[brave marsh]

You have m, that is the derivative at t=2.

You also know that the tangent line goes through (2,2) since that's where you computed the derivative

[dark crypt]

ok so our m value is =-24

[dark crypt]

oh and (2,2) because its 2 grams of solid at 2 seconds right?

[brave marsh]

Yep

[dark crypt]

ok so out y_0 value is 2?(and our x value)?

[dark crypt]

so our equation is gonna be y-2=-24(x-2)?

[brave marsh]

Yep

And you want t = 3 in there

This is the estimation bit

[dark crypt]

oh ok! so
y-2=-24(1)

y=-22?

[brave marsh]

Wait

Why is it negative lol one sec

[dark crypt]

because m= -24

[brave marsh]

Yeah but like, physically this is the mass of solid inside the beaker

[dark crypt]

oh yeah i cant be like

negative mass

that would probably be really bad lol

[brave marsh]

Oh

They're very sneaky

The derivative is in gram/min

Not /sec

[dark crypt]

GOD DAMN IT

this is the second time they've done this btw

[brave marsh]

So you have to put in 2 + 1/60

Not 3

[dark crypt]

ohhhhh

121/60?

[brave marsh]

At least the rest of the work is fine

Yep

This represents one second past 2 min

[dark crypt]

ok so lemme just send in my working

so this is our equation, and we find out slope right?

do i plug in for t then?121/60?

[dark crypt]

so i'd get -24.30083...

now i plug this back into the equation of the tangent line formula?

[brave marsh]

No. you don't need to compute the tangent line again

[dark crypt]

oh mb

[brave marsh]

You can use the old one

[dark crypt]

so y =-24(24.3008-2)+2?

[dark crypt]

im still getting a negative number guh

OH MY GOD WAIT

nvm i still got it wrong its so joeover

[dark crypt]

so now
y= 24.3008(3-2)+2 would be the way tofind out new y right?

[brave marsh]

No. y = -24(121/60 - 2) + 2

[dark crypt]

oh mb

8/5

[dark crypt]

HOLY HECK THANK YOU I WILL NAME MY FIRST BORN AFTER YOU

.close

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[still temple]

need some help

i dont understand the algebra steps

how did they isolate f so nicely

very frustrating been stuck on this for 45 min

[marsh citrus]

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[frigid lintel]

anyone able to help real quick? need some sleep

[tired oxide]

why would you expect there to be any relationship between ABC and PQR at all

[frigid lintel]

that's what i was thinking

complete different

just wanted to make sure

[tired oxide]

the transitive property doesn’t guarantee what they’re claiming

[tired oxide]

yeah nw

i need some sleep too

,ti

[elfin berry]

The current time for smayday is 04:23 AM (EDT) on Wed, 26/06/2024.

[frigid lintel]

holyy

1:23 for me

[tired oxide]

lol

[tired oxide]

get some rest before it gets to 4 am

[frigid lintel]

will do, you too haha

.close

[marsh citrus]

Channel closed

Closed by @frigid lintel

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[marsh citrus]

Available help channel!

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[limpid bison]

How do i do 2c

in fact how do i even do 2b

actually forget 2b

how do i do 2c

[ivory turtle]

!status

[marsh citrus]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[still light]

Did you do a)

[limpid bison]

yes

-2(x+1)^2-1) +11

so

-2(x+1)^2 +13

[still light]

Right

So then what would the axis of symmetry be

[limpid bison]

erm the turning point is at -1,13

it cuts through the middle of the graph

at x = -1 tho

so its that

[still light]

Right

[limpid bison]

nice ok thanks

.close

[marsh citrus]

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[marsh citrus]

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[rocky steppe]

How am I supposed to know which graph is correct? I know that it is VERTICALLY stretched by a factor of 2 but I can’t tell the difference between any of them? I know the y axis is changing (some of them are at y = 2) but idk which is the right answer

[marsh citrus]

@rocky steppe Has your question been resolved?

[marsh citrus]

Channel closed

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[marsh citrus]

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[wet laurel]

I can't figure out how to do the required line integral in this question. I know how to do the Stokes' theorem part, but I can't figure out how to do the parametrization for the given region.

[marsh citrus]

@wet laurel Has your question been resolved?

[wet laurel]

.close

[marsh citrus]

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[marsh citrus]

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[rain walrus]

Hi everyone. I'm French and I want to improve my english for my studies and I need help in a math problem. I have a project and if someone are interested to correspond with me, that could be nice for me. Thanks 🙂

[tender lily]

what do you need help with

[lucid turret]

!2ada

[ivory turtle]

!da2a

[marsh citrus]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[lucid turret]

!2adt

[ivory turtle]

almost meolve

[lucid turret]

I am still learning.

Ok

[ivory turtle]

you'll get there eventually :)

[marsh citrus]

@rain walrus Has your question been resolved?

[marsh citrus]

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Available help channel!

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[ivory turtle]

!status

[marsh citrus]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[exotic walrus]

2

[ivory turtle]

what do you have so far

[exotic walrus]

I find slope

of both lines cuz paralle

then i try to put slope intercept form cuz reciprocal intercepts

[ivory turtle]

!show

[marsh citrus]

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

[exotic walrus]

x/1/a + y /1/a

sorry can not show i am on laptop

[ivory turtle]

which question?

[exotic walrus]

no mobile

13

[ivory turtle]

or is it multiple choice?

[exotic walrus]

yup that one

[tacit marsh]

What even is a reciprocal intercept

[exotic walrus]

.

i think

ya guys any progress

<@&286206848099549185>

[exotic walrus]

.

[dire wagon]

I dont even know

Though the answer is between a and b

Since they have same slope

Aka parallel

[finite glacier]

i would assume recipricol intercepts mean that x intercept is the recipricol of y intercept

you can just find the intercepts of A and B to see if the have this property

since C and D are not parellel to the original equation

[exotic walrus]

you are saying i put x/a and y/1/a

right?

or y= 1/x

[marsh citrus]

@exotic walrus Has your question been resolved?

[finite glacier]

im suggesting to literally find the intercepts of each of the 2 equations

like for example A has x intercept at +-2 and y intercept at +-1

these arent inverses so this isnt the answer

and continue

[exotic walrus]

.close

[marsh citrus]

Channel closed

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[little oriole]

reciprocal intercepts means that x.y = 1. So you can evaluate each option. the way to evaluate is to find the intercepts. The intercept in x is when y is zero and the intercept in y is when x = 0. If you evaluate them you will find x and y, and the product of these two values should be equal to 1. After you find that you need to evaluate if the slope is the same as the given line (that means that both are parallel)

[marsh citrus]

Available help channel!

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[still temple]

the reference angle i get is 20 degrees

but like their is no similar triangle for 20 degrees

any other way of doing this?

[marsh citrus]

@still temple Has your question been resolved?

[still temple]

.close

[marsh citrus]

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[still temple]

HELP

[still temple]

u could argue they are congruent

but they're diff size

[ivory turtle]

So maybe it cannot be determined then?

[still temple]

hmm

ig

because you could argue

it isnt to scale

ITS JUST TOO MUCH BIAS

or