help-0

[oak chasm]

Right, you figure out what you want to get.

You add x to it.

So, let's try corner cases.

It worked for 30.

We tried one x value.

So it can sometimes work.

We need it to be able to give us ANY number greater than 1.

And it's easy to get really high numbers.

You just add the really high number to x.

But what about really low numbers?

You can add a low number to x, but you can't subtract something from x.

So if you need to subtract something, you're stuck.

[median mauve]

ok

[oak chasm]

We're adding the number we want to get.

So, subtracting would mean the number is negative.

But we're not looking for negative numbers.

[median mauve]

yes

[oak chasm]

Also, n ≥ 2.

So, the lowest number we can add is 2.

[median mauve]

ok

[oak chasm]

What do we need to check?

We handled adding negative numbers.

Do we ever need to add 0 or 1?

[median mauve]

no

[oak chasm]

Why not?

[median mauve]

from a x+lo to x+hi?

[oak chasm]

Sorry?

[median mauve]

here

[oak chasm]

Yes, n = number we want to get + x.

Hmm.

My question was wrong.

The number we want to get is greater than 1.

So, it's at least two.

x is at least zero.

At least two plus at least zero is at least two.

And so we need to be able to get all ns that are at least two.

And we are.

Does that make sense?

[median mauve]

yes

[oak chasm]

OK, so we can get any n - x - 1 value we need.

So we can get any odd 3 or higher as our odd factor.

Does that make sense?

[median mauve]

yes

[oak chasm]

Something I've noticed is that you need to know what you need to show.

Like you explain why it's true, but you leave out some things that are needed.

[median mauve]

ya

ty

[oak chasm]

No problem.

[median mauve]

.close

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[tawny fable]

Hi there

[tawny fable]

I am having some problems with some questions in calculus

[tawny condor]

Go ahead and ask

Here to help :D

[tawny fable]

Thanks!

well, I proved the first one.

with the help of @queen thicket

but the thing is, that I really cant tell why the second one is not true.

I prove the first one using the defenition of limits.

and using the fact that it is continuous and the rationals density in R

and it looks like I could do it just the same with < and not <=

exactly the same.

so its either I have a problem in my writing, or I have a problem understanding something

[noble sinew]

limits don't hold strict inequalities is where it breaks down

just like 1/n>0 for all n but lim n->inf of 1/n=0

so our 1/n>0 turned in to lim n->inf of 1/n>=0 when taking limit

[tawny fable]

So shouldnt I get something weird in my proof?

[noble sinew]

You likely thought and used that limit holds strict inequalities

And just did a false proof

[queen thicket]

@noble sinew I believe he did it with epsilons and deltas <:)

You can post your proof here and we'll tell you where it breaks down if u want

It's probably right at the end

[tawny fable]

ill translate it to english and rewrite it

so couple of mins

🙂

it took me a little while xD

but anyways. please let me know where am I wrong.

[queen thicket]

last set of inequalities ur using $-(f(q) -f(x_0))<-\varepsilon$ which is not true it's the other way around

[ocean seal]

Celephinnor

[tawny fable]

what do you mean?

[queen thicket]

$-(f(q) -f(x_0))<-\varepsilon$ is equivalent to $f(q) -f(x_0) > \varepsilon$

[tawny fable]

I am subtracting

[ocean seal]

Celephinnor

[tawny fable]

but I am not adding, I am subtractiong

subtracting

[queen thicket]

yes and substracting reverses inequalities

[queen thicket]

as in this

[tawny fable]

but all the values are bigger than 0?

| | < epsilon

[queen thicket]

$|f(q) -f(x_0)|<\varepsilon$ means $f(q) -f(x_0) < \varepsilon$ AND $ f(x_0) -f(q) < \varepsilon$

[ocean seal]

Celephinnor

[queen thicket]

If you have $a<b$ and $c<d$ then you cannot do $a-c<b-d$

[ocean seal]

Celephinnor

[tawny fable]

hmm

[queen thicket]

but you can do $a-d<b-c$

[ocean seal]

Celephinnor

[queen thicket]

Substracting reverses the inequalities

[tawny fable]

and what if I knew that $f(q) >= f(x_0)$?

lol

[ocean seal]

RanZ

[queen thicket]

then you would know one of these is useless because trivially true

Told you last time whenever you're substracting epsilons with epsilons you did something wrong

epsilons never cancel out in analysis

[tawny fable]

damn

so many mistakes..

