precalculus

[calm whale]

the number infront of r^8

[sick whale]

Let me try this

Isn't it r^4r^2r?

[calm whale]

2 is a coefficient infront of r

so yes it is

but im not sure if the other ones are just r^4 and r^3

*hint: they are not

=tex -2r9r^216r^4

is probably what you're looking for

[sick whale]

Sorry i'm still not quite clear on what you mean by expanding them all out.

[calm whale]

like kdude

(x+2)(x+4)=x^2+6x+8

[halcyon gate]

isn't this wrong? shouldn't the directrix be y=a?

[clever inlet]

Yeah

x = -a would mean a sideways parabola, which this is not

[quartz garnet]

[clever inlet]

Actually

I think that's a generic example

Not the one for the question

Wait, that still seems wrong

[quartz garnet]

The directrix of $$x^2=4ay$$ is $$y=-a$$ yawns

I'm tired qwq

[quartz garnet]

Fockin- I don't function I guess X'd

[viscid thistle]

get the sleep'd 😴

[potent nest]

i dont belong here but... I WILL NEVER UNDERSTAND

i will never do precalculus nor with i ever do calculus

[quartz garnet]

:p

[viscid thistle]

** don't say that, you have the ability to learn! **

[halcyon gate]

yeah it was wrong

thank you, i just needed to make sure i wasnt the only one thinking that

please help with this, where and why did they do (4-2)/2 ?

[quartz garnet]

it's the average of 4 and -2

[halcyon gate]

why did they take the average though

[clever inlet]

cause the vertex

is halfway between focus and directrix

wait

that doesn't look right

[hexed ermine]

you're right

[clever inlet]

shouldn't the distance be 6 units?

[hexed ermine]

Yeah

4-(-2)//2

3+(-4)

-4,-1

[clever inlet]

is that site just confusing x and y lines

lol

[hexed ermine]

Oh yeah the site is messed up lol

[halcyon gate]

oh man

this isn't good for me

if it's the site is messed up

studying my final on the 18th

[clever inlet]

I don't think that's a good site to use

Try to find an alternative

[halcyon gate]

it's chegg

paid for it :c

[clever inlet]

Rip

If it's paid, then they should be able to be contacted

So they fix the issues

[halcyon gate]

no time for that :c

Im glad I asked though

Otherwise I will be lead so wrong

[clever inlet]

Probably just use Khan for now

[halcyon gate]

yeah

thank you

[hexed ermine]

Or if you can just learn it

If the directrix is y=, then the vertex will be the midpoint between the y-value of the focus and the diretrix, and same for x= with the x-value of the focus

[halcyon gate]

okay, thank you. I'll try remembering that for the next problem

[hexed ermine]

np

[viscid thistle]

@jolly turret For ur trig one, it's super easy if you memorise the unit circle

[still yew]

also mark the coords of the form (cos(x), sin(x))

[viscid thistle]

yeh

if sin(theta)= 1/2

then theta = pi/6

-> cos(theta)= √3/2

[jolly turret]

@viscid thistle are you talking about a question i asked earlier? Or do you mean for my upcoming test

Although i do know the unit circle with pi and 2pi

I just need to get down cos and sin and how they relate to eachother

[still yew]

if based on unit circle

Drawn my best

[viscid thistle]

For ur test and the questio

This is supposed to be the first thing u learn in trig right?

[halcyon gate]

help pls

how did they get the two point on the side?

(-1,4) and(-1,0) ?

is it just symmetrical ?

[willow bear]

those are the points on the parabola that are on the line through the focus that is parallel to the directrix

the distance from them to the focus is thus the same as from the focus to the directrix

[halcyon gate]

but just how did they get that?

did they just use the vertex and diretrix

to get x an y and solve it?

[willow bear]

how did they get what?

the focus is 2 units away from the directrix

[jolly turret]

You should study the sections in foci and dirtectrix

[willow bear]

so those points you're asking about are 2 units away from the focus in either direction on the vertical line through the focus

[jolly turret]

if you dont get how you get those you probably didnt study well enough and shouldnt be tackling this problem

[willow bear]

because the directrix is vertical

[halcyon gate]

Ohh okay nevermind I understand, I got confused. I know that

[jolly turret]

You should take a refresher just to make sure

[halcyon gate]

yh

reviewing

thank you

@jolly turret @willow bear

[jolly turret]

I didnt do much but no problem

[halcyon gate]

made me think

good enough

lol

[jolly turret]

Hey guys im trying to study something but im not exactly sure what to search to find it. So what i want to study is how cos , sin, and tan relate to each other but i cant seem to put into words what i would search up. I know how you get each of them. i just want to see how they relate to each other because ive had problems where they say cos theta = 0 tell me what sin theta equal or something like that

[floral rose]

https://www.youtube.com/watch?v=ovLbCvq7FNA

@jolly turret

go to the 12:40 mark

somewhere around there

[jolly turret]

nice vid

[calm whale]

those guys had a really nice vid on tensors

[floral rose]

eugene khutoryansky is god

[calm whale]

KhanAcademy is god 🤔

@somber trellis Griff I'm just wondering, can we have a chat-room section with maths resources somewhere below read-me or something
good youtube channels khan academy etc not sure if there's a wiki we can steal from somewhere

[somber trellis]

huh, that's a decent idea, I'll look into that later tonight

[calm whale]

Thanks

[jolly turret]

i second KhanAcademy

[jolly turret]

Guys i took an assesment test after studying for quite a bit at least two weeks where all i did was study. So i take the assesment and it had good questions there were like 3 questions that i didnt really know too well like the exponents under radicals where you simplify. I take the test and it says i got a 64/100 and said i needed to study up on every subject i kid you not every subject says "needs review" mind you like 5 of the questions were fraction questions and even those said needs review. Do you guys think that test was a sham. I get that i might have missed some but i have been studying practicing on the placement tests for other colleges and was getting like 36/40s so i dont see how a 64/100 ever even happened with similar content. It seems like the kind of thing that tells you you had a lower score so they can rope you into taking their classes for those subjects

[blazing raven]

doh. where was the exam given and what was it for?

