help-10

[fickle cave]

ye?

$$CEC \theta = \frac{1}{cos \theta}$$

[warm shale]

♡ PandaPanda2By4 ♡

[fickle cave]

so can you explain how this works again

[glacial verge]

do you mean sec?

[fickle cave]

yeh

sec

mb oops

$$sec \theta = \frac{1}{cos \theta}$$

[warm shale]

♡ PandaPanda2By4 ♡

[fickle cave]

wouldn't the answer be 5?

[glacial verge]

what is the value of cos?

[fickle cave]

uhh

rn I did a graph that is

3, 1

right angle triangle btw

rn I'm finding the

sec theta =

[glacial verge]

cos is a/h
so, sec should be h/a

[fickle cave]

yeh

that's right

so

[glacial verge]

what's the h?

[fickle cave]

squareroot of 10

which is 5

[glacial verge]

how is 5*5=10?

[fickle cave]

wait uhh

I mess taht up?

lemme check again

3.16227766

mb

[glacial verge]

the a is 3 right?

[fickle cave]

ye

[glacial verge]

so, what would be sec theta?

[fickle cave]

would be

3.16227766/3

[glacial verge]

yeah, and you should keep the sqrt(10) like sqrt(10) instead of computing it

[fickle cave]

alr so

would be

sqrt(10)/3

[glacial verge]

yeah

[fickle cave]

how come for

cot

the answer was 3

but for this 1 I had to type sqrt(10)/3

[glacial verge]

what is the tan?

[fickle cave]

yes

1/tan =a/o

[glacial verge]

yeah, what's the value of the tan theta?

[fickle cave]

3/1

[glacial verge]

that's the cot
but sure, it's because 3/1 is equal to 3

[fickle cave]

true lol

omg haha

alr alr lol

just noticed

[glacial verge]

and because your thing requires the simplified answer, the answer is 3

[fickle cave]

for cos would be sqrt(10)/1

so I just write squrt(10)

correct?

sqrt*

[glacial verge]

cos is h/a right?

[fickle cave]

I mean

CSC

cosecant

1/sine=h/o

[glacial verge]

oh, yeah, it is

[fickle cave]

what about this 1?

do I just write sin of 4/5 ?

for my answer

[glacial verge]

cos is a/h=4/5
so, we can find the a, h, and o from that

[fickle cave]

oh ok I get it

so a=4 and h=5

then I use pythagorian to figure out the O?

[glacial verge]

the o, yeah

[fickle cave]

alr so it would be 20+25=55 and that's like sqrt 55 would be 7.416198487

alr got it

so sin would be sqrt(55)/5

ye?

[glacial verge]

that's the csc

[fickle cave]

oh

[glacial verge]

sin is the o/h right?

oh, wait no

[fickle cave]

the question confuses me lol

[glacial verge]

i don't think that h is right

[fickle cave]

oh

[glacial verge]

i mean o

[fickle cave]

hmm

sqrt of 4 and 5 is 20 and 25

so that' 55

right?

wait omfg

brain fart

I messed up

lol

mb

good eye sepdron

16 and 25

41

[glacial verge]

yeah, and subtracted right?

[fickle cave]

uhh

nope

I added

how come subtract?

[glacial verge]

because
a²+o²=h²

so o²=h²-a²

[fickle cave]

oh

alr

so the answer would be 9

sqrt of 9

sqrt(9)/

and uhh

I forgot

[glacial verge]

sqrt(9) is 3

[fickle cave]

4:27Am vibes

lol

yeh it's 3

but on the answer

[glacial verge]

ok, now we can solve for sin

[fickle cave]

yeh

sqrt(9)/5

ye?

for sin

[glacial verge]

you can simplify the sqrt(9)

[fickle cave]

alr so

3/5

alr now cot

[glacial verge]

yeh

[fickle cave]

tan

a/o

so 4/3

ye?

[glacial verge]

yeah, i think so

[fickle cave]

this 1 I think

I know how

3.4 btw

I do

tan-1 3.4

yeh?

[glacial verge]

yeh

[fickle cave]

73.6104

so

73.6

ye?

[glacial verge]

yeah

,w arctan(3.4)

[warm shale]

Results provided by WolframAlpha

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Requested by Sepdron#5047

[fickle cave]

I think 2 more questions

then I'm done

finna sleep lol

4:31am

this 1 is confusing my brain

so theres 2 right angle triangles?

1 for tan and 1 for cos

[glacial verge]

it's the same triangle

[fickle cave]

ohh

so I draw the

tan triangle

[glacial verge]

i think it's like this one but with tan

[fickle cave]

then use pythagorian

to find cos

yeh?

[glacial verge]

yeah

[fickle cave]

omg I'm so smart

jk lol

k gimme sec to do that

I got it

3/3.16227766

aka 3/sqrt(10)

correct?

[glacial verge]

yeah, that's right

[fickle cave]

wait if you read the question it says round to the first decimal

X_X

so I'm wrong then

???

[glacial verge]

nearest hundreds

so you need to calculate 3/sqrt(10)

[fickle cave]

oh

devide it?