[queen thicket]

Just 1 really

[tawny fable]

how?

I can't get rid of the | |

[queen thicket]

your only mistake is not reversing the inequality

wdym you can't get rid of the | |?

[noble sinew]

Well you are trying to prove something false - so ofc you are gonna get stuck at a point you are aware?

[queen thicket]

He's showing his proof for the first part

[tawny fable]

yup

[tawny fable]

in order to reverse the inequality I have to really understand why am I doing this

or to be able to explain that

[queen thicket]

You are doing this to show $0<g(x_0)-f(x_0) + 2\varepsilon$ or something

[ocean seal]

Celephinnor

[queen thicket]

Which in turn will imply $0 \leq g(x_0) -f(x_0)$

[ocean seal]

Celephinnor

[tawny fable]

well yeah

but we can not choose epsilon = 0 as I said.
and also, is that by addition?

I mean

ofc I want to prove that, I just don't get how to continue from where I am if I cant use substraction

[queen thicket]

You can use substraction

Just get the inequality in the right order

Instead of 0 on the right you'll just get 2epsilon

If u understand this step we can finish the proof by choosing adequate epsilon

I think the problem is you don't understand what $|f(x)-f(x_0)|<\varepsilon$ means

[ocean seal]

Celephinnor

[queen thicket]

Can you tell me what $|a| < b$ means ?

[ocean seal]

Celephinnor

[queen thicket]

You can make a drawing or pick numbers to check is u want

[tawny fable]

I think

you're right.

I am having some trouble understanding it

I went back to our lecture notes.

[tawny fable]

and I see now that it is just a shortcut to say that:
$f(x_0) - \varepsilon < f(x) < f(x_0) +\varepsilon$

[ocean seal]

RanZ

[tawny fable]

which means that f(x) is reaching f(x_0) at a certain point

since we can pick any epsilon we want

[queen thicket]

Yep

[tawny fable]

doesnt it mean we can change it to lim x--> x_0 of f(x_0) = f(q)?

[queen thicket]

No because there's no x in f(x_0) <:)

[tawny fable]

hmm

yeah

[queen thicket]

taking the limit as x->whatever of f(x_0) is just f(x_0) <:)

[tawny fable]

because it is constant

yeah

[queen thicket]

yep

[tawny fable]

So by saying ill get 2 epsilons on the right side

do you mean using the other side of the inequation?

I just don't understand exactly what does it mean "use the subtraction correctly"

[queen thicket]

Yes I mean using the other side of the equation

Then multiply by -1 which reverses the inequality

and then ADD

Substracting inequalities is a pretty good way to make mistakes

I advise you to only ever ADD inequalities

If u need to substract just multiply by -1 before adding

[tawny fable]

but when multiplying by -1 how does it change the absolute value?

[queen thicket]

No no i mean multiplying by -1 in this

After you got rid of the absolute value

[tawny fable]

something like this?

[queen thicket]

the idea is good

but the execution is wrong

because you've just made a loop

where you choose epsilon = g(q)-f(q)/2 before having defined q which depends on delta which depends on epsilon

this is highly illegal

<:)

[tawny fable]

true..

I just really dont understand this I guess..

i mean

I understand what is not legal

here

thats fine

but what should I do to make it work?

[queen thicket]

I'll write u something nice for u to see

do you mind me using $\forall$ and $\exists$ or do you want plain old english ?

[ocean seal]

Celephinnor

[tawny fable]

no its fine

just if you could write it by steps it would be great 🙂

[queen thicket]

yeye

Part 1

[lone heart]

@tawny fable Has your question been resolved?

[queen thicket]

Part 2

Sorry for the delay I fucked it up and had to redo it lul

[tawny fable]

thanks man

but what about the -2epsilon?

[queen thicket]

part 3 coming

[tawny fable]

lul

https://tenor.com/view/winnie-the-pooh-i-love-you-gif-10837742

[queen thicket]

part 3

[tawny fable]

so when you say it wouldnt work for >

it is because if you were assuming toward a contradiction it would be assuming it is <= 0

[lone heart]

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[tawny zodiac]

suppose we have a bag of marbles, the bag contains 5 red marbles, 3 yellow marbles, and 2 blue marbles. We draw the marbles two times without replacement.
what is the probability of getting at least 1 yellow marble? Show your work in percentage form, rounded to the nearest hundredth.

my work:
3/10 + 3/9
= 27/90 + 30/90
= 57/90
= 0.633333
= 63.33%

I keep getting 63.33% but the answer is 53.33%?

