help-39

[cursive wraith]

but I can create the line segment that connects 1 to 2

[1,2]

and 1.5 is on that line segment for example

[glass bolt]

Ah so the points between 1 and 2 would be out of the set

Thus not convex

[cursive wraith]

exactly

[glass bolt]

Okay

Tyvm

.solved

[pearl pond]

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[midnight haven]

help

[midnight haven]

i have a question

what software do people use to write maths papers

like

research sort of things

where you can quickly type mathematical symbols

like integral sign and stuff

[compact veldt]

latex

[midnight haven]

is that an app

[compact veldt]

coding language

overleaf would be the software

[vital estuary]

overleaf is what people use

[midnight haven]

oh

[compact veldt]

theres also

MATLAB

[midnight haven]

overleaf or latex?

[compact veldt]

search up overleaf

[midnight haven]

ok

[compact veldt]

the coding language is latex tho

[vital estuary]

overleaf is a compiler of latex

[midnight haven]

ok

thanks

,close

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[midnight haven]

.reopen

[pearl pond]

✅

[midnight haven]

@vital estuary @compact veldt

this is very much giving coding

is this what is usually used?

[vital estuary]

yes

[midnight haven]

also do you think i should learn to code this summer

[vital estuary]

for the most part, you can write like normal

see the visual editor at the top

[midnight haven]

i really enjoy maths but for literally everything i am doing atm the question sheets end with, write a code to test X..

[vital estuary]

i think you should learn to code anyway

[compact veldt]

there's a tutorial

"learn overleaf in 30 minutes"

[scenic aurora]

I mean, there are also WYSIWYG editor for latex too but I will say coding is better

[vital estuary]

you dont really need to code too often

[compact veldt]

https://www.overleaf.com/learn/latex/Learn_LaTeX_in_30_minutes

[vital estuary]

you can import images if you need to

[scenic aurora]

you can also use rich text editor in overleaf. that can help you adjust. You can import the template or write the preamble for your doc once and forget about it

[midnight haven]

thanks guys

what language to you recommend that i learn to code in?

[compact veldt]

a simple noob example i made with latex

[scenic aurora]

coding for what?

[midnight haven]

nice

[midnight haven]

general life

most of these assignments allow any programming language

[compact veldt]

java for usefulness
python for simpleness

[scenic aurora]

python would be easy language with a lot of tutorial

[midnight haven]

do you have any resource recommendations?

like youtube channels

[compact veldt]

snakify for learning python

[midnight haven]

is that free

[compact veldt]

i think so

[midnight haven]

okay

[scenic aurora]

I think CS50 is also good one

[midnight haven]

what is this?

oh that harvard course?

[scenic aurora]

Free Harvard course

[midnight haven]

basically i am applying to university in october

to study maths

should i spend my time between then and now learning to coe

code

[undone current]

Hey where can we get that free coarse

Is it online

[scenic aurora]

Learning code for mathematicians is beneficial, but not like you have to learn now. Feel free to learn if you want to but you can also learn that in college too

[midnight haven]

thanks

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[hybrid basin]

Hello

[hybrid basin]

How do I know if proj or perp has a shorter distance to point P?

Or do I just calculate both and compare

[pearl pond]

@hybrid basin Has your question been resolved?

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[lucid moth]

This is part 1 of a question (and its answer)

This is part 2, which is the part that im stuck on

Im guessing that we ignore everything and just take whats inside the square root (13t/6)

And set it equal to something

But idk what

[unborn abyss]

what is the question asking you?

[lucid moth]

What do you mean sry

[unborn abyss]

what does the question mean? can you explain it in your own words?

[lucid moth]

Get the smallest positive max of t

[unborn abyss]

that doesn't sound right, shouldn't this mention d(t)?

[lucid moth]

what is d(t)

I guess I dont really understand the question

Or what it means by 'distance is maximal' anyways

[unborn abyss]

didn't you just calculate that?

[lucid moth]

Ah yeah

True

[unborn abyss]

I think it means a local maximum

[lucid moth]

Ah ok

That was what I thought it meant but I got a bit confused haha

[lucid moth]

Get the smallest positive value of t where theres a local max?

[unborn abyss]

sounds good to me

you're right that you can ignore the square root and that will save you a lot of time

[lucid moth]

And I set it equal to something right?

But not sure what

[unborn abyss]

how would you normally find the value of t where there's a local maximum?

[lucid moth]

Set it equal to 0 to find where cos = 1?

[unborn abyss]

okay sure you could do that

[lucid moth]

Yeah?

[unborn abyss]

that's more clever than what I would normally do involving derivatives

[lucid moth]

t=0 is the only solution...

Which isnt a solution

[unborn abyss]

how

[lucid moth]

13t/6 = 0

[unborn abyss]

t=0 is one solution yes

but what was it you wanted?

[lucid moth]

No

[unborn abyss]

something about cosine

[lucid moth]

Wym?

[unborn abyss]

you're trying to maximize that function right

[lucid moth]

yes

[unborn abyss]

you identified that you only needed to maximize that cosine term

and noticed that cosine capped at 1

and cos(u) famously is only 1 exactly when u=0 and nowhere else 😔😔😔

[lucid moth]

There was a nicer way to say that but i get your point

2pi right?

so t = 12pi/13

It was correct! Thanks 😄

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[sinful frigate]

can i take here

(1000,0100,0010,0001)

[pine jay]

I think it might be useful to think of a), b), c) and d) as a certain transformation applied to the list of vectors you were given, so along as these transformations (which seem to be linear) don’t squish down to a dimension or so, then the list is still linearly independent (if I haven’t missed anything)

Or I guess just by observation, can you rewrite any of the vectors in the list from the others?

