precalculus

[opal vault]

celestial bodies

under the infulence

of a strong gravitational force

ex: the sun's

follow a relatively eliptical orbit

[ripe meteor]

Thanks!

[opal vault]

hyperbolic

is when

it passes the escape speed

very easy to calc escape speed using energy eq

earths is around 11.7 i think

km/s

im not sure

fun fact: keplers 3rd law is not entirely precise 😁

[chilly cypress]

can someone help me with this

ik the quality is bad

[stiff nebula]

You just need to figure out how many numbers can fit between 0.25 to 0.5

So try to experiment using trial & error for say 3/7, 3/8, etc until you each a fraction that is equal to 0.5

[vapid plaza]

Solving each half of the inequality yields x < 12 and 6 < x.

[solar olive]

yep, u may use newtons gravitational theorem to derive the relationship between the period and the radius

[thorn creek]

also in reality there are more than two bodies in most real systems, and the three-body problem has no analytic solution in the general case

[solar olive]

also that

[solar olive]

just for curiosity,
F_grav = F_centipetral
G M m/r^2 = m a
G M /r^2 = w^2 r
G M/r^2 = (2pi/T)^2 r
G M/r^3 = 4 pi^2 /T^2
T^2 ~ r^3

[fleet pendant]

$\begin{aligned}
F_c=F_g \
\frac{mv^2}{r}=\frac{GMm}{r^2} \
v^2=\frac{GM}{r}; \left(\frac{2\pi r}{T}\right)^2=\frac{Gm}{r} \
\frac{4{\pi}^2 r^2}{T^2}=\frac{GM}{r} \
\therefore \frac{4{\pi}^2}{GM}=\frac{T^2}{r^3} \
\end{aligned}$

[obsidian monolith]

anonymous.h

[solar olive]

ur right

[thorn creek]

you need some very basic calculus to do this rigorously

that said, in a two-body system the trajectories of both bodies will be conic sections

(provided that neither collides with the other)

[wind kayak]

1+1 = 11😎 xd

[thorny heart]

three sides and three angles = 90% geometry? sad

[warped cipher]

Bro this isn't a physics server lmao

[steady shore]

i mean physics is basically maths with a few extra steps

[fleet pendant]

it aint but i typed out the guy's steps

"just for curisoity"

[warped cipher]

Oh I didn't know that

[thorn creek]

we get a lot of physics questions regardless

[summer ruin]

no

it's a line such that the distance between the curve and this line approaches 0

I guess you wouldn't be asking that if you knew precisely what that meant, so no

[summer ruin]

<@&268886789983436800>

[opal vault]

no

an asymptote is something

that never meets

[summer ruin]

that's not true

sin(x)/x as x-> infinity has asymptote y = 0, which is crosses infinitely many times

[fleet pendant]

ya that's like the definition my teachers gave us when I was in grade 11, it's not accurate, cross overs can occur

[opal vault]

fine it converges

ig

[prisma ember]

anyone know how to find the limit of this?

x=/=5 basically, i was wondering if lim x--> 5- was undefined or 10?

[remote dune]

if approaching from the left side, yes you do use the formula given that represents the left side

[unkempt jolt]

The only reason that I didn't finish precalc with a d or f last semester is entirely due to the mercy of my professor
that is despite me allocating hours daily into studying precalc, going to tutoring centers, etc.
Concepts that people understood in minutes take me hours
I might be genuinely too stupid for college

Still have no idea how to simplify radicals

[uncut mulch]

post specific questions you're having difficulty with

[unkempt jolt]

I don't know anything confidently. I can only grasp everything very gently. Therefore, I don't really know anything and don't know where to start.

I do frequent here with specific problems, but that can only do so much

There are no good interactive courses online for doing math. Khan academy is mediocre, but others either require you to be part of an IRL class, or don't exist.

[uncut mulch]

khan gives a basic intro

is adequate for the basic stuff

if you have access to the content, just takes practice

[lavish adder]

.

[fickle cedar]

Is this true? $:x\left(x^2+1\right)^{-\frac{3}{2}}=\left(x^3+x\right)^{-\frac{3}{2}}$

[obsidian monolith]

Chuti | Argentina

[warped cipher]

no

[cyan wave]

can i get some help

[stone robin]

chat is this real

[vapid plaza]

no

[jaunty nebula]

For the upper half and lower half

notice that these are both functions

so think of how you can isolate y so that it's in the form y = f(x)

[cyan wave]

whats the point of this

[summer ruin]

you solve the inequality (x-3)(x+2)(x+5) < 0

even though first 3 lines are unnecessary

[willow bear]

can be helpful

[cyan wave]

wut

[gritty wren]

When Susan was born her aunt was 18 years old. Now her aunt is three times as old as Susan. What are their ages?

helpppp pleaseee😭😭😭

<@&286206848099549185>

[viscid thistle]

susan is 9 and her aunt is 27

@gritty wren

[gritty wren]

yes I got this but I don’t get it

[viscid thistle]

3x= aunt's age

[gritty wren]

if aunt is x + 18 and then Susan becomes 3x how is 3x equal to x + 18

[viscid thistle]

3x-x=18

2x=18

x=9

[void walrus]

See,
If Susan is x years old, her aunt is (x + 18) years old.

So now there current ages follow the following relation as per the question,
Susan age = 3 * Aunt age
3x = (x+18)
x =9

[gloomy rover]

you set them equal and find which x allows them to be equal

then you have what susan's age is

and you can use x+18 to get aunts age

[viscid thistle]

aunt= 9*3=27

[gritty wren]

I love u guys for this but I am not getting it😭😭

so Susan is x and aunt is x + 18 right

so then in three years time

Susan will be 3x

Then wouldn’t aunt be 3(x+18)

[late heart]

Susan is 9
Aunt is 27

[gritty wren]

tyy I know this but I don’t get how if now Susan is three times older would the x +18 also multiply by 3

[dense mango]

bc to create an equation both sides must be equal

so you have to multiply the smaller age by 3

bc the older person is 3 times older

[gritty wren]

ohhhh

so the aunt = x + 18

but now

Susan is 3x

so

aunt now is

3x

Wait

I don’t got it

[viscid thistle]

no susan's current age is x

[gritty wren]

wait I get it now

tysm

[viscid thistle]

u welcome

[gritty wren]

like literally tysm all of u😭😭😭

[dense mango]

lol yw

[viscid thistle]

igcse?

@gritty wren

[gritty wren]

In aus

its our semester exam

[unique schooner]

Don't complicate it. It's simple,
aunt = susan + 18 years old
aunt = 3 times susan

Now you substitue,
susan + 18 years old = 3 times susan
-susan -susan
18 years old = 2 times susan
/2 /2
9 years old = susan

[civic mural]

hi

[manic skiff]

I need help

15x=2.6^56>-5-{-4.8^436/7*[14.097657124-43+81/43+43/43+-9803^5-(-0.87912*7^98-17%74^3-13^2=-18-)-41+86.42371/45-3113+56.9740.9-]-41/98+33/33+33/33-}-412=2091628628516.709920820. Solve for x

My brain stop work

[opal plover]

so like how did u get allat

[manic skiff]

I forgot, it just appeared…

[opal plover]

wtf

(what the fweak)

[manic skiff]

Don’t ask…

It just came

[opal plover]

oh

[manic skiff]

So I need to find an answer or else my math experiment doesn’t work

Those people

I’m scared of them

[willow bear]

what math experiment?