[lone heart]

@tawny zodiac Has your question been resolved?

[tawny zodiac]

<@&286206848099549185>

[weak prairie]

Yo just sent a photo

The opposite event to getting at least one yellow is getting no yellows at all, so we can take the probability of no yellows away from one to get the probability we get at least one yellow

You get it ?

[tawny zodiac]

oh yea, im getting 53.33% now

[weak prairie]

Nicee

[tawny zodiac]

just one more question, why does my previous method not work?

[weak prairie]

Because that’s calculating the probability you get a yellow marble in your first go, and a yellow marble in your second go ( should be 2/9 not 3/9), I’ll show you how to do it that way correctly

[tawny zodiac]

ohh ok

I kind of get it now

thanks so much, I am going to close this now.

.close

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[red sluice]

So i have to make a frequency table but numbers 1-5 and 7-9 all have value of atleast 1

[red sluice]

And 6:0 so how do i put it in the list?

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@red sluice Has your question been resolved?

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[tawdry spindle]

Need some assistance

[tawdry spindle]

I need to find the error in approximating xe^x by x+x^2+(x^3)/2+(x^4)/6 on [-1,2]

@alpine sable this channel has just been claimed by me

go to the other ones. there is 3, 9, 11, and 23

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[glass lichen]

what about it?

[analog kettle]

for a the -t would be 4 right

not quite sure how to do b or c though

[glass lichen]

t=4 for a, yes.

not -t=4.

[analog kettle]

ok

how would you do b and c?

[glass lichen]

b you're solving V(t)=1000

c is self-explanatory

[analog kettle]

how do you solve b with v(t)=1000?

[glass lichen]

do you know logs?

[analog kettle]

a bit

[glass lichen]

so.. you'd set V(t)=1000, isolate for the exponential part, then take logs

[analog kettle]

?? how do you write the equation for that

[glass lichen]

you're given V(t).

and you set whatever V(t) is, equal to 1000

[analog kettle]

ohh so its the 10000(1.5)

[glass lichen]

yes.

[analog kettle]

how do you isolate the exp?

[glass lichen]

recall algebra.

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[topaz mesa]

Hey, I'm curious why x^3(mod6) is always x(mod6). Is there an name/theorem for this phenomenon?

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@topaz mesa Has your question been resolved?

[jagged imp]

I don't know if there's a name for it, but this is equivalent to proving that x^3-x is always 0 mod 6. x^3-x=x(x-1)(x+1)

thats the product of 3 consecutive integers, which must be divisible by 6

and is therefore 0 mod 6

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[crimson beacon]

Help

[severe sluice]

hint: try drawing a line parallel to the bottom line that goes through the middle point

[crimson beacon]

I cant visualize it

Ohh

Got it

Ilike this?

[severe sluice]

yeah

[crimson beacon]

140°?

Thanks!

.close

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[devout summit]

No... You mean x=140°?

[crimson beacon]

360-280?

[crimson beacon]

Yeah

[devout summit]

Is this 140?

[crimson beacon]

Its 80

I just figured out after I got the wrong answer

[devout summit]

Yes. So, x=80°

[crimson beacon]

Thanks

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[void swift]

I am learning what slices are, but I fail to see how to use them in the context of this question.

[void swift]

In other words... I haven't a clue how to get started on this question.

[red sluice]

.reopen

[velvet magnet]

Its a tetrahedron

So u can use the vector form of the formula for calculating volume of a tetrahedron

By taking the point vectors of a,b,c

[void swift]

But this is from my calculus unit

I want to know how to use cross sections and/or slices to solve this

[cursive badger]

triple integral

you want to figure out each of the bounds

so

0≤z≤c (bounds for outermost), gotta start somewhere

0≤y≤b-bz/c (so that when z is 0, y is maxed out, when z is max, y is 0, and they increase/decrease at the same rate)

and then 0≤x≤a-az/c-ay/b

so that when z and y together are maxed out, x is 0 (since when they're maxed it's the triangle on the yz plane)

and when they are 0, x is at its maximum value

combine all those bounds into an integral where volume is constant (no duh)

then evaluate the horrifying abomination you have spawned

[void swift]

Sorry, could you slow down a bit? What is the meaning of a triple integral, what exactly does it convey