[glacial verge]

yeah

[fickle cave]

then cos-

cos-1

right?

[glacial verge]

you only need to calculate cos theta, not theta

[fickle cave]

oh yeh right right

so 0.948683298

would be

0.95

ye?

[glacial verge]

,w 3/sqrt(10)

[warm shale]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by Sepdron#5047

[glacial verge]

yeah

[fickle cave]

alr

😄

alr

that's enough for today lol

time to sleep X_X

4:40Am

thanks sepdron

[glacial verge]

gn

[fickle cave]

c ya everyday until thursday

cuz I got test lol

hopefully I get to see you man

your the most helpful I've met

peaceee

Before you go

1 more question actually

mb @glacial verge

sorry

so I got this 1 wrong and was wondering why

[glacial verge]

where do you get those numbers?

[fickle cave]

oh wait

I used sin for finding a I'm so dumb

nvm

lemme try again

[glacial verge]

oh, i didn't see the little b

[fickle cave]

haha

sorry I figured out my mistake I think lemme try again

I'm so dumb haha

alright

it should be 27.8, 65.4, 140

am I right?

[glacial verge]

shouldn't the c also have a decimal point?

[fickle cave]

hmm

I thought it would be 140 because

it's 140.17

etc

but the rounding chart

not sure how that works but

I just round to the third number

lol

[glacial verge]

oh, i thought we're rounding to the nearest tenth

[fickle cave]

idk lol

[glacial verge]

but if we round to the nearest integer, yeah that's right

[fickle cave]

that chart confuses me lol

[glacial verge]

yeah, idk what it means too

[fickle cave]

gulp

alr but

other than the decimal

it's all right?

27.8, 65.4, 140

[glacial verge]

mhm

[fickle cave]

alright I'll try it out

yeh

it worked

140 worked

[glacial verge]

oh good

[fickle cave]

idk how the chart works it's weird asf lol

I got 95%

fk

I wanted that 100%

so sad

I missed 1 question

it was cus of rounding I had like 4-5 guys trying to figure out the chart too lol

[glacial verge]

i think the table is telling that if the degree is rounded to the nearest tenth, the accuracy of the trig functions is accurate to 2 digits after the decimal point

[fickle cave]

so

[fickle cave]

if I put 140.17 it would work

?

[glacial verge]

sorry

i didn't know what that button does

[fickle cave]

all good lol

no worries

I'ma sleep now hope to see you tmr?

[glacial verge]

cya

[fickle cave]

alr peace out thanks for everything

sleep well

[obtuse pebble]

@fickle cave Has your question been resolved?

[pliant raft]

A Shop has 800 parts, out of which 8 are defective. What percent are defective

[icy hill]

Probability?

[royal basin]

@pliant raft get your own channel

read #❓how-to-get-help

do not post in occupied channels

[icy hill]

@royal basin it's not occupied I guess😅

[royal basin]

it's occupied until the bot frees it.

[icy hill]

Oky

[royal basin]

it has another person's name on it

when you post in an unoccupied channel, the bot pins your message and puts your name on the channel

[icy hill]

Oh

I m new sorry

[obtuse pebble]

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[opal musk]

i need some help understanding logs

[opal musk]

specifically, is there a particular order in which the log laws need to be carried out?

kind of like bodmas or smoething

i dont fully understand how they work. for example.
-ln(2) + ln(4)
if you use the addition rule you get
-ln(8) = -2.079
but if u do the subtraction rule first
ln(4) - ln(2)
ln(4/2) = ln(2) = 0.693

so is there a partciluar order u need to do them in, to get the right answer?

[high lily]

ln(2) is just another number

the order of operations is still pretty much the same

[opal musk]

are both answers correct?

[high lily]

no

$-a + b \neq -(a+b)$

[warm shale]

ℝamonov

[opal musk]

i dont get it. all i did what switch the postions of the two values. i didnt factorise the - out or anything.?

[high lily]

in the first method you added the ln(2) and ln(4) first

violating the order of operations

which is what's represented in what I made above

[opal musk]

, so what should i have done first?

[high lily]

the second route you took is clear and valid

the first route you took wasn't valid for the reasons mentioned above

[opal musk]

im having trouble linking the way logs work with the idea that -a+b != -(a+b)

i get how the idea works with normal numbers, im just having trouble undrstanding that with logs

[spiral maple]

The rules don't magically stop existing in the presence of logs

-a+b still isn't -(a+b), regardless of how a and b look to you

[high lily]

ln(2) and ln(4) are just numbers

-1 + 1
you would NOT do -(1+1) here, (hopefully)
and that doesn't really change if you have logs or variables or w/e

[opal musk]

hmm, i think i get it

thanks for the help

!close

.close

[obtuse pebble]

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[obtuse pebble]

@celest zodiac Has your question been resolved?

[celest zodiac]

can anyone help me w this^

[glacial verge]

combine the fractions

[celest zodiac]

i was going to use the theorem the limit as theta approaches x, 1-costheta/theta=0

but i wasnt sure how to incorporate it here

[glacial verge]

i don't think you need that

[celest zodiac]

im honestly lost

[glacial verge]

can i see your attemt?