For d) for example

Let’s call u = a1 + a2, v = a2 + a3, w= a3 - a4 and z = a4 - a1.

Notice that u + z = a4 + a2,

And so adding w, we get that

u+z + w = a2 + a3

But this is just v

So we’ve managed to rewrite one of the vectors in the list in terms of the other

Hence d) does not form a linearly independent set

Try and do something similar for the others

Or possibly view them as transformations if possible

[pearl pond]

@sinful frigate Has your question been resolved?

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[unreal ridge]

I am confused upon this question

[unreal ridge]

I do know the basics of a sin function and how it works but I am confused about it being in degrees?

[sly dagger]

wasnt someone literally just asking about this question?

[unreal ridge]

who?

Like I have an idea to go about it

trying to find the maximum or higest point first

But I dont know what that would be since there is nessesairly no limit

[brisk lotus]

What is the maximum value of sin x?

[unreal ridge]

that is odd

No like I mean the maxmium point of intersect

so that I can work backwards from there

[brisk lotus]

Which is why you need to know what the maximum value of sin x is

[unreal ridge]

thats easy its sin 90

where that equals 1

[brisk lotus]

When does x/2024 equal 1?

[unreal ridge]

cause thats the maxmimum since no changes made to the amplitude

when x is 2024

[brisk lotus]

Perfect so we can deduce we have finitely many intersections

why?

[unreal ridge]

but sin 2024 does not equal 1

Wait is this supposed to be periodic

[brisk lotus]

what

[unreal ridge]

like if the range of sin is between -1 and 1

like sin 0 and sin 360 are equivalent

[brisk lotus]

I was about to get to that but I am glad you got to that yourself

[unreal ridge]

since both equal 0 one is just the phase shift 360 degrees forward

[unreal ridge]

thank you

[brisk lotus]

do you know the period of sin?

[unreal ridge]

yes I do

M T M B M

and its period unchanged is 360 degrees

[brisk lotus]

I was hoping for a number but thats fine too

2 pi

[unreal ridge]

No it wants degrees not raidans

[brisk lotus]

then 360 degrees

[unreal ridge]

yeah so its periodic and at one point they will interesect

[brisk lotus]

not at one point

[unreal ridge]

ok so there is more than one points it could intersect

[brisk lotus]

visualize x / 360

[unreal ridge]

ok

[brisk lotus]

and also sin x

[unreal ridge]

yep I graphed it out

So I see the crests and the troughs

at which degree

hold on would I have a common denominator

[brisk lotus]

only the number of intersections matter

how many do you see

[unreal ridge]

of x/360 and x /2024

[brisk lotus]

not really

but the 360 will be important

[unreal ridge]

but between interestcting x/360 and sinx its just a line

interestecting a sin function

over the x axis at y = 0

[brisk lotus]

Point is: you will intersect sin x at 2 points if my brain is working correctly

and im not seeing ghosts

[unreal ridge]

would the two points be the same except one set is negative

[brisk lotus]

thats not what I want to get at

[unreal ridge]

ok

[brisk lotus]

Thing is: for x / 360 (x >= 0) you intersect sin x twice

[unreal ridge]

alright makes sense so far

[brisk lotus]

Meaning for x / 720 you would intersect it how many times?

remember that 360 degrees makes one period of sin x

[unreal ridge]

would it be the same

since sin is periodic

[brisk lotus]

four times

[unreal ridge]

that would make sense

the two points

then 360 degrees forward of the 2 points

[brisk lotus]

The thing I want you to realize is that once you figured out how many intersections you got for one period you can calculate them for every period by just adding

[unreal ridge]

Ok I got that part

because of the periodicness

[brisk lotus]

for x / 1080 you would intersect sin x how many times?

[unreal ridge]

6

right

[brisk lotus]

Somewhere we will reach 2024

what we need to realize now is something very important

when do we intersect sin x

[unreal ridge]

ok

[brisk lotus]

Obviously x / 2024 is positive for positive x

[unreal ridge]

yes

thats what I wanted to talk about

sin90 to 180 is not possible

cause those are all negatives

[brisk lotus]

sin 90 is 1 and sin 180 is 0

[unreal ridge]

sorry i meant 180 to 270

[brisk lotus]

close but not completely right

[unreal ridge]

how

[brisk lotus]

180 to 360 is not possible

[unreal ridge]

thats the only part its in the bottom section

oh wait

nvm

cause 270 to 360 is where it uprises again

[brisk lotus]

for a half period we intersect, and for the other half period we do not

[unreal ridge]

yes now I am finally seeing that

cause one period goes from middle to top to middle

so that limits our options

[brisk lotus]

which makes it easy to tell the maximum number of intersections for positive x

[unreal ridge]

ok

this is where I am getting

I meant this sorry

[brisk lotus]

too complicated just take a look at what we already have

[unreal ridge]

ok

[brisk lotus]

we know for x / 360 we intersect sin x 2 times

[unreal ridge]

yes

[brisk lotus]

for x / 720 4 times and for x / 1080 6 times

how about x / 1440?

and by extension x / 1800?

[unreal ridge]

that is 8

1800 let me thinnk about it i believe its 10

for 1800

[brisk lotus]

ok now for x / 2160?

[unreal ridge]

umm 12

right

[brisk lotus]

ok but 2160 is bigger than 2024 no?

So why is that not a problem

[unreal ridge]

yes so we overshot it a bit

[unreal ridge]

wait is it because 2024 is in the period with 2160

[sly dagger]

x is in degrees

[brisk lotus]

its because 2160 - 180 is 1980

every x after 1980 can be fully ignored

[unreal ridge]

wait why?