[cursive badger]

ok so

a triple integral is just

three integrals

kinda like this

https://www.iti-global.com/uploadIMG/rfUploads/CADfix/CAM/slice-gen.png

[void swift]

So is it like
int int int x y z
(not sure what the thing I'm writing even means)

[cursive badger]

so as you go up (through your z bound)

you add up a 2d area, each infinitely thin

which are those slices you see

then each of those 2d areas can be found by integrating through changing y bounds (since as z increase, the y bounds will change) to find the width at each point

[void swift]

Looks like the layers a 3d printer would be printing

[cursive badger]

and that width changes based on both the length and height

mhm, but infinitely thin

since this is calculus

not rectangulus

just picked this image since it was the first approximation I found for how I see it

a triple integral is literally three integrals

$\int\int\int_Ef(x,y,z)dV$

[ocean seal]

Scythe

[cursive badger]

in this case it's

$\int_0^c\int_0^{b-\frac{b}{c}z}\int_0^{a-\frac{a}{c}z-\frac{a}{b}y}{1dxdydz}$

[ocean seal]

Scythe

[void swift]

$V=\int\int\int dxdydz$ ?

[ocean seal]

K004

[cursive badger]

more or less, yeah

you can consider it to be

$\int_0^c\int_0^{b-\frac{b}{c}z}f(y,z)dydz$

[ocean seal]

Scythe

[cursive badger]

where

$f(y,z)=a-\frac{a}{c}z-\frac{a}{b}y$

[ocean seal]

Scythe

[cursive badger]

and then you can turn the next integral into another function

[void swift]

My understanding in English:

The volume is controlled by the product of the 3 "sliders" of x, y, and z.

[cursive badger]

yup

[void swift]

Wow, those are messy aren't they

[cursive badger]

exactly

they are evil

demonic

only matched by surface integrals

[silver viper]

btw $\iiint_E$

[ocean seal]

kiid

[cursive badger]

omg tysm

you're my new favorite person

but yeah, you're integrating areas along a line (the z axis), each of which are found by integrating lines along a line (the y axis), the length of which are found by taking the difference between points (x axis)

I'd draw and send a picture of a cross section

but not in a good situation rn to do that

[void swift]

Here is the solution for the question. I couldn't make heads or tails of it before.

[cursive badger]

ummmmm

your teacher is on drugs, I'm afraid

no lucid or sane person would use a fourth variable to solve this

it makes so much more sense and is simpler to simply use a triple integral

the hell

[void swift]

I couldn't understand what they were doing, felt dumb reading it

[cursive badger]

dw, I don't know what's going on either

except for the first line! it's the equation of the plane

which was given in the problem

[void swift]

I will try using the triple integral approach. Hope it won't annihilate me too badly

[cursive badger]

it's just lots of work

REMEMBER

when evaluating a dx integral, TREAT ALL OTHER VARIABLES AS CONSTANTS

z stays z. ln(y^3) stays ln(y^3)

they do not change, only the variable being integrated by is modified to find the antiderivative and get a final answer

[void swift]

I will keep that in mind!

Thank you Scythe!

[cursive badger]

glad I could help 😭

I'd walk you through it but I gtg

gl!

[void swift]

It's no worry, you already helped a lot, thanks!

[cursive badger]

check out #multivariable-calculus for help if you get stuck

cya!

[void swift]

👍

.close

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[rich ivy]

in which case would it be best to use each of these options?

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[shadow stream]

would the support be from 0 < U < 3?
i kind of tried this:
1 - 0 = 1
3 - 0 = 3
1 - 1 = 0
3 - 2 = 2

and chose the greatest and smallest number

[shadow stream]

i'm not too sure if im going the right direction

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[shadow stream]

<@&286206848099549185>

.close

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[tidal condor]

graph the solution set of each given system of linear inequalities.

[tidal condor]

<@&286206848099549185>

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[cunning raven]

please help I'm dying. I mainly want to understand why cos for x, and sin for y, as well as why my typed answer is marked wrong.

[cunning raven]

please ping me

[weak prairie]

The curve you’ve defined - as t moves from 0 to 2 pi, moves anti-clockwise around the circle

So you really want sin(t) to start at 2 pi and go to zero so change it to sin(2pi - t) and cos(2pi -t)

[cunning raven]

how did you see that it moves anti-clockwise?

Actually I kind of see why now

[cunning raven]

but why is x cos and y sin?