[celest zodiac]

sry its prob rlly wrong

[glacial verge]

$(1-cos(x))^2$ isn't $1-cos^2(x)$

[warm shale]

Sepdron

[celest zodiac]

what would i have to do in order to fix it

[glacial verge]

think of a way to make $a-b$ to $a^2-b^2$

[warm shale]

Sepdron

[celest zodiac]

you just squared the a-b

[glacial verge]

that would be a²-2ab+b²

another way to find it is to factor a²-b²

you multiply it by its complement, (a-b)(a+b)=a²-b²

[celest zodiac]

so multiply 1/1-cosx by 1/1-cosx?

[glacial verge]

$\frac{1\cdot(1+cos(x))}{(1-cos(x)))\cdot(1+cos(x)}$

[warm shale]

Sepdron

[celest zodiac]

wouldnt the denominator equal to 1+cosx^2 which isnt the same as 1+cos^2x

[glacial verge]

it is the same, it's just notation

$cos^2(x)$ is actually $(cos(x))^2$

[warm shale]

Sepdron

[celest zodiac]

but you said this

[glacial verge]

you see the outer parentheses?

it's like $(a-b)^2$

[warm shale]

Sepdron

[timid silo]

It's the whole expression squared.
$$(a-b)^2 ≠ a^2 - b^2$$

[warm shale]

Pencil

[celest zodiac]

.close

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[vestal surge]

Can someone help me with this

To get the 16 i think its 4y/3

but I dont know how to get the q in there

[topaz bridge]

$\log {\frac{a}{b}}=\log a-\log b$

[warm shale]

tomzizek

[topaz bridge]

1st use this

[vestal surge]

so is it (4y/3) - 1 ?

[topaz bridge]

yes

[vestal surge]

okay

thanks for your help

have a nice day

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[vestal edge]

Hi

[vestal edge]

I have a device that uses 87.8kWh on average over a 24 hr period

How do I work out how much this will cost me in power bills?

[obtuse pebble]

@vestal edge Has your question been resolved?

[obtuse pebble]

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[turbid cedar]

x(t) = 3 sin (t) − sin (3 t)

[turbid cedar]

y(t) = 3 cos (t) − cos (3 t)

how can i draw them

[sage geode]

On paper or in desmos?

[turbid cedar]

if i was asked to draw it

so in a paper

[sage geode]

I guess you'd need to find (x, y) for some values of t between 0 and 2pi

And then just connect them

[turbid cedar]

do i also need to find the (x,y) that they touch its other?

[scarlet gale]

Real numbers don't touch each other.

[turbid cedar]

what do you mean

[sage geode]

wdym

[sage geode]

That's just the point where y = x, not really significant

[turbid cedar]

im asked to find the area below the two graphs

and then draw them

[sage geode]

Below y(t) and x(t) separately?

[turbid cedar]

is says find the area that is surrounded by the x(t) and y(t)

[sage geode]

I think then they'd mean the area of the shape made by (x, y) satisfying those equations

[turbid cedar]

yes

[sage geode]

Btw the shape looks like this

I guess you'd need to find x in terms of y and then integrate it from y = 0 to y = 4

(also don't forget to multiply the integral by 4 because of the symmetry)

[turbid cedar]

.close

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[mint atlas]

The product of two consecutive integers is 8 more than twice times their sum. Find the integers

[mint atlas]

How do i set this up?

[tardy epoch]

If you an arbitrary integer n, what's an expression for the consecutive number?

[mint atlas]

i wrote down n+1

[tardy epoch]

What's the product of these two numbers

[mint atlas]

n^2+n

[tardy epoch]

Keep going

Their sum, twice times their sum, twice times their sum plus 8

[mint atlas]

well for that part i wrote 2(n+(n+1))+8

alright thank you i understand now

.close

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[raven spire]

Yayyyyyyyy

[obtuse pebble]

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[distant plinth]

hi

[distant plinth]

I kinda found something a little bit weird

it makes sense

but like

[raven spire]

[distant plinth]

I feel like it still feels a little bit off to me

so maybe one of yall can help me with that

this is kinda weird

how it went from a non smooth function, to a smooth function

like

I don't get it man

[timid silo]

If you look closely

That smooth function is just a sine wave being vertically shifted upwards by 0.5

[distant plinth]

no it's not

it's close though

[timid silo]

Hmm

Okay that's really weird lol

[distant plinth]

in any case

so that basically means

that as long as the function is continuous

its anti derivative will be a smooth function?

[balmy mortar]

smooth means infinitely differentiable iirc

So not.

However, a continuous function should have continuous anti-derivative by FTC if I got it right

[distant plinth]

interesting

well

well what I meant is like

aight aight

but what i mean is like

for example

if somebody asks for point of inflection, for example at point (1, 1), and asks (is that a point of inflection for h(x) (which is the green function here).

without giving us the graph of h(x)

do we just know that h(x) will be continuous?

and that

point (1, 1) is indeed a point of inflection?

no matter what

@balmy mortar

[balmy mortar]

I'm not sure what you mean.