[jolly turret]

it was online and it was for math

for precalc style of things

for order of operations it said "not ready" then i take its little remdial lesson then says mastered. Seems more like a scam if i wasnt ready i wouldnt have been able to easily pass their remedial exercises

[blazing raven]

Well what level math?

Like Uni or HS?

[jolly turret]

its precalc along with geo/algebra so its more high school level math but for a placement test intended for college

[blazing raven]

Yeah. My suggestion is ... if there is any other way to place out using a more standardized test like the SAT General Math or ACT or SAT Subject test or CLEP go for that instead.

[jolly turret]

The thing is the test wasnt an official test it was a practice test from the same company where my college gets its tests

It was practice but it was quite anoying because i forsure didnt get that many wrong because i knew at least 90% of the content so there is no reason to get a 64% even if i got 2 or 4 incorrect

Its just odd that it doesnt even show you what you got wrong it just gives a score and what you need work on and for some reason mine said everything including adding and subtracting fractions like come on ive been adding and subtrscting fractions since second grade

[blazing raven]

Well it is just a suggestion if even possible at your college.

Not going to say who is right or wrong here :). But yeah. If there are other ways and opportunities to painlessly get the same result, look into it.

[jolly turret]

Hey guys how do you solve problems like these without a calc.

2sec(theta) - squareroot (3)
Solve for theta

[willow bear]

2sec(θ) - sqrt(3) = 0?

or what

@jolly turret

bc what you've posted there is an expression, not an equation

[jolly turret]

Yeah =0

[willow bear]

are you sure you meant sec and not cos?

[jolly turret]

im pretty sure it was sec but i might be wrong

[willow bear]

bc as it is, that becomes sec(θ) = sqrt(3)/2, and |sec(θ)| ≥ 1 for all θ

so that one has no solutions

[jolly turret]

say that it was cos how would you tackle that problem

[willow bear]

cos(θ) = sqrt(3)/2

this is one of those special values

[jolly turret]

does it have to do with the unit circle?

[willow bear]

i mean it's typically presented as being on "the Unit Circle™", as that sometimes comes with the easy angles marked on it

[jolly turret]

so it is 30 and 330

[willow bear]

in degrees and if you're instructed to only solve over the range [0, 360°], yes

[jolly turret]

oh i see cool then almost got into a calc 1 class missed it by 3 points and like 4 of the questions invovled working without a calc figuring out problems like that one

[willow bear]

oops, those are meant to be π/6, not π/3

you should know these ☝ off the top of your head

[jolly turret]

I havnt messed with the unit circle for like 2 months now so im a bit rusty but ill be sure to go back and get a refresher

[calm thicket]

I didn't notice that for sine and cosine it's just the square root going up and down respectively...

And of course tangent is just being weird as per usual

[deft shoal]

Tangent isn't being weird. 😃

$$\frac{1}{\sqrt{3}}$$

[calm thicket]

🤔

[deft shoal]

$$\frac{\sqrt{2}}{\sqrt{2}}$$

[deft shoal]

$$\frac{\sqrt{3}}{1}$$

[deft shoal]

😄

[calm thicket]

😮

[deft shoal]

$$\frac{1}{0}\rightarrow$$ undefined.

[calm thicket]

das cool

[deft shoal]

The first answer was rationalised, and the divide by 2 can be ignored since it's a constant in the ratio.

[calm thicket]

yea

[deft shoal]

It's the definition of tangent, in any case.

[calm thicket]

=tex \lim_{x\to\infty^+}\frac{1}{x}=\infty

[severe verge]

r u ok

i think that limit should be 0

:0

[neat trellis]

hmm

[willow bear]

what do you mean by $$\infty^+$$?

[floral rose]

positive infinity i assume

[hexed ermine]

Can someone please tell me how they got this answer and what they did?

[prime obsidian]

that should go before then

[prime obsidian]

ha is when x approaches infinity

so f(inf) is 3

[hexed ermine]

I understand how they got y=3 because you just divide the coefficients of the highest degree

[prime obsidian]

so then you just find the intersection of f(x) and y=3

[hexed ermine]

I was never taught limits

Is there some algebraic solution

[prime obsidian]

dont need limits for the second part

just the first part

[hexed ermine]

Yes I understand the y=3

I just wanna know how they got the (-4,3)

[prime obsidian]

f(x) = whatever

y = whatever

3 = whatever

solve for x

(x, 3)

[hexed ermine]

Oh I see

So you would multiply the 3 by x^2+x

[prime obsidian]

yeah

they wrote it out so it's like a cross multiplication

that's what the x is

[hexed ermine]

Ah I see thanks mate 😃

[prime obsidian]

np

[jolly turret]

cant seem to understand this

i understand the and problems but this or one has me confused

Does it work for infinity on both sides because its saying any number that can satisfy the first one then you also include the solutions that are included in the second one?