@manic skiff did you just type out some seemingly random symbol soup?

there's a few that don't belong, namely:

...(0.87912 * 7^98 - 17% * 74^3 - 13^2 = -18**-**) ...

this equals sign and this minus right after the eighteen

and then later there is another minus sign right before a closing bracket

[willow bear]

it looks like they've left now

so either a troll or 13

[gritty sage]

what is the congruent equation for y = (x - 4)^2 - 1 ?

what is the range for y = -3x^2 - 6x + 9 ?

send help 😭

[willow bear]

wdym "the congruent equation"

do you have the full text of the problem? @gritty sage

[gritty sage]

@willow bear

i think the x intercepts is (5,0) , (3,0)

[willow bear]

ok that did not help

[gritty sage]

omg 😭😭

[willow bear]

but at least we know it's the problem's fault

no idea wtf they want for either of these lmfao

[gritty sage]

oh my

anyone know any of these?

[marble leaf]

haha wtf is a congruent equation

frick I didn't mean to ping

sorry for the randomish ping

[marble leaf]

The domain of a quadratic is always all real numbers and the range of a quadratic is always y≥y vertex or y≤y vertex, depending on if the parabola opens up or down. In this case it opens up, so the range is y≥-18. The axis of symmetry is always the line x=x vertex, so the line x=4 in this case. The min/max value is just the y vertex, and u should be able to figure out if it is a minimum or maximum based on the graph.

[grave forge]

I think the congruent equation is the factorised form

So for first one just use the a² - b² identity

For the second u can easily factorise

[grave forge]

Yes

[marble leaf]

oh that makes sense ig

[grave forge]

xd

[gritty sage]

thank uu

[untold spindle]

<@&268886789983436800>

[viscid thistle]

$x^3-5x+\frac{40\sqrt{5}}{27}=0$

[obsidian monolith]

john.1970

[viscid thistle]

how can i solve this polynomial?

[wind grove]

Can someone help me solve 2^x = e^(x+1)

[summer ruin]

e^(x+1) = e * e^x

[wind grove]

How did u do that?

[summer ruin]

that's just how exponents work

[viscid thistle]

Well you could use the long division method

If you know how to use it

Or you could use long division

Assuming you know what the root is simplified

I think you’d have to simplify the top first 40 root 5 then do it

Using synthetic division could help once you find all of the terms

Because it’s to a degree to 3 then it makes sense

If you plan to use synthetic division you’d swap a 0 in

@viscid thistle

i'm here

Yeah, so can you simplify the root thing

what do you mean by simplify ?

Like simplify the top

It’s a 3rd degree polynomial

Do you understand that first of all? @viscid thistle

Do you understand you cannot use the quadratic formula as well

You want to solve right not factor? @viscid thistle

yes

Okay

like i just need the values of x that hold this equality

but i don't how to do that

Yes, you’re solving for X correct?

You’re not factoring right?

you suggest doing horner methkd and euclidean division but i don't see how is that applicable

[viscid thistle]

no

wouldn't be the same if i'm factoring ?

like i will have to find the values of x for this equality

You’re not factoring though

So you cannot use any factoring methods

You need to simply isolate x

I don’t think you can factor

If you do then you’d withhold like 4 answers

You’d have to check if some are extraneous

@viscid thistle okay you need to use synthetic division.

Once you use synthetic division I assume you’d know the factors

Once you find the 2 factors you’d just solve for it

The last term you’d have to simplify

You’d essentially go 1, 0, 5, last term whatever it is

Then use synthetic division to solve

That’s the best way I can think of

can i ask you a question@viscid thistle

[viscid thistle]

Yeah

you mean using the synthetic division on an arbitrary element a?

I mean you wouldn’t know what to use for the last term unless it is simplified

cause you’d have to use either a fraction, or a whole number whatever works

It’s to the 3rd degree you cannot follow up with a 2nd degree

Essentially using long division method, or synthetic works if you know all the terms

@viscid thistle let me explain it more simply. Do you understand how a polynomial works as in x3, x2, x

Basically if it goes x3, x, (term) you have to use something called synthetic division, and replace the x2 with zero

Assuming you use the method

[untold spindle]

<@&268886789983436800>

[warm patio]

Should i go back and review matrices / conic sections or can i go into more advanced math classes without knowing

[clever parcel]

In my example, why is it said that equality occurs at a=2, b=c=d=0. It can also occur at a=b=c=d= fifth root of 8, or some other random combinations of a b c d.

And when you have to prove an inequality by splitting it up into smaller ones, those smaller inequalities may equal at 2 or more numbers, then which one do I choose? In my example, the equality cases for those helper inequalities were either 0 or 2. I could pick one number equal two and the rest equal 0 because I was working with a concrete number (32). If I was given something non-specific, I probably would have thought that equality occur at a=b=c=d=2.

[untold spindle]

<@&268886789983436800>

;-;

[autumn wadi]

Me when I walk into precalc after 10 years and the first thing I see is a mod ping:

breathes in "the good ol' smell of #precalculus "

[grave forge]

Sad

[late geyser]

a=b=c=d=fifth root of 8 doesn't satisfy the assumed equality.

[clever parcel]

eh? why?

[late geyser]

,w compute 4(8^(1/5))^4

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by zybikron#0

[late geyser]

it's not 16

[clever parcel]

its 32

and to the fifth power btw

[late geyser]

read the first line of your problem.

[clever parcel]

ye?

[late geyser]

your choice of a,b,c,d don't satisfy that inequality. So those values don't work for a,b,c,d

[clever parcel]

do you mean "equality"?

if that then i see

[late geyser]

either one

21.1121 is not less than 16

or equal to

[clever parcel]

i see i see

[clever parcel]

can you help me w this too?

[jolly sparrow]

hey, i wanted to ask what seperates precalc from algebra?

[opal vault]

not much

imo its pretty useless

just know ur trig and e shit

[viscid thistle]

hey guys

quick question

what all does one need to know before going into calculus

because I know algebra 1, conic sections, I think most of algebra 2 (still a bit weak on probability and statistics), and trig

is that enough?

[warped cipher]

ever started with logarithmic and exponential functions?

[stark stream]

anyone know if this is correct

[restive plinth]

How to do this question?

[marble mauve]

You have to use number theory

Basically you have 3 digits you have to place

And 7 numbers

The restraint is that you can't start with zero(Problem a)

So the possibilities for the first digit is 7

The second digit also has 7 possibilities(1 number is used up)

The final digit has 6

The rest is fairly simple

[vapid plaza]

This aint number theory tho🗿

[celest glade]

It kinda has to use number theory it seems though

But yeah this is something like finding the number of cases problem

[vapid plaza]

I guess the only thing slightly “number theoretic” Is the fact that even numbers have even last digit

[craggy solstice]

f ( x ) = 3 − 21 · e − x , x ∈ R .

Task: The straight line with the equation y = 12 x + 25 is a tangent to Kf . Describe how to obtain the coordinate of the corresponding point of contact

I dont know how to describe it

wait is this the wrong chat?

[opal vault]

not much, just be very thorough with ur algebra knowledge, it would help if u knew algebra 2 but not much. also u need to have extensive knowledge in trig, trig identities, the graphs, the equations, all of that. also make sure u know logs and exponential stuff, that always helps but if u have a good book, it should teach u a lot abt logs itself.

calc is more like another idea like algebra is, so instead of having like addition and subtraction, u have like derivatives and integration

so in my opinion u dont need to know a looot of knowledge to dive into calc just make sure u grasp the ideas rather than memorizing the identities

cause there are too many to mem

[jaunty crow]

In the function (x-4)^2
The interval will be [0, infinity) or (0, infinity)?