'Is (1, 1) a point of inflection for h(x)?'

There is no indication h has to be continuous.

[distant plinth]

but you just said that a continuous function should have a continuous anti-derivative

and that's true even if the function is not differentiable right?

[distant plinth]

So ye?

@balmy mortar

[balmy mortar]

yes.

[distant plinth]

Interesting

That's kinda weir

[obtuse pebble]

@distant plinth Has your question been resolved?

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[lofty ruin]

Hello! Is 18:3(3+3) equal to 18:3x(3+3)? If we put it in a calculator the result for the first one will be 1 and for the second its 36. Why is this?

[fierce siren]

what does : mean? are you sure you wrote it in your calculator correctly

[lofty ruin]

Oh sorry in Hungary we use ":" for division

so its "/"

[fierce siren]

oh, ok! so you have $\frac{18}{3(3+3)}$ and $\frac{18}{3 \times (3+3)}$

yes, those would be equal

[warm shale]

ALIAS

[fierce siren]

it just seems to be an issue with your calculator

can you show me a picture of how you wrote it in the calculator?

[lofty ruin]

Sure

[mental wren]

Thats a casio thing I think

[fierce siren]

ooohhh, i know why it's 36

[mental wren]

Is it a casio?

[fierce siren]

or at least why the calculator thinks that

it's thinking this:

[lofty ruin]

its a no name calculator

[mental wren]

Because casio does this for user convenience I think, its in the manual

The order of operations is different without the ×

[fierce siren]

it's thinking this

not this

[mental wren]

And this is the reason for the differing results

[lofty ruin]

So it is a calculator related problem or the implied multiplication takes the precedence?

[mental wren]

Its does implied multiplications first it seems

So you can see it as a problem or as a notation thing

As I said, casio does this for convenience

[lofty ruin]

Alright, thank you for the help!

.close

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[timid silo]

i need help with finind area

[short spruce]

probably best not to ghost ping helpers, huh?

[timid silo]

i forogot the rules and and i needed help but i forgot so then i deleted message

[obtuse pebble]

@timid silo Has your question been resolved?

[mental solstice]

Hint: draw perpendicular lines from O to AB and from O to BC, then draw the line OB. Then consider just that smaller square.

[obtuse pebble]

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[sinful kraken]

which values does this series converges to ?

here is my work

value should be 1/7

[haughty coyote]

you wrote 4^n = 4^(n-1) / 4, but it's *4

[quaint glen]

The first term isn't 4

[sinful kraken]

ok

true

but how do i find the value of the sum

[quaint glen]

By correcting the first term and doing what you did again

[haughty coyote]

you got the formula, just use the right starting value

[flat anvil]

note that

$\frac{(-3)^{n-1}}{4^n} = \frac{1}{-3} \cdot \left(\frac{-3}{4}\right)^n$

[warm shale]

M.E.G. Yottachad

[flat anvil]

right just multiplying top and bottom by -3

so then you can just use the geometric series equation for this one

b/c |-3/4| < 1

[sinful kraken]

i need n to become n-1 so i can use my sum formula

[flat anvil]

oh

then you can say taht

[flat anvil]

$\frac{(-3)^{n-1}}{4^n} = \frac 14 \cdot \left(\frac{-3}{4}\right)^{n-1}$

[warm shale]

M.E.G. Yottachad

[flat anvil]

right i just factored out a 4 from the bottom

[sinful kraken]

how did you factor 4 out ?

oh

nvm

thats so smart

thank yo uso much

@flat anvil

.close

[obtuse pebble]

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[timid silo]

where did i screw up?

[zenith dove]

redo your cos^2 integral

[timid silo]

oh oops, the last term should be zero

[timid silo]

but that doesn't change much, no?

[zenith dove]

are you sure you did the integral correctly?

[timid silo]

uhh

no

the first part should be plus

wait

no

oml im so sorry

[zenith dove]

how did you go from line 2 to 3?

[timid silo]

yeah so that hsould be 1, not cos theta haha

got it

.close

[obtuse pebble]

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[obtuse pebble]

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[sterile spire]

For 5 a how am I supposed to find the phase shift to complete the equation

[slow sonnet]

,rotate

[obtuse pebble]

@sterile spire Has your question been resolved?

[sterile spire]

<@&286206848099549185>

[tardy epoch]

The last sentence

[obtuse pebble]

@sterile spire Has your question been resolved?

[sterile spire]

@tardy epoch I’m still confused

[obtuse pebble]

@sterile spire Has your question been resolved?

[tardy epoch]

The rung's height is zero when time is zero

[obtuse pebble]

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[surreal basin]

Hello

[surreal basin]

For this problem it says to find the discontinuities

I know that the denominator can’t be zero or negative

So for the next step can I divide by 2?

[flat anvil]

???