[willow bear]

errr let's see

why don't you try simplifying both of these a bit

[jolly turret]

y < 5

y > -6

[willow bear]

you'll get y < 5 for the first one, and y > -6 for the second one

yup

so the solution set of the first one is (-∞, 5) and that of the second is (-6, +∞)

[jolly turret]

i see i thought they capped each other but i guess it keeps going on both lines

[willow bear]

their union is the whole real number line

[jolly turret]

Thanks for the info

[hexed ermine]

Is Pre-calculus more of an algebraic course or a Calculus course? 🤔

[viscid thistle]

uhm

** no **

Its goes to complex algebra concepts

but covers, trig and some parts on limits and Newton-Rasphone polynomial solutions.

Algebra, Complex Algebra, Trig, some calculus

[jolly turret]

from what ive done in pre calc i havnt really seen much limits more like algebra on steroids and a little bit of trig

[hexed ermine]

Yeah same

[viscid thistle]

I was about to follow up that there is a differences

I do have some cool PDF's using assets from pre-calc sullivan books

however, the Algebra in pre-calc is called "College-Algebra"

[frigid hearth]

Find the value of sec(x) for angle x whose terminal side is at point (-2,-4)

What's a way to do this?

[hexed ermine]

Make a triangle from the point, the center and it's terminal side

[brisk root]

I seem stuck on a problem. I've exhausted my options it seems.

Number 2 and I'll send my work in a moment

[hexed ermine]

secant is defined as, 1/cos(x)

or just the reciprocal of cos(x)

[neat trellis]

the 1 + 1 becomes 2 and you can factor 2 out of the numerator

then you should be able to cancel terms

[hexed ermine]

cos is adjacent/hypotenuse so 4/sqrt(20)

So the answer is the reciprocal, which is sqrt(20)/4 or 2sqrt(5)/4

@frigid hearth

[brisk root]

I'm incredulously blind to the obvious. Thank you

[neat trellis]

np

[frigid hearth]

oooo i got it now thx @hexed ermine I just switched up the numbers on the sides of the triangle

[hexed ermine]

Yeah no biggie lol

[frigid hearth]

wait the actual answer is -sqrt5 @hexed ermine , so idk what happened

[hexed ermine]

Oops yeah my bad, hypotenuse over adjacent, 2sqrt(5)/-2 or -sqrt(5)

idk what I was doing lol

[hexed ermine]

Use the equation A=Pe^rt

You have the amount, time in years and the initial amount so it will read 2005=400(e^r(10)

divide by 400 giving you 5.0125=e^r(10)

Take the natural log ln(5.0125)=r(10)

Giving you 1.6119=r(10)

divide by 10

.161 which is 16%

[echo scaffold]

I did it and got it wrong :/

hold on

Question 2

[hexed ermine]

Oh shit I forgot to round to the two decimal places

And how they got the time to double is they took the natural log of 2 and divided it by the percentage (.1612)

[echo scaffold]

Round two decimals

[hexed ermine]

Ok now for this one you use the same formula 1705=200(e^r(10)

1705/200 = 8.525=e^r(10)

Take the natural log giving you ln(8.525)/10 = r

which is .2143

So percentage wise its 21.43%

Now to get time to double you take the natural log of 2 and divide it by the percentage

so ln(2)/.2143

3.2344 yrs

rounded its 3.23 yrs

[echo scaffold]

Wow thank you

[hexed ermine]

Np I remember doing alot of these in alg2

[echo scaffold]

this stuff doesn't make sense to me

[hexed ermine]

Haha just memorize the formulas

[echo scaffold]

Took my math final today and now just catching up on homework

and it's pretty frustrating since I dont need the class now

[hexed ermine]

Wew just took my final for precal

I feel stupid a bit lol

There were two parts that my mind went blank on

Partial Fraction Decomposition and Inequalities with a number line

[still yew]

How inequalities with a number line?

I would of thought of it as an easy section

[clever inlet]

Example?

[still yew]

I would give one, can't rn

[hexed ermine]

Ill give one

Straight from the test

Like this

I know it's easy, I was in the moment and I forgot

[fast tinsel]

man I never had any struggles with math ever, 'till we got to trigonometric functions and now algebra is getting to a stage where I can't find any solution without looking at the solution key to see how every step is done

We need to prove this one ^

but after trying so much, I'm doubting that it's even a correct statement?

[viscid thistle]

start from right hand side

hmm

(cosx - sinx)(1/sinx - 1/cosx)
= cosx/sinx - 1 - 1 + sinx/cosx

its correct

= cosx/sinx + sinx/cosx - 2
= cos^2x/sinxcosx + sin^2x/sinxcosx - 2
= (cos^2x+sin^2x)/sinxcosx - 2
= 1/(sinxcosx) - 2
= cscx * secx - 2

[fast tinsel]

ooooh

I see

thank you very much

[viscid thistle]

no problem

[viscid thistle]

@willow bear in the caption for this room, by expelled does it mean you aren't aloud to talk about?