[main rapids]

hello guys i need a bit of help with this question and im not too sure on how to go about it

this question as well

[opal vault]

it gets hella confusing once u get to integration in my opinion

im talking abt when u learn stuff like u sub or integration by parts

trig sub especially

[main rapids]

i can do basic differentiation and integration

im just having some issue understand what is going on

applications for me is kinda hard

no as in this topic is applications of ... because its not just the thing by itself but your doing it in an irl scenario, like normal calculus is basically having the eqn and just having to do the math, you need to understand what they're saying

[opal vault]

related rates?

calculus normally does not have word problems

[opal vault]

that comes when u understand

the concepts

[main rapids]

mhm

[inner stag]

how do you understand them

Im 22 but no

[opal vault]

yeah im not one with math either

unfortunately

but i mean that if someone gives u a calc prob, u should understand what it is asking for, and be able to apply the right tools to solve it

and some theorems/methods for example finding volume, should be kinda instinctive and something u can derive everytime rather than mem

[cyan wave]

can i get some help with this

[viscid thistle]

yep

I know ln and how to convert to exponenetial form

will I need to know about polar coordinates?

[burnt nymph]

I'm trying to test out Descartes Rule of Signs with R. This function afaik should have 3 sign changes, and with it 3 possible x-intercepts or one positive x-intercept:

#### Function 1 ####
fx1 <- function(x){
  (-4*x^7) + x^3 - x^2 + 2
} # 3 sign changes (from pos to neg), 1 pos x int.

When I plot it, it gives me one positive x-intercept, which looks like it matches:

#### Check Intercept ####
ggplot()+
  geom_function(fun=fx1,
                xlim=c(-1,1))+
  geom_vline(xintercept = 0,
             linetype = "dashed")+
  geom_hline(yintercept = 0,
             linetype = "dashed") # only crosses once as expected

However, when I evaluate f(-x), I get unexpected results

The negative should be coded like so:

#### Evaluate Negative Fx ####
fx1.neg <- function(x){
  (4*x^7) - x^3 - x^2 + 2 # change all odd powers to -
} # 2 sign changes, 0 negative

And when plotted:

ggplot()+
  geom_function(fun=fx1.neg,
                xlim=c(-1,1))+
  geom_vline(xintercept = 0,
             linetype = "dashed")+
  geom_hline(yintercept = 0,
             linetype = "dashed") 

The relationship is opposite of the last result, however, the x-intercepts don't match

If using Descartes rules, shouldn't this have either 2 or 0 negative x intercepts?

[warm jay]

abc for apple juice

[burnt nymph]

?

[fierce anchor]

is a definite integral treated as a cosntant

[charred frigate]

it depends. if the bounds are constant, then yes.

$f(x)=\int_0^x2s\dd s$ is a definite integral which is not a constant

[fierce anchor]

OHHH SHIT TRUEEE I FORGOT THAT THE BOUNDS CAN BE PARAMETERS AND SHIT

SO FUCING REAL FOR HTAT

then

how would you differentiate something like that

because i was gonna differentiate this integral but then i thought nothing would happen because it would be treated as a constant, but i have a parameter t in the bounds

wait maybe i just integrate then differenatiate

with respect to t

[obsidian monolith]

nixxy nilpotent (raving lunatic)

[fierce anchor]

TRUE

HOLY FUCK

THAT MAKES SO MUCH SENSE

[charred frigate]

if it's just the parameter by itself, it's easy

$\dv{x}\int_a^xf(s),\dd s=f(x)$

[obsidian monolith]

nixxy nilpotent (raving lunatic)

[fierce anchor]

YOURE SOO FUCKING RIGHT

THANK YOU NAMASTE

[warped cipher]

yes

[thorn hare]

how do i add vector

:(

nvmd i figured out

[glass dove]

just asking, but aint that calculus?

[vocal current]

dawg that was yesterday

ion think they gon reply

but who knows

[glass dove]

@vocal current @crystal sky if $f(x)=\frac{x^2}{2} - x+7$ then what is $f(3)?$

[vocal current]

17/2

[obsidian monolith]

Belgian

[glass dove]

how

[vocal current]

plug in x

[glass dove]

replace the x with 3

[vocal current]

ye

[glass dove]

3^2= 9, and 9/2 is 4.5

then 4.5 - 3 is 1.5

and 1.5

+7= 8.5

[vocal current]

yeah

[glass dove]

ooooooh

you just expressed it the hard way

lmao

[vocal current]

well i like fractions better sometimes

[glass dove]

lol

can you plot $f(x)= 5x-3$ on a graph?

[obsidian monolith]

Belgian

[vocal current]

are you doing hw

[glass dove]

no

its vacation

so why would i do homework lol

[vocal current]

idk

[glass dove]

lol

but can you graph it?

[vocal current]

yeah

[crystal sky]

17/3

btw

[celest glade]

Fyi fractions are much more frequently used than decimal points when it comes to calculus

But as long as the value is correct it works

[glass dove]

This is precalc

[celest glade]

Yeah but we use precalc in calc :)

Not to get too picky about it tho

[opal tree]

Decimals kinda stinky

[dense grail]

yoo

[wise marsh]

What question u answering because for the one above that's wrong

[crystal sky]

oh

I did not notice

[wise marsh]

U just put a 3 where a 2 should've been

Nothing bad just typo

[crystal sky]

can you slove this question

3x=1 slove

[wise marsh]

Bruh ur just trolling

[wise marsh]

x= -1/12

[crystal sky]

Nice

[viscid thistle]

Focusing on the epsilon delta definition of a limit, why is the derivative of the function multiplied by delta equal to the value of epsilon?

[hushed sphinx]

That doesn't even make sense.

[viscid thistle]

But it works like that

I think,m

?*

I might have been unclear

Ok so this is the limit, ignore the language

[viscid thistle]

You take the derivative of the function and multiply it by delta and for some reason it gives the value of epsilon, I think. I'm not sure if I.may be wrong

This is the classic way anywaus

[hushed sphinx]

If you're looking at a correct limit, there will be some pairs of epsilon and delta that have the property in the limit definition. The definition requires that for each epsilon, there must be at least one delta that works-- but if there is one delta that works, there will be many that work, and they won't all have the same ratio to the epsilon they were chosen for.

[viscid thistle]

What does that conclude

[hushed sphinx]

That it makes no sense to say "the derivative of the function multiplied by delta equal to the value of epsilon".

[viscid thistle]

Ok but how should I call that process anyways and why does it occur?

[stark shuttle]

State how many complex and how many real zeros the function has: f(x) = x^3 - x + 3

[viscid thistle]

?

[fierce anchor]

i think so but idk where to put it

im a high school student in australia and we dont call it ap calculus, or caluculus or whatever, it goes in levels of standard, advanced, ext 1 and ext 2

atleast in the state that i am in..

how come there is no claculus chat

dont you guys also have a calc curriculum?

[lucid trench]

it's in the early university section

[fierce anchor]

section is missing for me 😭

[celest glade]

The delta-epsilon states that for any arbitrary ε, that satisfies |f(x)-L|<ε, if δ exists that satisfies |x-a|<δ for any ε, the limit value of f(x) approaching a is L

In other words, you can't conclude that δ or ε is relevant to each other or something as there can wxist multiple values of δ for one value of ε

In other words, it is not worth it to multiply by δ or ε

[floral ore]

log3 base 4 should be positive right?

a book I was referring to wrote that it was negative

which did cause me confusion

[willow bear]

log_4(3)?