[brisk lotus]

ah wait there is some stuff I missed

let me quickly take a look

[unreal ridge]

no worries take your time

this help is helping me a lot

[sly dagger]

this would be it

[unreal ridge]

do you know the way to get to this answer

[brisk lotus]

ah yes for it to not intersect two times, the denominator would need to be smaller than a quarter period of sin x if that makes sense.

Visualize x/1889

[unreal ridge]

ok I see it

[brisk lotus]

here we do not intersect the period 1880 < x < 2160 anywhere

[unreal ridge]

ok i see it

[brisk lotus]

thing is, as soon as we look at x/1890 we do

[unreal ridge]

oh yeah

[brisk lotus]

and for anything > 1890 we intersect 2 times

[unreal ridge]

that makes sense

[brisk lotus]

So 12 is still the answer as 2024 > 1890

[unreal ridge]

ok

[brisk lotus]

But is it

[unreal ridge]

but some of these interceptions would not make sense

[sly dagger]

ur still missing the fact that x is in degrees

[brisk lotus]

Depends tbh

[unreal ridge]

cause 90/2024 does equal 450 over 2024

[sly dagger]

when x is 360 degrees x=pi therefore y=pi/2024 at x=pi

[unreal ridge]

3.14 degrees

[sly dagger]

radians

[brisk lotus]

If we get x in degrees and we use sin we would also look at sin x in form of degrees no?

[brisk lotus]

idk if what i wrote makes sense

[unreal ridge]

like this

[sly dagger]

y isnt in degrees tho

y is in radians

[unreal ridge]

there a way to convert desmos to degrees

[sly dagger]

yes

x degrees *(pi/180) = radians

[brisk lotus]

sin(90°) = 1 = sin(pi / 2)

where is y in radians

[unreal ridge]

i found a way to make it into degrees

[brisk lotus]

Does your exercise specifically state that sin shall only be used for x in radians?

[sly dagger]

because lets say x=2024 degrees, then sin(x) is not 1, but y=x/2024 if y were in degrees then y=1 which contradicts as y=sin(x) which is not 1

[unreal ridge]

yeah

[brisk lotus]

it still intersects twice though

[unreal ridge]

all it says sinx in degrees and x/2024 in degrees interesct

at what points

cause 0 is one of them

as 0/2024 = sin0

[sly dagger]

what?

[unreal ridge]

it does 0 is one them and 175.039

[unreal ridge]

is the other one

[brisk lotus]

I don't see the issue ngl

[sly dagger]

anyways sin(x) is a function that returns radians?

[unreal ridge]

no i made return degrees

[brisk lotus]

sin(x) returns a value between 1 and -1 regardless of input

[sly dagger]

nvm im dumb but my poiunt still stands

x at 2024 degrees y=sin(2024)=0.72726665034 but y=2024/2024= 1

[unreal ridge]

yea that point makes sense

but both at 175.039

are equal

same with 0

but the problem is quite simple I dont have a formula to calculate 175.039

[brisk lotus]

but sin(2024°) is -0.694658...

sin(2024 rad) is 0.72726665...

[sly dagger]

right i put it online which auto assumed radians, but none the less thats worse

[brisk lotus]

why is it worse

[unreal ridge]

ok thats not important

[sly dagger]

because y=x/2024

[unreal ridge]

we got the two points we just need a formula to show so

[sly dagger]

if y were in degrees then y=1 at x=2024

but as u just worked out its negative

[unreal ridge]

yes at x as 0 and 175.039

there are the same

[brisk lotus]

so your point is that f(x) = sin x = x/2024 is what we are looking at?

because y = sin x and y = x/2024?

[sly dagger]

correct

[unreal ridge]

yes

[brisk lotus]

yeah I belive those are two different functions

[unreal ridge]

and if x is 0 or 175.039 there are equal

[sly dagger]

u need their outputs to be in the same units

u cant work y in degrees for 1 and not for the other

[unreal ridge]

they are

both in degrees

[sly dagger]

but they arent, x is in degrees

everything else beyond that is just a regular number

[brisk lotus]

But -1 <= y <= 1 regardless

y has no unit y is a number

[sly dagger]

correct

but 2024 has no unit whilst x does

[brisk lotus]

rather the y coordinate of a point on the unit circle etc etc i guess its clear what i mean

[sly dagger]

i dont get how i can explain this to u so u can understand

[brisk lotus]

So your problem is that I am not dividing x/2024 but rather x°/2024

[unreal ridge]

ok but how would that differ

like how can degrees divide something

[sly dagger]

take y=sinx and y=x/2024, for those 2 equations to intersect then sinx=x/2024, but notice that if x is in degrees this doesnt hold true for 99 percent of values

[brisk lotus]

What

[sly dagger]

yes?

[brisk lotus]

I mean yes obviously it doesnt hold true for 99% of the values you could also say for 99.999...% whatever values

The real question is

What happens if I divide something with a unit by something without a unit

what unit is 2024 in

[sly dagger]

well it still remains in that same unit system

if u divide degrees by any non unit number its still in degrees

360°/4 is still 90°

[brisk lotus]

I hope we are not unnecessarily overcomplicating things here :

[unreal ridge]

we probably are

[brisk lotus]

because how would you plot y if y were in degrees

[sly dagger]

well it changes the answer by a large amount

[sly dagger]

y isnt in degrees tho

[brisk lotus]

if i divide x/2024 it is

going by what you just told me

[sly dagger]

if x and y were both in degrees like u said, then it would look like y=x/2024

[brisk lotus]

which also implies sin x never equals x/2024 because they are in completely different units

[sly dagger]

where they are both unitless

[brisk lotus]

one has none and the other is in degrees

[sly dagger]

u can convert degrees to radians

[brisk lotus]

yeah well same argument

than I would have none unit vs radians

[sly dagger]

the whole point of radians was to convert degrees to a standard unit of length that we could measure

[brisk lotus]

but it wouldn't change the fact its a unit

sin(x) give a unitless answer

x/2024 doesnt by your argument

[sly dagger]

radians are UNITLESS

they arent a proper unit

u can't pull out a radian lenghted ruler

because radians dont have a length

[sly dagger]

u can convert x so that it is in radians

[brisk lotus]

That would make for a more complicated answer ngl

@unreal ridge is this a university question?