[willow bear]

oops

that should be "Examples"

[viscid thistle]

Yeah I was confused

[willow bear]

i think autocorrect fucked up at some point

[viscid thistle]

Yea

Just making sure

[mossy pagoda]

doing Calc but forgot some stuff in precalc - but how is ln(125) = ln(5^3)

i know the log rule, and i know how to prove it

ln(5) * ln(e^3) = ln(125)

but i dont see how u can convert them to each other thru legitimate means

(or 3ln(5) = ln(125)

[mental maple]

ln(125) = ln(5^3) simply because 125 = 5^3

nothing fancy happened yet

the general rule is that ln(a^b) = b ln (a)

[mossy pagoda]

dam it seems so weird

ok thanks

[mental maple]

what seems weird?

[mossy pagoda]

o wait i mean how is 3ln(5) = ln(125)

[mental maple]

that's exactly the rule i just quoted

[severe verge]

thats one of the logarathm laws!

[mossy pagoda]

ye i know the rule

[mental maple]

ln(125) = ln(5^3) = 3 ln(5)

[mossy pagoda]

i know i know

i just wanna clarify to myself how it works

[neat trellis]

when you "ln" a number you can move the exponent down to become the coefficient

[mossy pagoda]

cuz it just doesnt seem right

[mental maple]

wow many people

one at a time please

[mossy pagoda]

i know the rule i kow the rule 👺

it just feels weird

[neat trellis]

true math is weird

[mental maple]

not really

math should be logical

[severe verge]

i disagree

[neat trellis]

logic l o l

[mossy pagoda]

how do u reverse engineer it tho

[severe verge]

well

to a certain extend 😛

[mossy pagoda]

from 3ln(5) to ln(125)

[mental maple]

=tex y = ln(x) ; \iff x = e^y

[mental maple]

right?

that's the definition

[mossy pagoda]

ye?

[mental maple]

ln is the inverse of the exponential

=tex \ln(e^x) = x, ; ; ; ; e^{\ln(x)} = x

[mental maple]

these two things work as inverses

is that clear?

[mossy pagoda]

how do u combine 3 and ln(5) to become ln(125) tho

[neat trellis]

you can move the 3 to become the exponent on the 5

[mental maple]

=tex \ln(a^b) = b \ln a

[mossy pagoda]

👺 i dont like that answer

[mental maple]

im using this rule from right to left

[severe verge]

well how about this

[neat trellis]

that's the simple explanation but it's a rule of ln like wizact put

[severe verge]

e^(x+y) = (e^x)(e^y)

[mossy pagoda]

okok its fine - im trying to find it thru the more basic definitions of log tho

[severe verge]

now take log of both sides

and u get

x + y = ln (e^(x+y)) = ln ((e^x)(e^y))

[mental maple]

@severe verge this rule follows from (e^a)^b = e^(ab)

you're thinking of the sum rule

[severe verge]

though i wrote it a bit wonky. maybe i shouldve written it in a way so that (e^x)(e^y) is easier to make sense of

ya i know!

i didnt specify which one i meant to do ~ ,~

[mental maple]

sure

[mossy pagoda]

okok u know what its fineeeeeeeeeeee my exam today aint abt this anyways LOL

[fast tinsel]

why is the domain of arctan(tan(x)) = ]-pi/2,pi/2[ and the domain of arctan(2x) = R?

[viscid thistle]

arctan(tan(x))?

[willow bear]

the domain of arctan itself is R

[viscid thistle]

what why

[willow bear]

because tan(x) runs over all real numbers as x runs from -pi/2 to +pi/2

[viscid thistle]

i mean why would someone write arctan(tan(x)) isnt that just x

[willow bear]

it's only equal to x when x is between -pi/2 and +pi/2

[fast tinsel]

well apparently not

[willow bear]

otherwise, it's equal to x - πk for some integer k depending on where x is

the domain of arctan(tan(x)) is R \ (π/2 + πZ)

[fast tinsel]

hmm I see

okay that should help me further, ty!

[viscid thistle]

help
Find sine (-) if cot (-) = 3/7 and cos (-) < 0

[dense zealot]

Ok

What is cot.

It's 1/tan right

So 1/tan = 3/7

So tan = sin/cos

So cos/sin = 3/7

Correct?

@viscid thistle

Ur not even on... ;-;

[viscid thistle]

srry

im back

[dense zealot]

Ok

So

7cos(-)=3sin(-)

We can use the identity sin x = cosine (180-x)

7cos(x) = 3cos(180-x)

Uh

:/

Cos x is less than 0

So that means it's in the 2 or 3 quadrant

Ok another indentity!

Cos(180-x) = -cos(x)

So 3(-cos(x))/cos(x) = 7

What

Waittt

I quit

[viscid thistle]

oh 😦

[dense zealot]

Cuz then -cos(x)/cos(x) = 7/3 and there's something wrong

So I obviously messed up

Oh I see where I messed up

Sin(x) = cosine(90-x)

So 7cos(x) = 3cos(90-x)

So cos(90-x)/cos(x) = 7/3

[viscid thistle]

ah

i already got it

[dense zealot]

Mk

[willow bear]

(-)

was that an attempt to approximate θ?

next time you can just type theta (or t)

[quartz garnet]

Pffhaha

That's pretty genuine though

(-)

[calm thicket]

(-)

[fair valley]

sinθ = θ

[viscid thistle]

no

unless (-) = 0

[willow bear]

[viscid thistle]

sin((-)) = (-)

[fair valley]

sin(z) = z

sin( 🍆 ) = 🍆

[viscid thistle]

🥖 = 🍆
🥖 = sin(🍆)

[untold wave]

baguette

[viscid thistle]

the axes are french bread and eggplant btw

🤔

[novel monolith]

"the axes are french bread and eggplant btw", there's a sentence I'll never see again

[dense zealot]

@novel monolith hi

By looking at this, you saw that sentence again ^^

[novel monolith]

@dense zealot , reported for bullying

[dense zealot]

xD

[gleaming schooner]

How would I do this problem?