<@&286206848099549185>

I'm so confused

what is the difference between using sin and cos

but can you help me visualize the difference btween using sin and cos

for y and x

I get how the start is affected

you assume t 0

t=0

that makes sense

now what if i switch cos and sin

what will happen

how do you understand this

I get that

but how do you visualize the cos and sin

and decide on which one to use and how to format them

but how do you know to place cos for x and sin for y

oh

how about sin?

ok I'm starting to get this, but what about the difference between behavior in sin and cos and how that affects the parametric graph?

isn't there a difference to using them since cos and sin behave differently even though being in the same range as each other?

is there any way to make desmos show the direction of parametric graph

no i think they meant clockwise?

and x = h+bsint, y = k+bcost traces the circle centered at (h,k) clockwise?

what if i want to start from halfway?

basically yeah

do I just change t's domain?

so basically sin and cos are in charge of clockwise/counterclockwise

and h,k are in charge of center

and b is in charge of radius

t is in charge of position then?

that's all i need to know for this?

if I have more questions do I start another thread?

[whole geyser]

you can use sin(t+(pi/2))

[cunning raven]

isn't that more of a painful way of formatting instead of just changing t's domain?

[whole geyser]

t domains remains 0 to 2pi

[cunning raven]

this is still a bit confusing to me

[whole geyser]

about clockwise you can use - t instead of t

[whole geyser]

sorry if i get you confusing.

[cunning raven]

I think i get how this whole thing works if we just start from the beginning of the circle

but starting halfway or other stuff like that I'm still confused

beginning as in middle right

then can you explain c to me? I followed the video provided and am still confused

i don't get how negatives affect the graph

[whole geyser]

sin(-t) =-sin(t)

you can use it.

[cunning raven]

parametrics is so confusing

why is this relevant in the function?

oh wait

I think I messed up in reducing the function of h and k to that of only determining the center

i think they also have a role in where you start the circle

<@&286206848099549185>

wtf

can we get the mods

I'll see if she continues the clownery

Still confused about what to do for halfway

I'm pretty much all caught up for starting at base (1,0)

but still confused about (0,1),(-1,0),(0,-1)

what should i consider when writing for different positions?

I get how to do it at (1,0) [base]

x = cos(t) y = sin(t) counterclockwise, center (0,0), starting at (1,0) 0<=t<=2pi

[whole geyser]

its start from (1,0)

[cunning raven]

typo

x = sin(t), y = cos(t) would be starting at (0,1) then?

[whole geyser]

yes

[cunning raven]

but how do you determine direction then?

this is so confusing

[whole geyser]

counter clockwise if t is between 0,2pi

[cunning raven]

but you still start at (0,1)

asfisafsdflkfaksdfj;asldfj

[whole geyser]

you shift the beginning point

[cunning raven]

what are the valid forms to answering this question?

yeah

I'm dying this is really damn confusing

[cunning raven]

but x = sin(t), y = cos(t) would be starting at (0,1)

how is it that switching sin and cos will not affect the starting point?

but when t = 0 x = 0

how are you starting at (4,1)

well in this case for a) t is starting at 0

so wouldn't x=0+4sin t, y=1+4cos t set the starting point somewhere else?

[whole geyser]

@tough hatch can you check this?

[cunning raven]

god I am so confused

if you just set t to negative that will reverse the direction??

[whole geyser]

yeah

[cunning raven]

Ok so basically sin and cos are for starting points then

and t is for direction

[whole geyser]

not necessarily.

[cunning raven]

ah rihgt\

you need to use them with negatives as well

so like halfway for a circle of radius 1 would be

x = -cost, y = sint

and if I want it clockwise i set it to
x = -cos(-t), y = sin(-t)

[whole geyser]

for an halfway circle you must have half domain.

[cunning raven]

oh god

[whole geyser]

i mean, if your full circle is 0 till 2pi,
then your halfway is 0 till pi.

[cunning raven]

You know what I'll just fiddle with this thing https://www.geogebra.org/m/cAsHbXEU

I have more questions so I don't want to be too fixated on this topic

how do i create a slider

yeah I see

I'm also kind of struggling with optimization

[whole geyser]

@cunning raven watch this

nope, its a picture

[cunning raven]

i watched this video to get a better understanding of optimization https://www.youtube.com/watch?v=cUMvwG7wvzM

hello?