[unreal ridge]

no

[sly dagger]

the question can be solved at any level

[unreal ridge]

techinically its a course question

if you took grade 11 math you should have the skill set to solve it

[brisk lotus]

The thing I am currently looking at involves some ridiculous values

If I were to transform x to radians with x° * (pi / 180), x / 2024 will only be ~ 1 at x = 115968

[sly dagger]

yes

the solution isnt that extensive

[brisk lotus]

which means I have come across ~18456 periods

which means I intersect sin x 9228 times

for positive x

negative is the same amount

so I intersect sin x 18456 times and crossed 36912 periods doing so

if this is the answer I am eating a broom

[sly dagger]

im getting around 728641 intersections

[unreal ridge]

i got 23 interesections

[pearl pond]

@unreal ridge Has your question been resolved?

[pearl pond]

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[lavish basin]

hi

[lavish basin]

.close

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[sweet grail]

Getting a couple different answer from different calculators. Can anyone confirm this is the right answer?

[quiet tendon]

L = lim (ln(z))^(1/(z-e))
ln(L) = lim 1/(z-e) * ln(ln(z)) = lim ln(ln(z)) / (z-e)
that's of the form 0/0 so using l'hopital's you get
ln(L) = lim 1/z * 1/(ln(z)) = 1/e * 1/(ln(e)) = 1/e * 1 = 1/e
then L = e^(1/e)

so that's correct

[sweet grail]

Thank you I appreciate it.

.close

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[summer imp]

!status

[pearl pond]

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

[charred dome]

Drawing graph

[charred dome]

I am stuck on part a and b

Using the fundamental theorem of calculus

For part 2

[summer imp]

Well for a) they are referring to a theorem you've hopefully seen before.

What does it say?

[charred dome]

it just means that whatever f(x) in the integral is the derivative of g(x)

I just dont get how to show the second part

[summer imp]

Can you find an antiderivative of 2 + sin(t) then?

[charred dome]

that would be 2t - cos(t)

[summer imp]

Ok. And then you want to plug the bounds of the integral inside of that right?

[charred dome]

yes but its from 0 to x so what do i do ?

[summer imp]

You can still plug in x and plug in 0.

[charred dome]

so then 2x - cos(x) - 1

[summer imp]

Yes, and you now want to differentiate that.

[charred dome]

ok

so thats 2 + sin(x) lol

[summer imp]

Indeed.

[charred dome]

so F' is 2x - cos(x) - 1 or is that just f?

oh wait thats the same thing

so is F = 2x - cos(x) - 1
or is F = 2 + sin(x)

and F' = 2x - cos(x) - 1

[summer imp]

g'(x) = 2 + sin(x). The point of the exercise is to use both parts of the theorem see derivatives "undo" integration.

F = 2x - cos(x) - 1 or F = 2 + sin(x) don't matter, since any antiderivative of the form F(t) = 2x - cos(x) + C will work.

[charred dome]

how can F = 2x - cos(x) -1 and F = 2 + sin(x)?

is F the antiderivative? or F' ?

[summer imp]

2x - cos(x) is AN antiderivative of 2 + sin(x).
2x - cos(x) - 1 is also an antiderivative of 2 + sin(x).

It doesn't matter which one you choose, hence why when you do (or will do) indefinite integrals you'll be asked to add a +C at the end, signifying that any constant does the trick.

I don't know exactly what the two parts of the theorem they're referring to are.

The goal is just to see that $\frac{d}{dx} \int_a^x f(t) dt = f(x)$, whether it's by realizing that $\int_a^x f(t) dt = F(x) - F(a)$, so $\frac{d}{dx} \int_a^x f(t) dt = \frac{d}{dx} (F(x) - F(a)) = F'(x) = f(x)$ \
OR straight away seeing that derivatives undo integration, so $\frac{d}{dx} \int_a^x f(t) dt = f(x)$.

[jolly parrot]

Azyrashacorki

[charred dome]

i can show you

oh so F is the anti derivative

and F' is the original function

so the anti derivative is F = 2x-cos(x) - 1 and then F' = 2 + sin(x)

[summer imp]

Yes exactly

[charred dome]

im having trouble understanding what the purpose of what we just did was

did we just get the derivative of the function?

cause there's another question right after this one that asks to get the derivaitve using part 1 of the fundamental theorem of calculus

$y = \int_{0}^{tan x} e^{-t^2} dt$

[jolly parrot]

wakamole

[charred dome]

so how would i do that?

[summer imp]

You showed that derivatives undo integration, essentially.

[summer imp]

Can you actually show me part 1 rq? Just want to make sure I'm not going in too deep hahah

[charred dome]

yea sure

[summer imp]

Okok

So let's say we have $g(x) = \int_0^{\tan(x)} e^{-t^2} dt$.

[jolly parrot]

Azyrashacorki

[summer imp]

Can you see that this is just $h(\tan(x))$, where $h(x) = \int_0^{x} e^{-t^2} dt$?

[jolly parrot]

Azyrashacorki

[charred dome]

yea

[summer imp]

So really it's just a composition of functions. Do you think you can apply the chain rule to it?