[willow bear]

what've you tried so far?

[gleaming schooner]

I have so far

=tex \frac{2(t+h)^2 - (t+h) - 2t^2 - t}{h}

[glass venture]

you want =tex

and a \

[willow bear]

\frac

[willow bear]

=tex \frac{2(t+h)^2 - (t+h) - {\color{red} (}2t^2 - t {\color{red} )}}{h}

[willow bear]

you missed these

=tex -(a-b) \neq -a-b

[gleaming schooner]

What do ya mean 😮

[willow bear]

you have some expression

and you want to subtract 2t^2 - t from it

=tex (...........) - 2t^2 - t

[willow bear]

this is NOT now to do that

like say you had 1000

and you wanted to subtract 5-3 from it

5-3 = 2, so you'd get 998

but

1000 - 5 - 3 = 992, not 998

subtraction is done left to right

so...

[glass venture]

Another way to see why what you did isn't right which might help:

[gleaming schooner]

Hmmm

I thought I set my problem up similarly to how you set yours up

[glass venture]

=tex A - (x-y) = A + (-1)\cdot(x+ (-1)\cdot y) = A + (-1)\cdot x + (-1)\cdot(-y) = A - x + y

[gleaming schooner]

What's the difference between what you wrote and that 😮

I'm a bit confused

@willow bear

[glass venture]

the last part of the top should be +t, not -t

that's the error

[gleaming schooner]

Ohhh

I missed that

[glass venture]

so when you have -(something), you have to remember to distribute the minus to every term in parenthesis, that's where you messed up

[gleaming schooner]

Ahhhh

I understand

Alright, gonna finish working this out and see if I can get the same answer that you got.

Thanks

[glass venture]

👍

[gleaming schooner]

Alright, I got your answer 😃 Thanks @willow bear @glass venture

[languid wind]

if logA base b = mlobB base y =c, then find logY base a in terms of c & m

*mlogB

[willow bear]

=tex \log_b(a) = m \ \log_y(b) = c \ \text{Find }\log_a(y)

[willow bear]

this?

@languid wind

[languid wind]

yes

wait

no logb(a) = mlogy(b) = c

[willow bear]

huh?

[willow bear]

can you post the problem exactly as it is stated in your book/worksheet/whatever?

[languid wind]

find y in terms of c and m

*loga(y)

[willow bear]

...it's very likely that they actually meant what i said

and it was just badly typeset

[languid wind]

can it still be done the way it is typed?

[willow bear]

it can but the answer will be a bit uglier

anyway

this all hinges on one property of logarithms

=tex \log_x(y)\log_y(z) = \log_x(z)

[willow bear]

do i need to explain why that is true?

[languid wind]

1 minute pls

[willow bear]

so...?

[languid wind]

is the solution m/c^2?

[willow bear]

1/(mc)

[viscid thistle]

Oh goodie

Logaritms

[meager timber]

Not goodie

[viscid thistle]

Logarithms are great though

[narrow hawk]

Logarithms in precalculus??

[dense zealot]

Logarithms r alg tho

[viscid thistle]

pre-calc is advanced algebra???

or

fuck, I don't know ._.

[patent beacon]

Yeah

My school called it "advanced functions"

It turns out they actually discussed the elementary functions. Go figure

[severe verge]

same

[plucky locust]

Guys I forgot something really simple.

Like what do you do when you have to integrate a variable with power -1

like u^-1

You can't make it u^0

You did something with ln right?

[severe verge]

ya

antiderivative of u^-1 du = ln(u)

[plucky locust]

what was the concept behind it?

[severe verge]

well u know the power rule

[plucky locust]

yeah

[severe verge]

to do ∫u^a = 1/(a+1) u^(a+1)

[willow bear]

du!!

anyway

[severe verge]

when a=-1, u get 1/0

:0

[willow bear]

=tex \int \frac{dx}{x} = \ln|x| + c

[severe verge]

and dividing by 0 is a big nono

so ln happens instead for the exact case of -1

o ya i forgot absolute value

[still yew]

Why there is absolute power, your integrating a negative exponent

Value*

[severe verge]

it makes more sense cause ln(x) goes into complex numbers if x is negative

but the area under 1/x is completely real

plus odd powers make odd functions

and integral of odd functions are even!!

[plucky locust]

Guys.

How would I integrate sin(lnx)

[willow bear]

sin(ln(x))?

[plucky locust]

yep

[willow bear]

oh jeez

[plucky locust]

would I just directly write cos(ln(x))?

with a negative

and the derivate of ln(x)

[willow bear]

no

definitely not

that's very very wrong

[plucky locust]

yeah...

[willow bear]

anyway

try making the substitution u := ln(x)

then you can write x = e^u

so dx = e^u du

[plucky locust]

Yeah,.

Oh, I think I can do this now

[willow bear]

also integration is definitely calculus, not precalculus

[plucky locust]

oh, I don't really know.