[floral ore]

yep

[willow bear]

yeah, it is positive. can you show the page from the book?

[floral ore]

alright

Honestly, could be a misprint

or maybe the solution was more checking whether x was greater or less than 1

[willow bear]

yeah could be either of those things

[viscid thistle]

Yeah I see, but what would be the name of the process then or how does it work?

[celest glade]

https://youtu.be/kfF40MiS7zA

Here's a 3B1B video that I find them explaining well how we should interpret epsilon-delta arguement visually , from 4:52
Epsilon-delta arguement is more like "do we know that δ will exist if I give you any values of ε, no matter how I give the value very small?" and it's more focused on the EXISTENCE of δ, NOT the value of δ

So dividing δ by ε or multiplying it or such is just not meaningful in any way

Since that approach is more focused on finding the value of δ or ε

So in the case of derivatives:

For any values of ε (examples: letters written in red, or letters written in blue), we can see that the value of δ EXIST and can be anything,
if the limit of (f(a+x)-f(a))/x exists at x=0

This is what epsilon-delta arguement is

We don't care about the values of δ as long as it exists, hell even it can be something small as 0.0000001 or as large as 100 as long as both values makes if 0<|x|<δ, L-ε<(f(a+x)-f(a))/x<L+ε

[celest glade]

So your answer to the question

It may or may not

We will never know if δf'(x)=ε

[celest glade]

Because, this whole thing (which is epsilon-delta arguement) does not show any relevance with what you gave: δf'(x)=ε

[viscid thistle]

Holy shit.

Woah thanks a lot let me just read it

[viscid thistle]

I see, so that's the classical way on proving if a limit exists right?

[celest glade]

It is a definition of what limits are

[viscid thistle]

The only part I don't understand whenever I encounter the definiton by using the graph is the reason why they add a range/radius to the limit created by f(x) and the x value approached

[viscid thistle]

Why is it always mentioned if it may be small or large, what is its impact on it?

[celest glade]

"Approach" is not defined well mathematically, so Augustin-Louis Cauchy decided to define what it exactly is using ranges, showing that radius of range by δ and ε

[celest glade]

for example, let's say ε=1
If when 0<|x|<100 , L-1<f(x)<L+1, δ can be any value between 0 and 100 or 100

As long as ε=1

So yeah

in this case ε/δ can be any value

That doesn't mean anything meaningful when we try to define the limit though

[viscid thistle]

I see

[celest glade]

Because, x and δ or ε is not related

Once again, we're only interested on the existence of δ when we're given a specific value of ε

Btw irrelevant question but do they teach epsilon-delta in your country in highschool?

[viscid thistle]

What would happen if delta is a value not included in that range? the limit does not exist?

Let me ask it again. The limit only exists if the value of delta is within that range?

[celest glade]

No

Like I said

[viscid thistle]

kinda

[celest glade]

We're only interested in the existence of δ that makes the following satisfied

[celest glade]

It does not matter if there is a value of δ that does not satisfy the equation

The important fact here is that there EXISTS a value of δ that makes the thing satisfied

[viscid thistle]

What causes delta to exist?

[celest glade]

Idk, the graph of how the function looks, and the value of L and ε

[viscid thistle]

Is it just the allignment with the L-ε and L+ε values?

[celest glade]

It depends on the function

[celest glade]

Not necesarrily

I'm not sure how to explain things when it gets to philosophical things but

The important thing is

[viscid thistle]

Oh

[celest glade]

We found that delta exists

So we can say the limit exist

If delta does not exist for a certain value of ε, we say the limit doesn't exist

Think of limit as a property of a function

And epsilon-delta is a way to describe that property fully mathematically

[viscid thistle]

Oh I see

[celest glade]

Please do review epsilon-delta several times so you don't get confused about it

It's quite confusing when you learn it for the first time, and I was as confused as fuck as you were when I first learned it as well

[viscid thistle]

Yeah I'm on that

[viscid thistle]

Oh yes

[celest glade]

If you're still unsure those youtube videos are for you to explain things visually

[rough moon]

why is the area of a parallelogram defined by vectors (a,d) and (b,c) ac-bd

/cross product

[amber scaffold]

Hi

[summer ruin]

https://youtu.be/GwZuASZt0e0

[rough moon]

ok

[summer crag]

|A×B| = |A| |B| sin(angle b/w A & B) , area of a parallelogram with sides A and B is base times height, sin(angle b/w A & B) = height/adjacent side(either A or B).

So the denominator term in sin(theta) get cancelled with either A or B giving height times base

[rough moon]

i see, thank you

[cyan wave]

guys i need help like asap can someone dm me

[summer ruin]

can you guess the third root?

[cyan wave]

no 😭

or like idk what ur even asking i havent payed attention at all

[summer ruin]

if -6 -4i is a root of a polynomial, then what would be another root?

[cyan wave]

uhhh

4i?

[summer ruin]

no

[cyan wave]

bruhhh

[summer crag]

conjugate pairs

[cyan wave]

english pls

[summer crag]

comlpex roots are always in conjugate pairs

[cyan wave]

what does that mean

[summer crag]

conjugate of a complex number z = x + iy is x - iy

[cyan wave]

bro what

u lost me

[summer crag]

you understand this ?

so for example 1+3i is the conjugate of 1-3i

[cyan wave]

oh yeah

that makes sense

[summer crag]

now can you guess the other root ?

[cyan wave]

no 😭

yes

[summer crag]

this ?

[cyan wave]

no

or like

kinda

WAIT

IM SO DUMB

I JUST RE READ THE QUESTION

I thought -6 and -4i were 2 seperate zeroes 😭

[summer crag]

hmm so can you now guess the third root ?

also you understand that (x-4) is a term in the given polynomial of degree 3 ?

[cyan wave]

yeah yeah

[summer crag]

great you're good to answer it then

[cyan wave]

wait so (x-4)(x-6-4i)(x-6+4i)!?

im so smart

[summer crag]

just multiply the last two terms which have complex numbers

and write it in quadratic

[cyan wave]

done n done

tysm 😭

🙇‍♂️

[muted turret]

Rational question. BC workbook

[willow bear]

,w 1/6 + 1/8 - 1/10

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by ann_dec#0

[willow bear]

,calc 120-5*23

[obsidian monolith]

Result:

5

[willow bear]

@muted turret did you just want confirmation regarding this work & answer or was this written by somebody else and you wanted an explanation of what they did?

[granite basin]

9

[muted turret]

This was my work. Just showing this work as a sample

The things right

[viscid thistle]

what have you tried?

[untold spindle]

<@&268886789983436800>

[tardy glen]

Hello guys, do someone have the demonstration of this pls ?

[hushed sphinx]

The simplest is to write y^x = e^(x·ln(y)) and use the chain rule.

(But I don't think this is meaningfully __pre__calculus).

[tardy glen]

Yeah sorry i don t have acces to the calculus channel

OK thanks for the information i will do some research on the chain rule

[hushed sphinx]

Hmm, it shouldn't need any special privileges to see it, but Discord does some funky thing where it sometimes hides channels it thinks you're not interested in. You can try clicking "Channels & Roles" and select the "browse channels" tab to get a fuller list.