[charred dome]

$e^{-tan^2x}$ ? to that?

[jolly parrot]

wakamole

[summer imp]

We have $h(x) = \int_0^x e^{-t^2} dt$, you want to compute $(h(tan(x))'$.
The chain rule gives $h'(\tan(x)) \cdot (\tan(x))'$ right?

[jolly parrot]

Azyrashacorki

[summer imp]

But we know the derivative of tan, and we also know what h' is from Part 1 of the theorem

[charred dome]

im not getting it im kind of bad with chain rule

why do we want to apply chain rule?

again?

[summer imp]

Well it turns out that the function we're looking for is the derivative of h(tan(x)).

[charred dome]

can we do it a different way?

im just not thinking of it like that for some reason

[summer imp]

And this function is composed. All that Part 1 says that that $h'(x) = e^{-x^2}$.

[jolly parrot]

Azyrashacorki

[charred dome]

ok

[summer imp]

The issue is that I don't think this one has a nice antiderivative. And whatever you do, you'll need to use the chain rule at some point.

[charred dome]

so we substitute in tan(x) and then subtract 0?

like h(0)?

or is thtt for part 2 only?

[summer imp]

That's part 2, and we'd need an antiderivative, which we don't.

[charred dome]

oh

yeah

so like cah you shoe me what to do then and i can ask questions

im just not getting it at this point

[summer imp]

Yes that's fine.

[charred dome]

like my equation is y = the integral

not like g(x)

does that matter at all?

[summer imp]

Not really, it's just a name for the function at the end of the day

So we have $h(x) = \int_0^x e^{-t^2} dt$. Why do we use that? Because that's the only function we have information about from the theorem's first part.

[jolly parrot]

Azyrashacorki

[charred dome]

true

ok and it says like wahatever f(x) is the derivative of h(x)

[summer imp]

Yep

[charred dome]

so then h'(x) = f(x)

but im looking at the answers for the odd questions and they plugged in the top interval to the variable

[summer imp]

It will come

Now, the function we actually are after is $(h(\tan x))'$, since $h(\tan x) = \int_0^{\tan x} e^{-t^2}dt$.

[jolly parrot]

Azyrashacorki

[summer imp]

So far so good?

[charred dome]

yeah so far

[summer imp]

Ok, so using the chain rule, we know that $(h(\tan x))' = h'(\tan x) \cdot (\tan x)'$.

[jolly parrot]

Azyrashacorki

[charred dome]

yea

[summer imp]

But we know that $h'(x) = e^{-x^2}$, so we actually know that $h'(tan(x)) = e^{-\tan^2 x}$.

[jolly parrot]

Azyrashacorki

[summer imp]

And then we get this extra factor at the front since $(\tan x) ' = \sec^2 x$.

[jolly parrot]

Azyrashacorki

[summer imp]

So at the end of the day, we get $y = (h(\tan x))' = h'(\tan x) \cdot (\tan x)' = e^{-\tan^2 x} \cdot (\sec^2 x)$.

[jolly parrot]

Azyrashacorki

[charred dome]

wow

i am getting it

the chain rule applies to so many things

we could also have done it that we just get the derivative of $e^{-tan^2x}$ which is also chain rule ? or is that not the right way to dfo it

[jolly parrot]

wakamole

[summer imp]

Hum, you'll notice that if you actually compute the chain rule in there you'll get a -2tan(x) term out front which we don't want.

The reason why this part of the Fundamental Theorem is useful here is that it turns out this function (e^{-x^2})has no antiderivative, at least not in terms of general functions we're used to working with.
So you couldn't even look for an antiderivative in the first place.

[charred dome]

oh yeah youre right

is it usually using the chain rule to find the derivative in this form? for these types of questions?

Like for any of these ones?

[summer imp]

Yeah it's all the same principle

[charred dome]

its confusing becuase for 16 h(x) already has x as the top part

so would this be the quotient rule then?

[summer imp]

Well I used h(x), but it could be anything.
You should always try to write it back as a composition of the integral with just x (from part 1) with the upper bound (sqrt(x)).

[charred dome]

wait i just read online somewhere that you do h'(x) * d/dx the upper bound

is that a shortcut?

[summer imp]

Well it's just this but replace tan(x) with a general function f(x).

[charred dome]

oh yeah

that is it

but if the lower bound was x and theupper bound was like 2x youd subtract the lower bound from upper

or is that part 1?

im kinda confused on what part 1 is ... like when does it become part 2

i think i get it though

$\int_{-5}^5 e\ dx$

[jolly parrot]

wakamole

[summer imp]

Part 2 requires finding an antivderivative. Notice that part 1 says that the lower bound can be any a. It doesn't change that g'(x) = f(x).

[charred dome]

how does this even make sense lol

oh yea thats right you need the antiderivative of upperbound- antiderivative of lower bound then derive that

so thats the same thing as the chain rule?

[summer imp]

Yeah it will still be h'(x) * d/dx (upper bound)

[charred dome]

oh thats right but if we do that then we arent derivnig at the end

are we?

so wait F(b) is chain rule

and F

is

the derivative of f(x)?

oh wait i mean antiderivative of f(x)

and b is the upper bound

[summer imp]

It that's really your integral you should be able to just use part 2 though.

[charred dome]

how, theres no x..

is it just e * b - e * a ?

[summer imp]

e is just a constant, so yes it's just eb - ea

[charred dome]

and you just leave it like that right?

im just a little confused becuasee when we did the very first problem you had me take the antiderivative of f(x) first, then plug in the bounds, subtract F(b) - F(a) and then derive it to get the answer.

but now i feel like we're not getting the antiderivatine of anythng and we're just doing d/dx f(x) * b - d/dx * a, am i right or am i going crazy here lol

[summer imp]

It depends what they're asking for.
If they want the derivative, then yes you have to differentiate.