I'm pretty new to calculus.

sorry. I'll ask all questions there from now on

[somber trellis]

@severe verge here's my reasoning for why you should never ever use ln|x|

the complex logarithm is multivalued

since e^(2pi i) = 1

now the reason why using ln|x| works is that when you use it, it's an integral from a negative x value to another negative x value

which avoids the singularity at 0

you can show pretty easily that ln(x+iy) = ln(x) + i arg(x+iy)

[quartz garnet]

[somber trellis]

so when you use the negative x values as inputs, your argument doesn't change and it cancels out and you can just use Re(ln(x+iy)) = ln(|x+iy|)

but that kinda.. obscures the fact that in reality you're using a complex logarithm

if you wanna avoid it entirely use a change of variables

but don't use ln|x| because it obscures the reasoning behind why/how you're doing what you're doing

all imo

[quartz garnet]

uhh....

[severe verge]

i dont know much complex number stuff :0

i barely know any ;o

[somber trellis]

I can show you some! they're super helpful

[quartz garnet]

=tex {color{red}f:\bb R\to\bb R,}~{\color{white}\frac{f'(x)}{f(x)}=x},~{\color{gray}x\in\bb R}

[willow bear]

@severe verge well you know the definition right?

[severe verge]

umm ya

[granite stirrup]

Rendering failed. Check your code. You can edit your existing message if needed.

[quartz garnet]

Damn

[somber trellis]

which definition do you mean?

[quartz garnet]

=tex i

[somber trellis]

I thought you meant for log

lol

[willow bear]

definition of complex numbers

that's what i was asking her about

well i mean there's two equivalent definitions

or constructions rather

the "pairs of numbers with specially defined + and *" construction
and the "here's a magic number" construction

[viscid thistle]

can anyone explain the domain rules for me

like how to find the domain in (f/g)(x)

Generally, you're looking for an input set of values such that the output of the function isn't not in the original input set.

So things like, output of function being a divide by zero or being a bunch of solutions - not just one. Are usually the two things that you're looking for in simple real functions.

[willow bear]

@viscid thistle the only "rule" is you start with R as your domain and you look for numbers that break the formula defining your function

[viscid thistle]

thank you !

[hoary yoke]

can anyone help with trig identity?

double angle identity?

[willow bear]

are you familiar with angle sum identities?

specifically

=tex \sin(x+y) = \sin(x) \cos(y) + \cos(x) \sin(y) \ \cos(x+y) = \cos(x)\cos(y) - \sin(x)\sin(y)

[hoary yoke]

im familiar with them

and i saw Blackpen Red pen doing vid on it

to prove it

[willow bear]

okay

then try rewriting sin(x+x) and cos(x+x)

[hoary yoke]

=tex \sin(x+x) = \sin(x) \cos(x) + \cos(x) \sin(x) \ cos(x+x) = \cos(x)\cos(x) - \sin(x)\sin(x)

huh

[willow bear]

\\

but anyway

\\, not \ \

[willow bear]

so

=tex \sin(2x) = 2\sin(x) \cos(x) \ \cos(2x) = \cos^2(x) - \sin^2(x)

[hoary yoke]

2 sinxcosx?

huh?

[willow bear]

multiplication is commutative, alex

[hoary yoke]

2(sinxcosx)

[willow bear]

sin(x)cos(x) + cos(x)sin(x) = sin(x)cos(x) + sin(x)cos(x) = 2 sin(x)cos(x)

@hoary yoke

[hoary yoke]

yes

im alive alive

[willow bear]

how is this not obvious??

[hoary yoke]

it is

im just used to do it with paranthethis

so i took it as 2SINa times cosa

:/

[willow bear]

same thing???

multiplication is associative

(a * b) * c = a * (b * c)

[hoary yoke]

how is 2sina times cos

and 2 times cos times sin same thing

or is it

wait im dumb

or just dum

wait

im alseep

ye ok so anyways

so

=tex \cos(2a) = \cos^2a + \cos^2 a -1

[willow bear]

no

[hoary yoke]

sry

[willow bear]

that is correct then

=tex \cos(2x) = \cos^2(x) - \sin^2(x) = 2\cos^2(x) - 1 = 1 - 2\sin^2(x)

[hoary yoke]

my mom threw away my trig papers i think

and im still mad about it

and she ripped up the papers about sinx/x being 1 as limit of x approaches 0

without permission and im still mad about it p-;

[willow bear]

and she ripped up the papers about sinx/x being 1 as limit of x approaches 0

huh what why

[hoary yoke]

idk

:/

i legit dont know

and she said she thought it was nothing

[willow bear]

yikes

[hoary yoke]

and i have to rewrite my whole thing about it

...

does cos2a

mean 2cosa

:/

[willow bear]

no

those are two very different things

[hoary yoke]

hm

ok

[willow bear]

...

2 cos(0) = 2 * 1 = 2
cos(2 * 0) = cos(0) = 1

[willow bear]

👀

[hoary yoke]

o.o

[viscid thistle]

Here it is

[willow bear]

okay

so you know that sin(x) and cos(x), by themselves, oscillate between -1 and 1, right?