[timid silo]

Imagine f(x) = sin(x)

[flat anvil]

First it should be >=

Second you should notice that only some values (so not intervals) work

You can’t divide by 2 and cancel the inside

[timid silo]

You have $1 - f(2x) \le 0$

[warm shale]

n/c

[timid silo]

Clearly $f(2x) \neq 2f(x)$ because $f$ is not a linear map

[warm shale]

n/c

[surreal basin]

So

I can’t divide by 2

[timid silo]

Think of a simple example. Is sin(pi/2) = 1/2 sin(pi)?

[surreal basin]

I don’t know

That looks more complicated

[timid silo]

Just. compute them?

[surreal basin]

Oh

They’re not equal

[timid silo]

Right, so if f(2x) = 2 f(x), then it has to be true for all values of x

You just found a value (x = pi) where they are not equal

[surreal basin]

True

Would I have to use inverse sin?

[timid silo]

Okay, so you have $1 - \sin(2x) \le 0$

[warm shale]

n/c

[timid silo]

How can you simplify this further?

[surreal basin]

I can send sin(2x) to the other side

[timid silo]

Right, so do that

[surreal basin]

But then what

If I can’t divide by 2

[timid silo]

Think about what sin(anything) can equal

Maybe draw a graph of sin(x)

[surreal basin]

But

This one is 1/sin

[timid silo]

We're talking about solving the inequality, aren't we?

[surreal basin]

I’m trying to find the discontinuities

[timid silo]

Okay, let's take a step back. Why are we solving that inequality?

[surreal basin]

Because the square root has to be > 0

[timid silo]

That is true, and what else? Specifically, what causes a discontinuity?

[surreal basin]

0 being in the denominator

[timid silo]

Exactly, so right now we're looking at both of those cases.

When is 0 in the denominator?

[surreal basin]

When the square root makes a 0

[timid silo]

Right, so what equation do we need to solve to see when that happens?

[surreal basin]

The sin?

[timid silo]

Be specific, write it out here

[surreal basin]

Do I need to include the square root in the inequality?

I’m solving for 1-sin(2x)

1-sin(2x)>0

[timid silo]

Okay, so you're solving for when it's positive, which also works

So why don't you simplify that inequality

[surreal basin]

I made it

1 > sin(2x)

[timid silo]

So actually, let me tell you something I did before

Before, we were solving $1 \le \sin(2x)$

[warm shale]

n/c

[timid silo]

This is because if we solve this inequality, we solve for what we don't want to happen

That is, we don't want the square root to be negative, and we don't want it to be 0

You, on the other hand, are solving for when neither of those things happen

But that won't help us with what we're trying to do, because we're trying to find cases where it goes wrong

[surreal basin]

How

Should I flip the sign?

[timid silo]

Okay, so what happens if sin(2x) = 1?

[surreal basin]

A zero

[timid silo]

Right, and what happens if sin(2x) > 1?

[surreal basin]

It’s continuous

[timid silo]

What do you mean?

[surreal basin]

There isn’t a discontinuities if sin (2x) is > 1

Wait

Doesn’t sin (2x ) have to be < 1

[timid silo]

It does, but I was getting to that

My point was if sin(2x) > 1, then the inside of the square root is negative

Which is the other case that we don't want

Thus $1 \le \sin(2x)$ covers both cases: denominator being 0, and the inside of the square root being negative

[warm shale]

n/c

[surreal basin]

Ohh

Got it

But

Now I have 1 >= sin (2x)

[timid silo]

No, you need $1 \le \sin(2x)$

[warm shale]

n/c

[surreal basin]

No it’s 1 <= sin (2x)

Yea

Now I divide by 2?

[timid silo]

Didn't we go over that already?

[surreal basin]

Oh

So what do I do

I can use the other sin

The inverse sin

[jolly ginkgo]

Would you steal my cookies?

[timid silo]

Okay, how about we consider 1 = sin(2x)?

[surreal basin]

Yes.

Isn’t it going to be 1/2 = sin (x)

I really have no clue what the next step is

[timid silo]

I really suggest reviewing some khan academy algebra videos

https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:foundation-algebra

[surreal basin]

Pls give me a hint

I’ve solved a million problem like this before

[tardy epoch]

,w plot y = 1 and y = sin(2x) for -20 < x < 20

[warm shale]

Results provided by WolframAlpha

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

Requested by riemann#2529

[raven spire]

[raven spire]

What's the point of discontinuity for this function?

[surreal basin]

For sin(2x)?

[raven spire]

no, for f(x)

[surreal basin]

I don’t know

I don’t see a discontinuity

[raven spire]

Is sin(2x) = 1 allowed?

[surreal basin]

Yea

Why not

[raven spire]

Put sin(2x) = 1 in the f(x)

[surreal basin]

Oh

It doesn’t touch 1?

[tardy epoch]

you can't divide by zero, right?

[surreal basin]

Yea

[tardy epoch]

1 / 0 is bad. 1 / something is bad if something=0 . solve for when something = 0

can you figure out what something is?

[surreal basin]

Something is < 1

Wait

[tardy epoch]

something = ?