If not then no need.

[charred dome]

ooh

ok tham

thanks

wait so is the integtral of this e * (5 - (-5)) ?

i found the answer is that but why not e(5) - e (-5) ?

nevermind i got it... constant so it's c(b-a)

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[arctic roost]

is the local max and abs max (16,4.8)

[arctic roost]

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[wet apex]

Hey

[wet apex]

Can anybody explain

How is suddenly

Pls

Im stuck

[slender viper]

do it step by step

[slender viper]

you do additions and subtractions from left to right

[tardy bridge]

so obv the $210\sqrt{2} - 210\sqrt{2}$ cancels, then for the $\sqrt{3}$ terms, you can group them tgt to get $\sqrt{3} \cdot (35 - \frac{10}{3} + \frac{1}{3})$

[jolly parrot]

PhenomPlasma

[slender viper]

but you can reorder $a-b+c=a+c-b$

[jolly parrot]

Element118

[tardy bridge]

i think you just made an arithmetic error tho

[slender viper]

or you can do $a-b+c=a-(b-c)$

[jolly parrot]

Element118

[wet apex]

Huh

But still

How is it 32

The answer is 32

[pearl pond]

@wet apex Has your question been resolved?

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[plain vessel]

for this can i just do 1/10 = 1/5 + 1/i ? its a concave mirror so can i just do 1/10 - 1/5 and take the inverse to find i?

[pearl pond]

@plain vessel Has your question been resolved?

[pearl pond]

@plain vessel Has your question been resolved?

[pearl pond]

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[naive hemlock]

,w integral from 0 to 1 (x-1)/(1+x^2).ln(x)

[jolly parrot]

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[naive hemlock]

What

,w int from 0 to 1 (x-1)dx/(1+x^2).ln(x)

[jolly parrot]

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[naive hemlock]

Na

,w int from 0 to 1 (x-1)/[(1+x^2).lnx]

[jolly parrot]

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[naive hemlock]

What

Bro

,w int from 0 to 1 (x-1)dx/[(1+x^2).(ln(x))]

Aaah

,w int from 0 to 1 (x-1)/(x^2+1)*ln(x)

[jolly parrot]

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[naive hemlock]

No

Bro

,w int from 0 to 1 (x-1)/[(x^2+1)*ln(x)]

[jolly parrot]

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[naive hemlock]

Ok

[next dove]

wolfy was literally high on something

[naive hemlock]

I got it

.close

[pearl pond]

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[vital estuary]

do this in #bots next time

or #latex-testing

[pearl pond]

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[wraith badger]

I graphed both 1/cos(x) and 1/sin(x+(x/2)) on my calculator but they dont coincide (like what the answer shows for cosec (x+x/2) and sec (x)😭

i also dont understand how sin(x+pi/2)=cos(x) I literally can't figure out angle relationships

[wraith badger]

this was the original question

[shy oak]

should the graph be like this

red is sec(x)
blue is cosec(x)

green is [-2π=<x<=2π]

do you know sine addition formula?

⬇️

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

sin(π/2) = 1
cos(π/2)= 0

so...

sin(x+(π/2)) = sin x • 0 + cos x • 1 = cos x

@wraith badger

[wraith badger]

I’m not sure but I got something similar on my calcualtor

Idk why the question shows them like together

[wraith badger]

OH YES THESE I KNWO THESE

Bro I had no idea we had to use that to figure it out

I SEE NOW

Thank you so much

[shy oak]

wc😊

[pearl pond]

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[wraith badger]

.close

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[verbal sigil]

converting from degrees to radians, do I find the highest common multiple between the degree and 180. e.g 150 degrees pi over 180? What would I do

[spare lark]

Degree * pi/180 = radian

150 * pi /180 = 150pi/180 = 15pi/18 = 5pi/6

Well yes, simplify the fraction

Till you cant

[verbal sigil]

oh my god

it was this simple

thank you!

.close

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[hybrid haven]

can someone explain how to get from the first line to the second line

[pearl pond]

Please don't occupy multiple help channels.

[hybrid haven]

i thought u were supposed to simplify

like factor out 2x firt

first

[regal herald]

they did do that
left them with 2x(2x^2+9x+9)
then factorise the rest of it in whatever way you choose, ac method, quadratic formula etc

[hybrid haven]

whats the ac method

i havent done it in too long so i forgot how to

i thought u just had to find a number that multiplies to 18 and adds up to 9

[worthy lance]

Why 18?

[hybrid haven]

cuz theres still a 2 in front of the x^2

[worthy lance]

Lets do this slowly

2x(2x^2+9x+9)

Lets factor inside the parenthesis

2x^2+9x+9 = 2(x^2+9x/2+9/2)

Do you agree?

@hybrid haven

So now we have 4x(x^2+(9/2)x+9/2)

We are gonna factor inside the parenthesis again, now that x^2 has 1 as coefficient

x^2+(9/2)x+9/2 = x^2 + (9/4)x + (9/4)x + 9/4 + 9/4

x^2 + (9/4)x + (9/4)x + 9/4 + 9/4 = x^2 + 2(9/4)x + 18/4

x^2 + 2(9/4)x + 18/4 = x^2+2(9/4)x + 72/16 + 81/16 - 81/16

x^2+2(9/4)x + 72/16 + 81/16 - 81/16 = (x+9/4)^2 - 9/16

(x+9/4)^2 - 9/16 = (x+9/4)^2 - (3/4)^2

(x+9/4)^2 - (3/4)^2 = (x+9/4-3/4)(x+9/4+3/4)

(x+9/4-3/4)(x+9/4+3/4) = (x+6/4)(x+12/4)

(x+6/4)(x+12/4)=(x+3/2)(x+3)

Gathering with the previous the final result is

4x(x+3/2)(x+3)

Which exactly the same as 2x(2x+3)(x+3)

[pearl pond]

@hybrid haven Has your question been resolved?