[viscid thistle]

Yep

[willow bear]

yeah so for example sin(x) + 4 would oscillate between 3 and 5

[viscid thistle]

Yee

[willow bear]

i hope you're able to find the minimum and maximum values for the other three things yourself

[viscid thistle]

But

I tried doing that

But it’s wrong

[willow bear]

it's not wrong, it's an incomplete answer

[viscid thistle]

In the back of the book it says

Somthing else

[willow bear]

what does it say?

[hoary yoke]

cos would be 2 and 0

[viscid thistle]

1sec

[hoary yoke]

right?

the d)

[willow bear]

no, cos itself still goes between -1 and 1

[hoary yoke]

no i mean answer to the d question

[willow bear]

yes

[viscid thistle]

@willow bear apparently that’s supposed to be it

[willow bear]

yeah, so you've got the maximum and minimum values (3 and 5) there

for a)

again

not wrong

just incomplete

[viscid thistle]

How do I find those ones

[willow bear]

well

recall the unit circle

[hoary yoke]

min value is 3 spins of unit circle

😐

yep

[viscid thistle]

Discord died

@willow bear

Crash

[willow bear]

sin(x) = 1 precisely when x = π/2 and -1 when x = 3π/2

[viscid thistle]

So

Yeah

This is really hard

[hoary yoke]

but we need -3pi over 2 right/

?

[viscid thistle]

yes

so

@willow bear i don understand 😭

[willow bear]

ergh

i really should be sleeping right now tbh but

@viscid thistle you're familiar with the unit circle, right...?

[viscid thistle]

kinda, not fully

nowhere near fluent

[willow bear]

@hoary yoke this belongs in #calculus and that's false anyway

[hoary yoke]

ok

thanks

[hoary yoke]

i mean i just understood unit circle yesterday

I just minus the points

if they tell me to find for example sin of 210 degrees

i would minus that 180 bcs it is closest to that full angle

and i would get 30

now sin of 30

at 30 degrees i draw right triangle

[viscid thistle]

i know that stuff, some of it anyway, but im not fully familiar with

[hoary yoke]

define fully familiar

[viscid thistle]

like solving for triangles and shit

finding lengths

anything that isnt

just the exact values

[hoary yoke]

like using trig to find triangles side lengths and angle measures?

[viscid thistle]

mmm nah

[hoary yoke]

u dont know that?

😐

[viscid thistle]

i said naa. i do

[hoary yoke]

oh

Pythagorean Trig Identity?

do you know that

[viscid thistle]

i remember using them and the quotient identities last year,

dont really remember tho

[hoary yoke]

its:

=tex \sin^2x + \cos^2x = 1

[viscid thistle]

ye i remember that

[hoary yoke]

thats the identity

what grade may i ask?

[viscid thistle]

i hate identities

this is pre calc

my question is

sad that no1 knows

what to do

[hoary yoke]

what ws your question again?

[hoary yoke]

can u find the critical point?

@viscid thistle so can u find the critical point of the sinx plus 4

[viscid thistle]

idk how to do that

[hoary yoke]

critical point is the point that when you take functions derivative it should equal to 0

so lets say

[viscid thistle]

OHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHh

wait

[hoary yoke]

ok

so

[viscid thistle]

i understand now!!!!!!!!!!!!!!!!!!

[hoary yoke]

derivative of sinx plus 4 is cosx

[viscid thistle]

THIS IS FUCKING EASY AS A DICK

[hoary yoke]

cosx = 0

dick isnt easy

[viscid thistle]

THANKS

[hoary yoke]

so continue

[viscid thistle]

well

u didnt really help but,

i thank you

anyway

i just look at the max values and min values according to that peanut, and change the peanuts

[hoary yoke]

cos is 0 at 3pi over 2 and pi over 2

BY UNIT CIRCLES

and there u go

ann found your 2nd min and max point

so gj

you solved your shit

[viscid thistle]

yeah

and im a dumb ass too!

im surprised i foudn this out

[hoary yoke]

it is pretty obvious

:3

[viscid thistle]

with the help

of u and ann

[hoary yoke]

oh

btw

wait

for me

https://www.desmos.com/calculator/p4uwbgpxaf

here

[viscid thistle]

yeah i just went on that

ok i know how to see the answers now,

BUT

hmmm

nvm no but

[marsh dew]

what would a series question classify as?

[viscid thistle]

algebra

?

kek

[marsh dew]

well off i go then

[hoary yoke]

byee

lol

[viscid thistle]

xddd

@hoary yoke how u so smarty

so ig its easier if i graph first but,

how would i find those answers without graphing

[hoary yoke]

i watch math videos

[viscid thistle]

same

[hoary yoke]

i remember blackpenredpen finding it with critical points

then i googled what is a critical point and then i solved it myself

then i told you

[viscid thistle]

yeah

its kinda hard 🕉

that question was enuf math for today

got me a head injury

[hoary yoke]

lol

[marsh dew]

why doesnt the tangent graph work in degrees?

why does it only work in radians?

[willow bear]

define doesn't work?

[hoary yoke]

uhm

o.o

[willow bear]

it's just stretched out by a factor of approximately 57

[marsh dew]

if you graph the tan graph in degrees and then in radians, they're a different graph shape

its not

[hoary yoke]

can u find cos30 divided by sin 30 without unit circle?

[willow bear]

what window are you viewing your graph in?

like, what ranges along the x and y axis are actually being displayed?

[marsh dew]

okay, im gonna put this differently

can you upload a tan graph here?