(You'll probably get to see it automatically if you switch from "pre-university" to "undergraduate" -- the topic division follows American tradition of considering calculus to be something first taught in college).

[tardy glen]

Yeah you are right, when i joined this server discord asked me what kind of things i wanted to see.
The thing is that in english and in french the words "college " and "university" don t have the same meaning at all, so my mistake was here

But fine now i can see all the channel

With the tips you said

So, thanks you for the help and thanks for being a good moderator
Have a great Day

[gray magnet]

What 3b 1b video

It’s explain everything thing’s pretty well

[dense blade]

can someone please explain to me what the average rate of change means.

[hushed sphinx]

It's pretty much defined as [amount of change in total] / [time it took for that change to happen].

This makes most intuitive sense if the time it took is a large number of units, e.g. seconds.
Then the average rate of chage is the amount of change that -- if that amount of change kept happening consistently every second -- would have led to the actually observed total change after the time period has passed.

[dense blade]

then why are the outputs subtracted instead added together

[summer ruin]

because subtraction measures change, while addition measures average

for a line segment [a;b] the point x = (b+a)/2 is the midpoint, while the value (b-a) shows how much you need to add to left point "a" to get to the left point "b" (in other words how much change needs to be made to "a" to get to "b"), indeed a + (b-a) = b

[dense blade]

ok ty

[hushed knot]

So, what kind of math does this involve? https://www.ccvx.nl/VoorbeeldTentamensEngels/VoorbeeldTentamensWiskunde - EN/Voorbeeldtentamen wiskunde B 1 - EN.pdf https://www.ccvx.nl/VoorbeeldTentamensEngels/VoorbeeldTentamensWiskunde - EN/Voorbeeldtentamen wiskunde B 2 - EN.pdf

I have a matriculation exam to pass if I want to go to university.

I'm 23, but better late than never.

[thorn creek]

this is a mix of precalculus and calculus by the looks of it

some analytic geometry

i saw trig but didn't look to see if it was straight trig or caluclus involving trig functions

[quick jolt]

could someone mind trying to explain how to find the coordinates of the terminal side of an angle on a unit circle given the arc length? my textbook tried but it didnt really make sense

[quick jolt]

speaking of trig...

[thorn creek]

i don't know of a good way to approach this without invoking calculus

there are ways to do it without calc, but i don't remember them

[quick jolt]

well do your best

anything is better than nothing

[thorn creek]

you will be better served by someone else's tuition

[quick jolt]

ok

ill go consult a guy with a random accent on the internet. dont let me down now pls (idk if youll get the joke or not but thank you for trying to help :P)

[summer crag]

Isn't the coordinate just (cos(theta),sin(theta)) , theta = arc length/radius = arc length (becuase it's a unit circle)

[burnt nymph]

This video series has been excellent so far, but a specific video today has been confusing me. I get that you factor the equations, use long division, etc. but I'm not understanding when solutions belong to "the real number system" and "the complex number system" here. I must have missed that in a previous video:
https://youtu.be/xuhk2kSVwe0

[thorn creek]

you understand that not all quadratic polynomials have real solutions, right?

all linear polynomials with real coefficients have exactly one real solution

all quadratic polynomials with real coefficients have exactly two solutions (up to multiplicity), but only some of those will have solutions that are real numbers

[burnt nymph]

Yes I understand all of that

I guess I'm more lost on what are the last steps here sometimes

I think he's done and then it continues going on until there is more

[thorn creek]

i guess i don't understand what your question is

[burnt nymph]

Felt like I forgot a lot along the way

For example this part:

[thorn creek]

in this context, it's probably easiest to uinderstand that all of the solutions of a polynomial are complex numbers

[burnt nymph]

I think he's done when its x = 5i and x = -5i

[thorn creek]

and that some of those (those with no imaginary part) will also be real numbers

[burnt nymph]

Wait nvm

I think I figured that out

[thorn creek]

at that point he's reduced the polynomial to a product of linear terms with complex coefficients

since each linear term contributes one root, that gives you a polynomial with four roots

[burnt nymph]

Yeah so that part I understand too

Once you have something like a five order polynomial, there should be 5 solutions right?

[thorn creek]

yes

that's the fundamental theorem of algebra

[burnt nymph]

So that part I understand and I get when you use long division and all that but I think the complex numbers bit is what messes me up

But I think I just figured out my issue

[thorn creek]

it is a bit confusing because of how we write complex nyumbers

[burnt nymph]

My brain just wasn't connecting the pieces til just now

[thorn creek]

(x-(a+bi))

[burnt nymph]

Yeah

I think before when it was focused on real numbers and when he began with complex numbers it was simple enough

Now that I'm doing this it throws me off

[thorn creek]

it's a bit easier if you think of complex numbers as a single number and ignore the fact that they're made of two "pieces"

[burnt nymph]

But ive also forgotten the initial steps we learned like Descartes Rule of Signs and Zero Whatever the Name is

Rational Zeroes Something

[thorn creek]

i can never remember that one either

[burnt nymph]

The joys of getting older. Miss learning this stuff when I was really young. Much harder to pick up now.

Anyways thanks for the discussion.

[thorn creek]

np

[burnt nymph]

Sorry I have one last question. Why is it that the left part with the real number solutions and the right part with the complex number solutions match, but he does an intermediary step where he subtracts 2 from zero and divides 1 by 3? Is this just to show the logic behind why he square roots the unfactorable x^2 + 9 ?

[thorn creek]

those are to demonstrate the values of the roots

a factor of x+2 corresponds to a root of -2 (x+2=0 => x=-2), and a factor of 3x-1 corresponds to a root of 1/3 (3x-1=0 => x=1/3)

similarly the two imaginary roots of 3i and -3i correspond to terms of (x-3i) and (x+3i), which when multiplied make (x^2+9)

remember the fully factored polynomial is zero exactly when any one of the factors is zero

[burnt nymph]

Ah right

[knotty sequoia]

I have a question: .
The graph of a function that is to the 4th root or with an even index number is:
B and C only
Broken
Continuous
Always straight

[glass dove]

Is there someone that wants to talk precalculus

[celest glade]

I didn't understand the question, (on the part B and C only)could you take a pic of your question

[celest glade]

So I assume B and C only is broken and continuous only?

And "Broken" in this context means piecewise

[zinc fulcrum]

there was a discussion about this question already and everyone decided it made no sense for like 4 different reasons at least

[knotty sequoia]

Yup

[celest glade]

Yeah ofc because it isn't specified what the function is

Let f(x) be x⁴ and (f(x))^¼= |x|

So it can be a piecewise function

While let f(x) be 2x² and it doesn't become a piecewise function anymore

Both functions are not straight

So I assume the answer is probably "continuous" unless someone pulls up a counterexample where (f(x))^¼ isn't continuous where it is defined

Whoever made this question needs to be bonked

"Broken" is not even mathematically defined well here

[thorn creek]

i also don't know what an "index number" is

or what it means to be "to the 4th root"

also is |x| "always straight"?

what does "always straight" even mean

[zinc fulcrum]

and what does broken mean

honestly one of the worst questions i’ve ever seen

“a function that is to the 4th root” kills me

[knotty sequoia]

An Index I believe is the number above the radical

[thorn creek]

honestly did not know anyone had given a name to that

[knotty sequoia]

Me neither

[quick jolt]

I was able to get a graph and find the solutions using a graphing calculator, but how would i solve that algebraically?