[pearl pond]

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[balmy shoal]

Hi I need help with this problem

[naive orbit]

and, what is the question

[balmy shoal]

Idk my professor just told me to solve this with no explanation 😔

He said this is the question

[tropic saddle]

put a bit more effort into your trolling

[balmy shoal]

Idk if my professor is trolling my guy

[spare lark]

He is

[balmy shoal]

Oh

Mb

How do I close this channel

[spare lark]

.close

[pearl pond]

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[spare lark]

Like this

[balmy shoal]

Ok thanks

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[hasty citrus]

help with logic

[hasty citrus]

say P is "throwing an egg at a wall" and Q "the egg explodes"

if I throw the egg and it explodes, P->Q is true

[midnight haven]

An egg can explode without it being thrown at the wall btw

[hasty citrus]

but when I dont throw an egg, whether it explodes or not, P-> Q is still "true"

so what does it mean when I say its "true"?

[midnight haven]

But the egg also might not explode when it’s thrown at the wall

[hasty citrus]

true as not proven false?

[pine iron]

I like to think of its this way

if I say if 'I have a cat I will name is bob'

if i don't have a cat

I am technically not lieing

hence the statement is till true

[midnight haven]

But as the egg might not always explode

You might call your cat Fred

[hasty citrus]

I think my issue comes down to what "true" and "false" mean here, because if I didnt do any relevant thing to the argument P -> Q then its odd to say it is "true", its just unknown

if I say "when I buy a pen, it'll grow legs and run away" then until I dont buy a pen, my statement is true?

ok, brought down to math:

0<|x-c|<d -> 0<|f(x)-L|<e

if both are true, the limit exists

if x is within delta of c, but f(x) is not within epsilon of L, the limit does not exist

but now, if x is not within delta of c, then it is meaningless if f(x) is within epsilon of L

I think im confusing "true" with "proven true", I guess I misunderstood that

I read about vacuous truth and I think that sums up my doubts

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[hybrid haven]

binomial probabilities dont use standard deviation right

[muted remnant]

Binomial random variables have standard deviation.

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[pearl pond]

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[brisk steeple]

Can the similarity of triangles be proved using the cosine rule [which is easy to prove]?

[austere quail]

What do you mean with the cosine rule?

Do you want to prove that cos(A) = cos(B) or something?

[brisk steeple]

To prove similarity of 2 triangles using:

[austere quail]

What in there do you want to be the same?

Like how would you want to prove it with that?

[brisk steeple]

oh

I want to prove the AA criterion.

using that.

[austere quail]

2 angles are the same?

[brisk steeple]

yes.

[austere quail]

because then you can just use soh cah toa?

[brisk steeple]

For all triangles.

[brisk steeple]

this is for right triangles, innit?

I just want to know whether it is possible. I highly doubt it isn't possible.

Oh

Yeah

[austere quail]

You could use zhz then?

[brisk steeple]

What is that?

full form?

[austere quail]

“If you know two sides and an angle of a triangle, that is actually enough to be able to say that those two triangles are congruent. What is important here is that the included angles are exactly the same size. Then we speak of the congruence characteristic ZHZ for triangles.”

[brisk steeple]

I just want to know that is it possible to prove : that the ratio of sides will be the same.

If two angles are same.

Consequently the third of course.

[austere quail]

yh then they will use the lowercase

[brisk steeple]

Given that the angle is between the two equal sides.

[austere quail]

the zhz like I said

[brisk steeple]

U mean SAS?

[brisk steeple]

z is side and h is angle?

[austere quail]

yes

[brisk steeple]

Okay......

I know SAS. = zhz

[austere quail]

After you’ve proven that the triangles are similar or congruent, you can say that the angles are the same

[brisk steeple]

ikr, but that's the thing I want to prove.

[austere quail]

I might not understand your question…

[brisk steeple]

My conjectures (which the former is proved):
"If two angles of a triangle are equal to the angles of another triangles, consequently the third angle, then the ratio of the corresponding sides will be equal. This can be proved using the cosine law."
For example, in triangle ABC and DEF, if ang A = ang D, ang B = ang E, then AB/DE = BC / EF = AC/DF.

-=-===--==Is the latter part of my conjecture true which is italicised?-==-==-==-=

[austere quail]

I think so, because you put the same stuff in it (from the zhz)

[brisk steeple]

Yes. I just needed a confirmation...., stamp?

Ok thanks.

.close

[pearl pond]

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[wild fable]

how do i do this if im missing 2 values ?

[muted remnant]

You're given the definition of X

The way P(X=3) is given, there is a direct implication of the distribution of each of the two spinners

[wild fable]

how is P(X=3) calculated, sum of two numbers

so the numbers 1,2 can add to 3 so is that 1/16 + 2/16? im confused

[muted remnant]

How do you get a sum of 3 using the spinners?

[wild fable]

if you spin a 1 and then spin a 2

[muted remnant]

What does it look like the probability of 1 on first, then 2 on second are?

[wild fable]

1/4 and 1/4

[muted remnant]

So you have 1/4 on every number for both spinners

Then how do you get a sum of 7 with the spinners?

[wild fable]

if you get a 3 and a 4

which is 1/4 and 1/4

[muted remnant]

Therefore, the probability is

[wild fable]

1/16 if u multiply them im assuming

[muted remnant]

I think that's an ok assumption

Then you can get the last number by calculating it, or just using 1 - everything else

[wild fable]

thanks

[pearl pond]

@wild fable Has your question been resolved?