[willow bear]

=wolf plot tan(x) between -2pi and +2pi

[granite stirrup]

Assumptions
Assuming "tan" is a math function. Use as 🇦 a color instead
Assuming "plot" is a plotting function. Use as 🇧 a word instead

Timeouts
Plot

Query made by @willow bear
Data sourced from Wolfram|Alpha: http://www.wolframalpha.com/input/?i=plot+tan(x)+between+-2pi+and+%2B2pi
Do more with Wolfram|Alpha Pro: http://www.wolframalpha.com/pro/
🐺 Try out the new =pup command! It's much more concise.

[marsh dew]

=wolf plot tan(x) between 0 and +360

[granite stirrup]

Timeouts
Plot

Query made by @marsh dew
Data sourced from Wolfram|Alpha: http://www.wolframalpha.com/input/?i=plot+tan(x)+between+0+and+%2B360
Do more with Wolfram|Alpha Pro: http://www.wolframalpha.com/pro/
🐺 Try out the new =pup command! It's much more concise.

[marsh dew]

shit..

[hoary yoke]

isnt that just extended?

[marsh dew]

no wait

[willow bear]

err that's in radians lmao

[marsh dew]

thats .... wrong

[willow bear]

this is what tan looks like if you do it in degrees

[marsh dew]

THERE SEE?!?!?

[willow bear]

same shape

stretched out

[marsh dew]

NO ITS NOT

[willow bear]

yes it is

tan(x degrees) = tan(πx/180)

[hoary yoke]

if u converted those numbers in radians

it wouldve been same thing

[willow bear]

it's Really stretched out

[marsh dew]

look at the gradient when it passes throught the x-axis

the gradients are different

[willow bear]

yes! that gets changed when you stretch it!

[marsh dew]

at 0

[willow bear]

that gets changed when you stretch it!

[marsh dew]

wut

[willow bear]

try plotting y = x

[marsh dew]

oh i get it

[willow bear]

then stretch that graph by a factor of 5 horizontally

[hoary yoke]

Ann you dont have anything to do in these weeks bcs christmas and new year are coming up right?

[marsh dew]

thank you for pointing out my mistake in a way that would get to my dumb ass head

[willow bear]

new year's day is Today™

[hoary yoke]

but christmas?

u have a free time

but

[willow bear]

christmas has already happened

[hoary yoke]

its in 7th?

o.o

[willow bear]

my family rarely celebrates orthodox christmas if at all

this year we did catholic

because Reasons™

[hoary yoke]

aw 😦

we only celebrate strictly orthodox

why am i not sleeping goddamn sleep ghost

[marsh dew]

okay

i have one last one

[hoary yoke]

its 3:30

[marsh dew]

from what ive gathered this server has very maths specific people, not a surprise, im very happy to have found all of you

but this question has been haunting me for a year now

i must ask it

[hoary yoke]

yes

[willow bear]

go ahead

[marsh dew]

what is the highest point on the graph y=x^x^-1

[hoary yoke]

x to the 1/x ?

[marsh dew]

show working

[willow bear]

hhh

[marsh dew]

yes

[hoary yoke]

doesnt it have an asymptote?

[willow bear]

lemme just try to differentiate that

[hoary yoke]

wait lemme check

[viscid thistle]

graphing tan function in radisn is HARD

its HARD

[willow bear]

=tex x^{1/x} = \exp\left(\frac{\ln(x)}{x}\right)

[marsh dew]

it does have a highest point, i can tell you that much

[hoary yoke]

yep it has a horrizontal asymptote

[willow bear]

=tex \frac{d}{dx} x^{1/x} = x^{1/x} \frac{1 - \ln(x)}{x^2}

[marsh dew]

it has a highest point though

[willow bear]

yup

got an extremum

at x = e

[viscid thistle]

yo uhm

tricks to graphing tan

[hoary yoke]

1.445 by the graph

[marsh dew]

so its e?

[willow bear]

e^{1/e} is the y value

== e^(1/e)

[granite stirrup]

1.44466786

[hoary yoke]

yep

1.445

[marsh dew]

my head.... i need an hour for the workings

[viscid thistle]

yeah i feel like a new tumor every time i math

[marsh dew]

so wait, how did you work it out?

[willow bear]

took the derivative

set it to zero

[hoary yoke]

http://www.dummies.com/education/math/calculus/how-to-find-local-extrema-with-the-first-derivative-test/

she found extremum

[willow bear]

SHE!

[viscid thistle]

whats the difference

between. 1st and 2nd

[hoary yoke]

im used to saying he ok?

[viscid thistle]

derivative

[hoary yoke]

...

2nd derivative means

you differentiate what you differentiated

...

😐

yep

[willow bear]

2nd derivative = derivative of derivative

n'th derivative = derivative of (n-1)st derivative, generally

[viscid thistle]

ann

were u the smartest in all ur classes

in math

[willow bear]

i... happened to be that, yes

[hoary yoke]

😮

[viscid thistle]

dammn

[hoary yoke]

tbh every, one, of, my, classmates, HATE math

and i know hate is a strong word

thus im using it

[viscid thistle]

i love math but im so shitty

at it

[hoary yoke]

u cant be shitty at something you dont need a talent in

[viscid thistle]

yes u can

[hoary yoke]

you can be lazy

not shitty

ok its alrdy 3:40

[viscid thistle]

u can be shitty.

[hoary yoke]

yesi can