Sorry my keyboard is not working

ok my keyboard works again

I'll send the work I did so far

did I mess up by calculating the cost

I think the optimization is asking for minimizing price so that's what I went for

[whole geyser]

its an open top, so you have 1 wl

[cunning raven]

is the next part finding the derivative?

what's the fastest way of finding critical number?

[whole geyser]

alright. have fun. see you later

[cunning raven]

second derivative test?

yeah my bad for bad wording

that's what I meant

yeah but the domain of w kind of debunks that

ignore waht i just typed

wait but isn't this the normal function

don't we find =0 for derivative

function with no derivative applied

but don't I find c'(w) = 0 for critical number

c(w) = 30w^2 + 270/w

this isn't c'(w)

this is so weird

the problem is w is a really nasty number

i calculated

I got 1.6509

ok

alright

[cunning raven]

wait is 3/cuberoot(6) actually correct?

yeah

but within the cuberoot you still have a 10

how did you take that out

I'm ad

bad*

how do I do the 2nd derivative test on c''(w)?

is w' 3/cuberoot(6)?

@tough hatch

after this do i just sub w back into cost equation

well i see that it's addition statement

and w itself is pos

ok

I think im correct to assume that it is minimum

what do i do next?

yeah

sorry for bad wording

ok

ah right

I got 245.3 rounded

aight

this one is actually tricky af

thanks a lot for the help so far

sorry if I'm being a bother

[tawny condor]

@cunning raven Do you know trigonometry?

[cunning raven]

yew

yes*

[tawny condor]

okay

Mark the small side as x, as in the picture

You know the angle is 60 degrees.

Now using x and L, find the area of the rectangle, and use that to find for what x the area is largest.

[cunning raven]

alright lemme work on it

this is what I have so far. I am stuck

[tawny condor]

You are making this too complicated

w is indeed L - 2x

but for the other side of the rectangle, just look at the small triangle on the bottom right

You have an angle, and a side. you should be able to use that to find the side of the rectangle

[cunning raven]

I'm sorry I don't get what you are saying

[cunning raven]

how do you use the triangle at the rihgt to solve for w?

[tawny condor]

tan.

opposite over adjacent

[cunning raven]

but that's h

the small triangle doesn't encompass w

I have no idea what a degenerate rectangle is

[tawny condor]

@cunning raven Use the angle, 60

and the side x

in the small triangle

to find the side you need

with trigonometry

@tough hatch I think you're just overcomplicating this

Idk

[cunning raven]

oh frick did you mean make a small triangle with the rectangle

[tawny condor]

Like

That's not what they asked about tho

They are trying to find the area with L and x

I didn't say you were

@cunning raven

In this triangle

[cunning raven]

that encompasses height

what???

[tawny condor]

find y

in terms of x

[cunning raven]

y is h

what?????

[tawny condor]

No

What's h?

[cunning raven]

h is y

did you see my drawing

[tawny condor]

Oh

wait

No

You used cosine

[cunning raven]

I'm confused about how to find the top and bottom of the rectangle

yeah mb

that was on me

should have used tan

[tawny condor]

You need to use tan

yes

so just isolate h

[cunning raven]

but i need w

i don't know how to get w

[tawny condor]

w is easy

subtract the 2 small sides, from the whole side of the triangle

[cunning raven]

isn't that just l-2x

[tawny condor]

Yeah

[cunning raven]

yeah but this is 3 variables

I need to reduce it to 1

[tawny condor]

No

L is a constant

they want the answer to have L in it

Just look at L as a number, not a variable

[cunning raven]

dfaq

[tawny condor]

It's like

instead of giving you a side of 3

they want to generalize it

so call the side "L"

Refer to L as a number

[cunning raven]

how are you supposed to do 2nd derivative test with that abomination?

[tawny condor]

You don't need a second derivative test

Only first

but okay

lets say

you need to take the derivative of

L * x

what would that be?

[cunning raven]

but isn't 2nd derivative test more clean

[cunning raven]

L

[tawny condor]

sometimes no

[tawny condor]

great

[cunning raven]

how do you test first derivative

[tawny condor]

I'll show you when you get there

k?

[cunning raven]

k

[cunning raven]

Are you sure this is what I'm supposed to do

[tawny condor]

the side is not x/tan(60)

Also...

tan(60) has a value

use your calculator

or just use ur head

An angle of 60 forms the golden right triangle

90 60 30

[cunning raven]

tan60 = sqrt(3)

iirc

[tawny condor]

yeah

now find h again

[tawny condor]

good job!

now

the only thing you need to do, is show that this is the maximum point, right?