[surreal basin]

It’s actually > 1

1 -sin(2x)

I don’t know

[raven spire]

$1-\sin(2x)=0$ contributes big time to the discontinuity in f(x)

[warm shale]

Ansh

[obtuse pebble]

@surreal basin Has your question been resolved?

[surreal basin]

I know 1 = sin (2x ) will be pi/2

I think

Or pi/4 when I divide everything by 2

Anywa

Thanks guys for helping me

I’ll throw in the towel

.close

[obtuse pebble]

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[timid silo]

anyone know what that means

full question btw

[quaint glen]

probably $(\log_5(18))^2$

[warm shale]

Sneaky

[quaint glen]

in fact almost certainly that

[timid silo]

so i got

(4a^2+4ab+b^2)/c

i also got (4a+2b)/c

and none of them were correct

any help

waittt i know what id id wrong

its

(2a+b-c)^2

idk its not working

can someone help

.close nobodys helping

[obtuse pebble]

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[brittle swan]

i ✨still✨ don’t get why this implies
‘integration is the inverse of differentiation’

[worthy drum]

if everything is integrated together, then you have everything, "F", if you differentiate everything from another, then you have something observable in F, but unique to other items belonging to F, i.e. "f"

[brittle swan]

i alright watched its animation

[brittle swan]

i dont get this at all

[ruby finch]

https://math.stackexchange.com/questions/664759/why-is-integration-the-inverse-of-differentiation

[royal basin]

what nonsense is this

[ruby finch]

The first answer uses an intuitive approach, might help you

[worthy drum]

[royal basin]

buxy, did you post this knowing it's a nonsensical non-answer to chromium's question?

[worthy drum]

I didn't, that is what I use to explain that integrations is the opposite of differentiations

[astral ivy]

Almost sounds like crank speak lol

[royal basin]

@brittle swan if you think of F(x) as the area under the graph of f between a and x, and you nudge x slightly, then the area changes by approximately a rectangle of height f(x)

there is an illustration that can be made to that effect

but i don't have the energy to do that

[brittle swan]

i know it

and i still don’t get it

[astral ivy]

@brittle swan are you trying to understand intuitively or rigorously why they’re inverses

[brittle swan]

rigorously

well, i don’t really get it intuitively either

[astral ivy]

Do you understand why the statement of FTC 1 means they’re inverses?

[brittle swan]

i get d/dx (integral) = f

not integral (d/dx) = f

[brittle swan]

this is my question

i can even prove it

[royal basin]

so you want to know why int[a,b] f'(x) dx = f(b)-f(a)?

[brittle swan]

wait

[brittle swan]

this is ftc2??

[astral ivy]

This is a corollary of FTC1

[brittle swan]

what

[astral ivy]

If you scroll down a little on the Wikipedia article, it says that

(do u know what a corollary is)

[brittle swan]

(no)

(something that follows directly from another thm?)

[astral ivy]

(Precisely)

Lol so I assume the “inverse” concept is referring to FTC1

Because FTC2 almost always is talking about the computation $$\int_a^b f(x)dx = F(b) - F(a)$$

[warm shale]

abs_0

[astral ivy]

FTC1 says that $$\frac d{dx} \int_a^x f(t) dt = f(x)$$

[warm shale]

abs_0

[astral ivy]

@brittle swan

[brittle swan]

yea

[astral ivy]

So wait what are you asking again lol

[brittle swan]

why integration reverses differentiation

differentiation is ‘instantaneous rate of change’

integration is ‘area under curve’?????

how the hell are they inverses lol

[astral ivy]

Ah so you’re asking about intuition

Good luck with that lmao

It’s one of the hardest things to get intuitively imo

[astral ivy]

Just play around with this idea on your own @brittle swan and it’ll make some sense

[civic zealot]

integration over an interval (meaning [a,b] where a and b are numbers, or maybe infinities) is area under a curve. Otherwise integration results in a function that is the antiderivative. That's what FTC1 says.

[brittle swan]

i watched the animation.

[astral ivy]

But it’s hard to get to a point where you just get it, without having to go “ok I’m adding up the rate of change of the area of the rectangle, which is the same as dadada” every time

@brittle swan did you watch the 3b1b video

[brittle swan]

yea

[astral ivy]

Personally it didn’t super help for me lol

Nothing did, it’s just very unintuitive

[brittle swan]

ok

[astral ivy]

Sorry that prob isn’t very helpful lol

[brittle swan]

this theorem is intuitively true

what isn’t is how it implies integration, differentiation are opposites

[astral ivy]

You intuitively understand why they’re inverses?

[brittle swan]

no

this is my question

[astral ivy]

Doesn’t that mean you get why they’re opposites?

[brittle swan]

i don’t

it should be obvious for you why this theorem implies differentiation reverses integration

not for me

(this is the motivation for ftc2 isn’t it)

[astral ivy]

You mean the statement of the theorem, right?

[brittle swan]

no

i know the statement of the theorem

i know why it’s true

i can prove it

[astral ivy]

Yeah

But you don’t get why it’s true?