[dense mango]

You have to know values of trig functions for 30 and 60 degrees

Tan x = √3 for x = 60° or π/3

But you're also adding the period

So you get π/3 + nπ where n is an integer

[viscid thistle]

-sqrt(3)

so x should be 2pi/3 + pi*n

but you gotta only find answers in the interval from -2pi to 2pi

[knotty sequoia]

apparently the graph was broken

[burnt nymph]

I have no idea what is being "proven" here
https://youtu.be/x5cWX-EyLEI

I love this guys videos, but he spent like 15 mins explaining how this works and I'm still completely lost about what he actually did

[wicked viper]

i think he's just trying to show how stuff gets proven by mathematical induction

so the two formulas on top are his lab rats to test mathematical induction on

based on the work on the board (too lazy to watch the whole thing lol) it looks like he's just plugging in consecutive numbers into the formula to find a pattern

[lucid trench]

probably discontinuous

[tacit prairie]

That 1+3+5+...+(2n-1)=n^2. It is right there at the top center for the entire video.

[celest glade]

And whoever made this question needs to be bonked for making the question bad

🔨🐢

BONK

[burnt nymph]

I can read

Im more looking for the intuition here

[burnt nymph]

Yes I understand that so far, but Im just lost on how this proves it

[charred frigate]

here is an induction explanation I got from a professor a few years ago

the process of proof by mathematical induction is like proving that you can climb an infinitely long ladder to the heavens. to climb an infinitely long ladder, you only need the ability to do two actions, and those actions alone can allow you to climb as high as youd like, given enough time.

the first action is climbing onto the ladder in the first place. if you cannot get on the ladder, how can you climb it? this action is the base case.
a simple verification that what youre trying to prove works for the smallest case which it is supposed to. usually just showing that plugging in n=0 or n=1 gives a true statement.

the second action is climbing from one step to the next. once youre on the ladder using the base case, you can simply perform this action over and over again until you're as high as you desire. this is called the inductive step.
to do this you suppose that it's true for some value n=k (this is called the inductive hypothesis), and show that the fact it is true for n=k means it must be true for n=k+1 (the inductive step). so as long as you can stand on one step, you can climb to the next.

[burnt nymph]

Okay I think that makes more sense. Thanks for the useful analogy.

[dense mango]

I didn't want to give the solution, just a clue...

[viscid thistle]

i was saying the problem said -sqrt(3) not +sqrt(3)

you just made a small error

[dense mango]

I did that on purpose, I wanted them to figure out what to do when tan x = -60°

[short saddle]

The two red spheres are tangent both to each other and internally to the green sphere; the blue spheres, all of different radii, form a ring around the point of tangency of the two red spheres. How many blue spheres are there, and how are their rays related to each other?

[hushed sphinx]

I think you have the best hope of calculating anything if you start by assuming the green sphere has twice the diameter of the red ones, so the red/green contact points are diametrically opposite -- in that case the blue spheres are actually identical.
Afterwards I think one can appeal to inversive geometry to argue that the number of blue spheres will be invariant as you make the red ones smaller.

What does "their rays" mean here?

[hushed sphinx]

Oh, or better yet: invert the whole configuration with respect to a sphere centered on the point where the two red spheres touch. Then those two spheres turn into parallel planes, and every sphere that touches both -- that is, green and blue alike! -- become spheres that just touch the two planes, and therefore all have the same size. It is now pretty easy to count the blue ones.

[willow bear]

their radii i think

but run through a translator

[dense mango]

How to evaluate limit c without l'Hospital

[slim steppe]

multiply by conjugate

[dense mango]

I got this

,rotate

[summer ruin]

continue

[dense mango]

Idk what to do now

[summer ruin]

think about why you can't evaluate this limit still

[dense mango]

I have 2 square roots in the denominator so I probably have to somehow get rid of them

But am I really supposed to multiply by conjugate again?

[summer ruin]

are they a problem?

[dense mango]

Hmm

[summer ruin]

well if you fixed one of the roots that way

you can just repeat?

[dense mango]

I can't figure it out...

[summer ruin]

well what did you do

[dense mango]

I'm stuck on this, idk what to do next

[summer ruin]

didn't you want to multiply by the conjugate?

[dense mango]

Ok I'll try

[visual torrent]

logx(a^b) = b multiplied logx(a)
right?

[summer ruin]

absolute value of a

[summer crag]

If f and g are periodic functions (from R to R) then can we say that f+g is also a periodic function (from R to R) ?

for some p,q,c in R⁺, f+g(x) = f(x+p) + g(x+q) = f+g(x+c) = f(x+c) + g(x+c) for all x in R ?

here f and g are periodic fn (from R to R) with period p and q respectively

[vapid plaza]

No because the periods may never “match up”. For example, f may have period 1 and g may have period sqrt2

[summer crag]

hmm can you give a example ?

[charred frigate]

,w period of sin(x)+cos(sqrt2 x)

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by nilpotentnix#0

[woven quail]

Could someone help me understand how and why DeMoivre's Theorim works? Why do I have to convert the equation to trigonometric form?

[finite void]

im very confused... my teacher said square root of a real number cant be negative...
x = sqrt(56-x)
x^2 = 56 - x
x^2 + x - 56 = 0
(x-7)(x+8) = 0
x = 7,-8
then my teacher said -8 is impossible bcuz:
-8 = sqrt(56-(-8))
-8 = sqrt(64)
-8 = 8
i always thought squart root of a real number is always +- and not only positive

(idk if im in the right channel lol)

[summer ruin]

sqrt(5) is positive as is for any positive number

sqrt(x^2) ? well, depends on what x is

if x is negative, e. g. x = -1, then sqrt((-1)^2) = sqrt(1) = 1 = |-1| = |x|

if x is positive such as x = 1, then sqrt(1^2) = sqrt(1) = 1 = |1| = |x|

therefore sqrt(x^2) = |x|

[finite void]

so what u r saying is sqrt(<a real number>) can only be positive but sqrt(x^2) like sqrt(2^2) = |2|?

[summer ruin]

it's simpler than that, x is a real number just like any other

the problem is that it's not determined whether it is positive or negative

[finite void]

so my teacher is right or not 💩 im too stupid to understand... i almost argued with him but no one else in the class seem to agree with me that sqrt(64) can be -8

[summer ruin]

the teacher is right because it's pretty important that you learn it this way to not make silly mistakes like sqrt((-8))^2) = -8

but you're also right, but for a more complicated reason

[finite void]

thanks btw

[summer ruin]

(the reason being the existence of 2 squares roots, 3 cubic roots, 4 4th roots, etc thanks to complex numbers)

[cyan wave]

could someone explain this to me

[short saddle]

radius dian er

[buoyant aurora]

im gonna be in this class nextyear

looks very hard

im doing alg 2 over summer

[willow bear]

,calc 0.34*60

[obsidian monolith]

Result:

20.4

[willow bear]

1°20' maybe? hold on

,calc acos(1/4.286) * 180/pi

[obsidian monolith]

Result:

76.507517683468

[willow bear]

@cyan wave where did you even get 1.34° from

ah, looks like you switched your calculator to radian mode

[cyan wave]

idek what i was doing

[viscid thistle]

Is this equal to 1/4? when x equal to 0

(evaluating continuity)

I multiplied by conjugate and applied division by x in both, and then lim x->0 sin(x)/x = 1 but Idk...

[viscid thistle]

how do u integrate lnx?