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[ivory sun]

how to tell if these series converge

[ivory sun]

this is not telescopic series right

so I cannot just check the lim of partial sum

other than that I think I can bring it to common denominator and then multiply by conjugate

[ornate creek]

hello

what is the integration of root(cos(x^2-1) * sin(x^2-1))

@ivory sun

[ivory sun]

@ornate creek hi you need to send your question in one of the unoccupied channels

Like #help-40

[wild fable]

integral test maybe

[pearl pond]

@ivory sun Has your question been resolved?

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[pearl pond]

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[quartz tulip]

Starlights Productions is hosting an annual outdoor concert. Average attendance is 40 000
people when tickets are $100. For every $5 increase in ticket price, 1 000 fewer people are
expected to attend the concert. What admission price will produce the most revenue? Explain how this can be
determined. BTW, the quadratic function in standard form is: R(C) = - 5000p^2 + 100,000p + 4000,000. Where R(C) = revenue concern and "p" = number of ticket price increases of $5.00.

who could figure this out as soon as possible pls

[pearl pond]

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[dusk moat]

i was wondering how it can be -8/17 if the ratio represents a length

[dusk moat]

the answer makes sense and i have no issue with solving but i just wanted some clarity on it

[quartz tulip]

<@&286206848099549185>

[haughty dagger]

do you know how the sign of sin cos and tan changes with variation in quadrants

[dusk moat]

yes

[haughty dagger]

thats how

[dusk moat]

okay

.close

[pearl pond]

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[quartz tulip]

.close

[pearl pond]

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[visual canyon]

for this i need this formula right

[knotty meadow]

Yes

[visual canyon]

ok great, now,

i need to plug in cos theta?

[knotty meadow]

Look at the signs too

[visual canyon]

which is -3/5

[knotty meadow]

Yes

[visual canyon]

so i would get (- 3/5)(2A)= 2(-3/5)^2 (A)-1

that feels wrong

am i pluggin in values correctly?

[knotty meadow]

Why did you write A again?

No wait

A is not seperate

It's Cos A =-3/5

[visual canyon]

oh

so

where would i plug in my -3/5

?

[knotty meadow]

cos 2θ = 2cos^2θ-1

[knotty meadow]

Put cos θ in RHS

[knotty meadow]

See A as θ

[next dove]

$2cos^{2}(A)-1=2[cos(A)]^2-1$

[jolly parrot]

77²

[visual canyon]

2cos^2(A)-1= 2 (-3/5)-1

?

[next dove]

yup you got it almost

but with a square

(-3/5)^2

[visual canyon]

oh

whoopsie

lets use that then

2cos^2(A)-1= 2 (-3/5)^2-1

now how do i condense that?

to get a proper answer

cant use a calculator obviously

[knotty meadow]

You don't need one

Just solve it

Keep it in fraction

[next dove]

!occupied

[pearl pond]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[visual canyon]

-0.28

2c0s^2(A)-1= -0.28

[knotty meadow]

bruh the options are in fraction

[visual canyon]

its A

i like decimals i just convert back

[next dove]

you're odd one

but anyways

[knotty meadow]

wow

[visual canyon]

you guys are judgy

ik its incovienient

[visual canyon]

looking goof?

d

-7/25= -0.28

so

unless ive screwed up in a major way

[next dove]

it's just the regional trait

[visual canyon]

is the thumbs up a yes im wrong or yes im right

[knotty meadow]

You're good

[next dove]

it's alright

[visual canyon]

yipee!

thank both of you!

.close

[pearl pond]

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[pearl pond]

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[ruby crown]

is there a specific way to simplify the numerator?

[vital estuary]

sus

[ruby crown]

no like why is it 49x3 instead of some other number like 1x147

[vital estuary]

the denominator should just be 8, no sqrt

[vital estuary]

49 is 7^2

you can split sqrt(49*p^4*3)

to sqrt(49*p^4)*sqrt(3)

[ruby crown]

thats not really my question

[midnight haven]

I assume it’s to help you

[vital estuary]

you want to make whats in the sqrt as little as possible

[ruby crown]

oh

[midnight haven]

Because the sqrt of 49 is obvious

The sqrt of 147 is not so obvious at first glance

But whoever previously spoke is right

The bottom should not have a square root

[ruby crown]

wait so why does 147 become 49x3 instead of any other possible two numbers that multiplies to 147

[midnight haven]

Well

49 is a square number

If you can find any other square numbers that fit into 147 be my guest

[ruby crown]

ohh

[midnight haven]

But you’ll end up with the same result, 7 sqrt(3)

[ruby crown]

thanks i get it now

so whatever number that can multiply in~~ blank~~ numerator should be a perfect square?

[pearl pond]

@ruby crown Has your question been resolved?

[pearl pond]

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[pearl pond]

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[midnight oar]

im hardstuck 2nd question i just dont get how to find the major axis

[midnight oar]

do i need to use c^2 = b^2 - a^2

i know that b=3

and then what do i do with this chord length of 10

[pearl pond]

@midnight oar Has your question been resolved?

[zealous shore]

We're given that the minor axis is 3, so we know that the equation of this parabola is $\frac{x^2}{a^2}+\frac{y^2}{3^2}=1$. Since the chord we're given is length 10 and at height 1.5, we know two more points on the ellipse: (-5, 1.5) and (5, 1.5), we can plug in either of these points for x and y to solve for a.

[jolly parrot]

DonDoesMath

[midnight oar]

oh thank you so much

i solved it then

:))

.close

[pearl pond]

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