[brittle swan]

i can’t see why it has to do with ‘diff = inverse int’

[astral ivy]

Oh

Inverse function means f(g(x)) = x

[brittle swan]

this statement and ftc1 are complete aliens to one another

to me

[astral ivy]

We’re working with “functions of functions” (I’ll call it a functional)

[brittle swan]

composite functions you mean?

[upbeat plinth]

inverses act as undo buttons

[astral ivy]

So if f,g are functionals, then f(g(s(t)) = s(t) for any function s

[upbeat plinth]

we start at f

integrate

then differentiate

ftc1 says the result is f

[ruby finch]

Wow, how helpful

[upbeat plinth]

ie differentiation undoes integration

[civic zealot]

Start with a function $f(x) = F'(x)$. Then $\frac{d}{dx}(\int_a^xf(t)\ dt) = \frac{d}{dx}(F(x) - F(a)) = F'(x) =f(x)$

[brittle swan]

it says F’(x) = f(x), where F(x) = int f(t) dt [a, x]

[warm shale]

Zybikron

[brittle swan]

so F is an antiderivative of f

but x is just some upper bound in F’s expression

[astral ivy]

F is the antiderivative. In words, it’s literally saying “the derivative of the antiderivative is equal to the original function”

[brittle swan]

so basicallt

‘antiderivative just so happens to be an integral with x acting as an upper bound’

‘and that means integrals are heavily related to derivatives!’

[astral ivy]

Oh you’re wondering why it has those bounds

Instead of saying $$F(x) = \int f(x)dx$$

[warm shale]

abs_0

[astral ivy]

?

[brittle swan]

no

[brittle swan]

from this

[brittle swan]

to this

i don’t understand

[astral ivy]

Do you understand this

[brittle swan]

yea

[astral ivy]

Okay

[astral ivy]

The thing that links

antiderivative is integral with upper bound stuff
to
integrals are inverses of derivatives
is precisely the Fundamental Theorem of Calculus

[brittle swan]

????

[astral ivy]

Sorry

Editing rn

@brittle swan ?

[brittle swan]

i don’t get this

[astral ivy]

You don’t get how those could be linked?

(This is rhetorical, you’re not supposed to)

(That’s why it’s called the fundamental theorem of calculus, why it’s so powerful)

(It links these two crazy things that seem totally unrelated)

[dreamy wolf]

integral of dy = y + C. do u understand how that's intuitive?
dy/dx = slope. do u understand how that's intuitive?
dy/dx * dx = dy. do u understand how that's intuitive?
integral of dy/dx * dx = y + C. do u understand how that's intuitive?

[brittle swan]

how

[astral ivy]

Bro ok hold up

@dreamy wolf the dy/dx * dx = dy thing always felt so hand wavy and sketchy to me

When dy/dx is presented in calculus, it’s basically a symbol

When did it become concrete values you can multiply by

[astral ivy]

How? That’s what the proof explains

[brittle swan]

they’re differentials?

[brittle swan]

the theorem says ‘if we have f, take the integral of it from a to x, then differentiate it, we get f back’

[astral ivy]

Exactly

[brittle swan]

how does it become

[dreamy wolf]

this is basically like doing u sub. it's hand wavy yes but it's legal b/c there is rigorous math behind it that lets u do that.
dy = dy * (dx/dx) = dy/dx * dx. this kinda thing u see in physics a lot

[astral ivy]

How do you get f back?

[brittle swan]

‘integrals are opposite of derivatives’

[astral ivy]

This doesn’t sound like opposites to you?

You take a function, integrate it up to x, differentiate with respect to x, and you get the original function back

[brittle swan]

no

[astral ivy]

They undo each other

[brittle swan]

one you differentiate with respect to x

one you integrate with respect to a and x

[astral ivy]

Broooo I hate that a lot lol

For applications I understand though

[dreamy wolf]

if u do A, and then do B, and doing B reverses when u did A isn't that opposites

[astral ivy]

You can’t go through 10 hoops to avoid small logical problems

[astral ivy]

It sounds like what you’re really asking is what these bounds are about

Would you still be confused if I said $$F(x) = \int f(t) dt$$ instead?

[warm shale]

abs_0

[brittle swan]

i guess?

[brittle swan]

isn’t boundless integration introduced after ftc2?

[astral ivy]

Yes, that’s precisely the issue

we need bounds

And what bounds do we choose? Well a is arbitrary and irrelevant (most people choose it to be 0), but the important part is the x upper bound

[civic zealot]

everything is with respect to x, there is a t in the integration as a dummy variable because having and the variable x as your bound, and as the variable in your function is.... problematic.

[brittle swan]

yea i know

[astral ivy]

@brittle swan Think about it, F takes in a number x and gives you the area under the curve from 0 to x

[brittle swan]

0?