[obsidian monolith]

square_pie

[viscid thistle]

$\int \ln{x}dx$

[obsidian monolith]

square_pie

[marble elk]

IBP

let u = ln x

and dv=dx/dx

[willow bear]

dv = 1 dx

[vagrant anvil]

math

[west swallow]

why range wrong?

[charred frigate]

you only look at the y values for the x values in the domain. i dont see any x values where the function has values of -100, or -100000000.

[woven quail]

your range is wrong for a couple reasons.
First of all, the function does not go down to negative infinity. It has a starting point at (-2, 0) and an ending point at (3, -5). This would mean that (-∞, y) would be a wrong range.
Second, your upper part of the range lands on the number 4, as you say in your range. But it includes the number 4, and doesnt exclude it. Therefore, you would want to use a "]" bracket instead of a ")" bracket after the number 4.

[cyan wave]

need help

[viscid thistle]

The diagram of the triangle??

[steep sapphire]

is a findable without diagram?

its 35 . √3 / 3

or just 35 . √3

if we cant find "a" I dont know any way to find c too

[knotty sequoia]

(-5,4)

[untold spindle]

Don't give out solutions, especially if its not even correct and has no explanation behind it.

[knotty sequoia]

No the range isn’t wrong. The range is between -5 and 4 because the graph is restricted to those Y values

[untold spindle]

The numbers you put are correct but the parenthesis are incorrect

[knotty sequoia]

I’m aware

[untold spindle]

Then why leave your answer as is instead of correcting it?

Actually that's besides the point. Point is, don't send the answer to a question in general. It helps the questioner more if instead you provide an explanation of their problem and give them nudges in the right direction.

[cyan wave]

can someone explain

nvm

[onyx saffron]

i need a bit of help

[ripe wave]

For the thing N sent?

[gaunt sand]

it touches the y axis and intercepts at 4 so that value is inclusive

in the range

so the [ bracket gotta be used to indicate that 4 is inclusive

and there is a restriction on the graph so it doesnt continue for forever so infinity wouldnt be right

[glass dove]

Im trying to learn functions

Are inverse functions like:

If $f(x)=2x\cdot(2-7x)$
Then
$f^{-1}(x)=\frac{2x}{2+7x}$

[obsidian monolith]

thatonebelgian61

[celest glade]

No

[glass dove]

I dont really understand it so

[celest glade]

Also f(x) does not have an inverse function in that case

Let's say y=-14x²+2x=f(x)

Look at the green locus

[gaunt sand]

Is it cuz it's not a function

Therefore it doesn't have an inverse

[celest glade]

Yeah

[glass dove]

I think i kind of get it

[celest glade]

If you say though y=-14x²+2x when x>=2 it has an inverse function, because it's surjective and injective(one-to-one)

Khan academy should explain better than me so check that

Sorry it seems I suck at explaining lol

[glass dove]

So lets say $$f(x)=2x+7$$,where
So $$y=2x+7$$
Then $$x=2y+7$$
Now we need to find x
$$x+7=2y$$
$$\frac{x+7}{2}=y$$
$$\frac{x+7}{2}=f^{-1}(x)$$

[celest glade]

You seem to get the essence

[glass dove]

Finally

[obsidian monolith]

thatonebelgian61

[celest glade]

Though there seems to be an error in your process

[glass dove]

Sorry for editing so many timss

[glass dove]

Where?

[celest glade]

x=2y+7 -> x**-7**= 2y

y= (x**-7**)/2

[glass dove]

On yeah

Mb

Was just a thought error

Whatever

Its vacation

We can make mistakes 😉

[celest glade]

But you do get how to find the inverse functions generally

[glass dove]

Yes

Also if one of the variables is $n^m$, then you need to turn it into $\sqrt[m] n$ at the end

[obsidian monolith]

thatonebelgian61

[celest glade]

depends though as you try for y= x², it doesn't have an inverse function(try graphing the locus for x=y²)

[gaunt sand]

Say for a question where like this:

Find the coordinates of the stationery points of the graph y=x^3-3x^2-24x

So Find the derivative

Dy/dx =3x^2-6x-24
3x^2-6x-24 = 0
3(x^2-2x-8)
3(x-4)(x+2)
X=4, x = -2

I plug it back in to find the two stationery points but how do I know whether it is a minimum or a maximum?

Or how would I figure that out

[celest glade]

To #calculus

[gaunt sand]

Oh yeah sorry

I forgot channel name

[warped cipher]

I mean you can use help forum if the help channels are full or smth

I do think they're both the same

I found comments like !status and !nosols don't work in help forum

It seems like mniip already said this #1026808241067405312 message

[digital pendant]

You can visualize the inverse like this:

Where you just change the direction of the arrows

[dense mango]

I have a question, why this

is not the same as this

I thought of this rule and I'm confused

[summer ruin]

because the expression under the root sign must be non-negative

[dense mango]

so basically if there is a possibility that the expression under the root is negative then I can't use this rule?

[summer ruin]

this rule already assumes that a>=0, b>=0

if a < 0, b < 0, then a/b > 0, yet the right hand side doesn't exist

[digital pendant]

Essentially the bottom function isn't defined for x>1/4 because then the 1-4x term is negative.

[dense mango]

oh I see it now

thanks guys

[astral moat]

cos 70

[cyan wave]

yeah i figured that it wanted me to add them

[burnt nymph]

In this part of a precalc video I'm watching, the inside of the cos function is multiplied by 3, so he adds another 2 pi to each alpha.

However, earlier he had something with a similar product on the inside of sin and added more to each alpha

Why is there a difference?

[summer ruin]

was this constained to some segment?

[burnt nymph]

I believe that he just said they were within each of the function's respective periods

Beyond that I don't recall any other commentary about constraining something to a particular segment

[summer ruin]

then these are just wrong

[burnt nymph]

Which one is correct then?

[summer ruin]

sin(x) = -1 is true for x = 3pi/2 + 2pi*k for any integer k

[burnt nymph]

So basically I just need to add as many 2pi as there are multiplications of theta?

[summer ruin]

no? this has no relevance to why the solution is wrong

[burnt nymph]

Oh hold up

This might not be the final part of what he did

Lemme check

[summer ruin]

writing sin(alpha) = -1 and then saying that alpha = 3pi/2 or 7pi/2 or 11pi/2 is wrong because it doesn't account for all solutions

[summer ruin]

because of this

[burnt nymph]

This was the full calculation of one of the problems

This was the other complete calculation

The solutions are supposed to be at the bottom

[summer ruin]

well as I was saying earlier this doesn't account for all solutions

[burnt nymph]

Hmm

Lemme see if there was a part where he clarified why he did this

Otherwise I wouldn't know where to begin with figuring out what is right

Does it matter if he is just talking about within the unit circle?

Idk im confused

[summer ruin]

not a single thing is right here

[summer ruin]

the unit circle is infinite, you can traverse it forever

[burnt nymph]

Okay I looked at the beginning and he said he is looking for all of the solutions between 0 and 2pi

Does that matter or nah

[summer ruin]

it does because then there's only finitely many solutions

[burnt nymph]

I guess this was what I was getting at but I didnt realize that til now

So for those two equations, what is going on with them?

I understand mathematically what he is doing for the parts where he is adding 2pi stuff to each part, I just don't get why they're different

[summer ruin]

you just make a change of variables 3*theta = alpha and solve the simplest possible trigonometric equation sin(alpha) = -1

after figuring out what alpha is you backsubstitute to find theta

[summer ruin]

I don't see where they're different

[burnt nymph]

For this one he is only adding 2pi

But in this one he is adding 2 pi and then another 2 pi

[summer ruin]

it doesn't matter how many you add

you want to find all theta that solve the equation that belong to [0:2pi)

so you pick more alphas than you could possibly need just in case, to not miss out on solutions

[burnt nymph]

I don't get it. You get to just make up how many you add? Why would he do it differently here if he could have just used one method?

[summer ruin]

because the point is not to memorize "what's the perfect amount to add"

if you want to figure out exactly how many needs to be added - that's more work

if you want to do it - you can

[burnt nymph]

I guess I'm just not understanding then how you are supposed to come up with these then

It felt simple enough when he covered it the first time but now this is just more confusing

[summer ruin]

because when you convert back to theta you will have to divide by some number, namely by 3

therefore all alphas that you have chosen will be scaled down by 3

what if you didn't cover all possible cases and all your thetas are inside the [0;2pi) interval? then you could potentially miss some solutions

if at least one theta appears outside this bound then you'll know for sure that you checked all cases

[burnt nymph]

I understand the why, im not getting the how

You are simulataneously saying "it doesn't matter how many you add" and also saying "but they need to be sufficient to meet [0,2pi]"

How do you achieve this then?

Its super confusing to me

[summer ruin]

you just pick enough

[burnt nymph]

So how do you know what is enough?

I think that is the meat and potatoes of my question

[summer ruin]

isn't that pretty intuitive that if you pick twice as much as you need you'll meet all possible demands?

you consider the interval 0 to 2pi, for alphas you pick 3pi/2, one that's 3pi/2 + 2pi and one that's 3pi/2 + 4pi

the third point is at least twice the required distance from first one

it's not a precise calculation, as I was saying if you want to do it precisely - that's more unnecessary work, but it's a rough estimate

of course if you guessed wrong you'll have to start over

[burnt nymph]

So why then does 2pi/3 not have the same calculation

Wouldn't it also be 2pi/3 + 2pi and 2pi/3 + 4pi

[summer ruin]

I mean if you want to be sure you can always include more

[burnt nymph]

Nevermind I think I figured it out after he explained another part

The key here is that you are doing some guess work on which of these solutions will eventually surpass 2pi?

I think you were basically saying that already but I'm just rewording it in a way I understand

In other words, the multiplier doesn't really matter until the end, you are just trying to figure out how many solutions you can use before later multiplying (here by 3)?

[summer ruin]

when you're working in terms of new variable alpha there's no multiplier at all

and you're not multiplying, you're dividing by 3

[burnt nymph]

Ah right dividing by 3

Okay I think I get it now

Thanks for your patience and wisdom Transparent

[topaz belfry]

Where from and how is 'e' derived? Because, as a high school student, I am aware of some of its special properties in calculus and logs, but how does its derivation allow/create the special properties?

[hushed sphinx]

Part of the wonder is that several of the many special properties can each be used as the basic definition and then the rest of them can be derived from it.

So if Alice defines it as "the number a such that f(x) = a^x satisfies f'(x)=f(x)" and Bob defines it as "the limit of (1+1/n)^n for n -> infinity", then each of them can eventually prove that their number satisfies the other one's definition. So there's no need to come to an agreement about which of them is THE definition.

(Meanwhile Claire defines it as the infinite sum 1/0! + 1/1! + 1/2! + 1/3! + ... (where 0!=1), and Dan defines it as the number a>1 such that the area bounded by the curves y=0, xy=1, x=1 and x=a is exactly 1).

An interesting fact is that each of these definitions can be viewed as really defining the same function exp(x), and then just say that e means exp(1).
(This is not equally obvious for all of them, though).

[dull tendon]

https://cdn.discordapp.com/attachments/1130035163649282058/1130035163905151026/doubtimage.jpg

Can someone point out the mistake in this

[thorny shore]

Also who uses z in u substitution.

Wait nevermind.

[dull tendon]

x for independent variable, y for dependent, next natural order is z :\

[dull tendon]

the correct answer is supposed to be 2/3 sin^-1 (x^3/2) + c

even my answer might be correct cause i can't find any mistake and, if i differentiate it I get back to the integral question

i guess it's just different because of inverse trigonometric transformations or something..

[opal tree]

,w graph -1/3 arcsin(1-2x^3) and 2/3 arcsin(x^3 / 2)

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by umbraleviathan#0

[opal tree]

They ain't equivalent whatsoever

[lusty phoenix]

Can someone show me how I can simplify a natural log I know there's a formula log a+ log b= log (ab) but I don't know if it's same for ln

[spice bane]

It is the same for ln

[sick summit]

how do I determine if this is an identity please?

[edgy tartan]

ln is just a log to the base e

[celest glade]

ln x= log x/log e= log(e, x).
Basically log with any base has the property you mentioned if the two has the same base

[peak delta]

can an augmented matrix have multiple row-echelon forms that give the same solution?

[summer ruin]

yes, but reduced row echelon form is unique

[snow flame]

I need help after this part.

How am I supposed to find the trigonometric functions using a calculator?

I know I can use this, except I don't know how to use it with a scientific calculator.

Please ping me so I see your message.

[dense mango]

You need to know that sin has a period of 2nπ where n is an integer, and also sin(-x) = -sinx

So your solutions would be -π/6 + 2nπ and -5π/6 + 2nπ

[thorn kestrel]

shouldn't the +-infinity instead just be undefined?

[dense mango]

I just noticed that lol

I thought it was undefined

Tf are infinities doing there

[thorn kestrel]

thats like saying 1/0 = +-infinity

[dense mango]

Yeah thats bs

[snow flame]

Yeah it should.

There's no proper mathematical sollution for it.

[snow flame]

Thanks, can you explain how you did that?

As well as how you would solve this problem with a calculator?

I'd really appreciate it.

[dense mango]

I think that solving it with a calculator is completely unnecessary

You just have to memorize values of trig functions for angles 30°, 45°, 60° degrees

And maybe 90° for sin and cos

@snow flame

[snow flame]

On my tests it's going to be useful though, right?

[snow flame]

What about 120, 135, 150, etc?

[dense mango]

You can use trig formulas, eg. sin(180°-theta) = theta

There is plenty of them though

But when it comes to solving trig equations with a calculator, I've never done that

[snow flame]

Really?

You just memorized the values?

[snow flame]

What do you mean?

[dense mango]

Something like this

I know it's not english but more or less you know what it's all about

[dense mango]

Yeah

[snow flame]

I can't see it clearly.

[dense mango]

Just zoom in

[snow flame]

Even if I zoom in, that's what I see.

[dense mango]

Weird

Maybe cause I'm on my phone rn

[thorn kestrel]

works fine for me

[snow flame]

Perhaps.

[snow flame]

Wait, nevermind it's fine.

I opened it in another tab.

[thorn kestrel]

just curious, what do the square brackets mean?

[snow flame]

So how would I know all of the angles for sinx = -1/2

Pure memorization?

[dense mango]

Normal parentheses but squareshaped to avoid confusion

[dense mango]

Yeah

[thorn kestrel]

okay i see

[snow flame]

How long did it take you?

[dense mango]

A couple of days

About 1 hour everyday

[thorn kestrel]

I guess you would have to memorise but i dont see why you cant just use it by itself

or at least memorise the important ones like 30, 45, 60, 90, 180 and figure out the others by formula

[dense mango]

Also, when did you start doing trigonometry?

[snow flame]

I'm learning Precalculus this summer.

I started the trigonometry unit like one or two weeks ago.