[astral ivy]

If you take the derivative of that function, you get f back

(Just think of it as 0 for now, the important part is that it’s irrelevant, you’ll see why in the proof)

[brittle swan]

yea i know why it’s irrelevant

[astral ivy]

Ok

I sorta think I get why you don’t get it

When we say “derivative of f” what are we saying?
It’s a new function f’ such that f’(x) is the slope of the tangent line to the graph of f at x

When we say “integral of f” what are we saying? It’s a new function $F(x)$ such that $$F(x) = \int_a^x f(t)dt$$

[warm shale]

abs_0

[astral ivy]

@brittle swan FTC1 establishes that f’ and F are opposites, according to these above definitions

[brittle swan]

the hell

[astral ivy]

Translating to colloquial language, you can say “derivatives and integrals are opposites”

Did that make any sense at all lol

[brittle swan]

not really

[astral ivy]

Dksichdiejf

[brittle swan]

(i’ve been asking this for 3 days)

[astral ivy]

Really

[brittle swan]

(yes)

[astral ivy]

:(

[raven spire]

oiiii

Help me :c

[astral ivy]

@raven spire use another channel

[raven spire]

It's relevant to the topic!

[astral ivy]

Oh

[brittle swan]

repulsive: starvation

[astral ivy]

@brittle swan I think part of it is that your question requires us to understand something you don’t get in your head

[brittle swan]

yea i guess

if it doesn’t bother you..?

[mental solstice]

maybe i should try?

though i have no idea what's already been discussed

[astral ivy]

I will say that I don’t really get exactly what you don’t get, like sorta but not really lol

[mental solstice]

or rather, what's already been tried

[raven spire]

The integral of a function is the area under the curve! ✓ I can see that visually.. and it's fine!
But ..
The derivative of the integral,i.e., the derivative of the area under the curve.... returns the function ;-; how do you visualize that?

[astral ivy]

Ok @brittle swan for this person can you just try to phrase your question in the most exact words you can

[brittle swan]

bottom text

[astral ivy]

It’s very difficult

[brittle swan]

and that maybe answers my question

[astral ivy]

I spent a while trying to force my brain to accept it, and it’s just plainly an unintuitive concept

[raven spire]

:p

[astral ivy]

Unintuitive**

[mental solstice]

is it really that unintuitive?

[astral ivy]

The tangent slope doing the opposite of the area under the function? I mean

Call me crazy but those sound like totally different things lol

[mental solstice]

it seems to me that between $\int_a^x$ and $\int_a^{x+h}$ where $h$ is small, all you're doing is adding an infinitesimal rectangle of height $f(x)$ and width $h$

[warm shale]

Camilleone

[dreamy wolf]

what i think about is a rectangle. length f(x). side length dx. integral u get the area so it's f(x)dx. derivative u divide by dx u get back f(x). and then i think about riemann sum

[mental solstice]

and that is exactly the rate of change of the area

[astral ivy]

But whenever you think about how they’re opposites, you must go through this thought process every time

At least, for me

Intuitive means I don’t have to go through any thought process at all, I just get it in my gut

[raven spire]

Oooooooooo omg that's smart >_<

[brittle swan]

like one way to view, is change of area

[raven spire]

and dumb of me at the samee time T_T

[brittle swan]

what about ‘area function when you input the tangent line’?????

wtf

[raven spire]

[mental solstice]

the what?

[brittle swan]

this explains d/dx (int), not int (d/dx)

[mental solstice]

i must have missed that

[raven spire]

How does the "derivative is the slope of tangent to a function at some point" comes into equation here

Does it have to do with, slope of tangent being a rate of change of function 👀

Yes I believe

[mental solstice]

oh i see

[astral ivy]

Yes, the slope of the tangent is “instantaneous rate of change”

(Really, rate of change as h, the distance between the points, approaches 0)

[raven spire]

Got that... now the int(d/dx) part plz... as chromium asked

[mental solstice]

the first problem with any argument of int (d/dx) is the introduction of a constant which leads to many antiderivatives

[astral ivy]

Changing areas makes sense… area under changes doesn’t

[mental solstice]

like, many functions all differing by constants have the same tangent function

i believe this is what's making the arguments hard to understand intuitively

so to understand this we have to go back to $\int_a^x$

[warm shale]

Camilleone

[mental solstice]

the derivative, as you know, is the rate of change function

so let's say you start with a function f, and then you differentiated it to find the derivative f'

the integral of f' can be understood as infinitesimal addition of all the little rates of changes at each point

[brittle swan]

huh..?

[mental solstice]

which means that now to get $\int_a^x f'(x) dx$, what you want is the total sum of all the little changes of $f$ from $a$ to $x$, which ends up as the difference between $f(x)$ and $f(a)$, i.e. $f(x) - f(a)$

[warm shale]

Camilleone

[brittle swan]

i can’t visualise this

[astral ivy]

Ohhh cause a rectangle with 0 with is the same as the height basically right

[mental solstice]

yeah

[astral ivy]

Wait no that doesn’t make sense

[mental solstice]

well

[astral ivy]

You’d be adding up infinitely many values

It would diverge

[mental solstice]

i admit it isn't that well-worded

what i should have said was

take for example a rate of change at a point $x$

[warm shale]

Camilleone

[mental solstice]

for h small, between x and x+h you basically add an infinitesimal "increase" of the function by that rate of change - of course if the rate of change is negative your function decreases, but that's not important