[cunning raven]

right so 2nd derivative test or 1st?

[tawny condor]

there is no 1st test or whatever

it's logic

look at the original area

and multiply out the brackets

You get

-2sqrt(3)x^2 + sqrt(3)Lx

this is a parabola.

[cunning raven]

ah

[tawny condor]

and since there is a negative in front of the x^2

it will face down

which means that the critical point must be a maximum

you can do a second derivative test if you want

Anyways, now all that's left is to calculate the sides of the rectangle when x = L/4

[cunning raven]

w = L-2x

[cunning raven]

is the final answer
x = L/4
y = L/2

[tawny condor]

y?

One side is L - 2x, right?

[cunning raven]

mbmb

[tawny condor]

L - 2 * L/4

L/2

So L/2 is one side

now find the other one

[cunning raven]

h = Lsqrt(3)/4

[tawny condor]

yup

question done!

[cunning raven]

ty

[tawny condor]

np! :D

[cunning raven]

One more question, how would you set up the equations for this problem?

[tawny condor]

I would mark this as x.

That way, you have the whole bottom side (x + 4) and you can find the height of the building to the ladder

(By using similar triangles)

Then, once you have the 2 sides, use Pythagoras theorem to find L.

You always need to mark a variable, find what you need the maximum/minimum of using that variable,

Find the derivative,

Find the critical point of the variable you marked,

And then you know for what value of it, what you needed is maximum/minimum.

[cunning raven]

don't you need theta to use similar triangles?

[tawny condor]

theta?

[cunning raven]

angle

[tawny condor]

Similar triangles. Not trigonometry.

[cunning raven]

how do you use similar triangles again?

[tawny condor]

Triangle ABC is similar to ADE

[cunning raven]

yup

[tawny condor]

the ratio of their length is always the same.

So:

AB/AD = BC/ED

which is also equal to AC/AE

But you don't need that part

[cunning raven]

I see that's cool

[cunning raven]

wait this is a bit wack

[tawny condor]

simplify what's inside the exponent

Open the parenthesis and simplify

[cunning raven]

is that even possible?

[tawny condor]

Why not?

(4+x)^2

x^2 + 8x + 16...

same thing with the other one

[cunning raven]

i tried simplifying it before

yeah it's the other one

[tawny condor]

Ok let me try then

[cunning raven]

that I'm having trouble with

the 4+x is nice

[tawny condor]

okay I see what you mean

[cunning raven]

yeah

I thought that it'd be more nice to just derivative them in square form

[tawny condor]

you did the derivative wrong

You need to multiply by the derivative of the inside

[cunning raven]

frick how does chain rule go again

yeah it was not done

I panicked after seeing its scale

[tawny condor]

okay thats

weird

I don't think you're supposed to get this kind of derivative.

Let me see where is the problem.

[cunning raven]

thanks I also think this is sus

[tawny condor]

That doesn't help.

What he wrote turns to this

[cunning raven]

holy frick that looks cancer

[tawny condor]

The real problem is the derivative of the inside.

Maybe this is how they want him to do it, if so they are evil...

I don't see a better way to do this.

OH

@cunning raven

There is something called thales theorem

Did you learn that?

[cunning raven]

never

[tawny condor]

@tough hatch It's missing the derivative of the inside

Okay so

[tawny condor]

Look back here

oh wait.

gimme a sec

I got this for the length

to be honest I don't know

yeah right

nothing cancels out bro

Oh

(4+x)^2 you mean?

The squared cancels.

So?

Tell me

What cancels out?

@cunning raven You'd have to do it that way. Just take the derivative and compare it to 0 I guess.

Oh so yeah

I mean cool

[cunning raven]

I'm so confused

[tawny condor]

But 1. that doesn't make it that much better

2. I don't think we even need to open the parenthesis there.

[tawny condor]

Take what you got there, multiply with the derivative of the inside

Don't touch what's inside the exponent, when you compare it to 0 it doesn't matter.

Anyways, I have to go

[cunning raven]

aight thanks for the help

[tawny condor]

good luck with that monstrosity

[cunning raven]

what does it look like?

I kind of have to somewhat expand for chain

what I have so far

how do you product rule thi

yeah

f'g + g'f

[cunning raven]

did you get the right side by expanding it?

how did you product rule?

[cunning raven]

I'm sorry would you mind showing me how you got here step by step

I'm kinda noob

also what do you mean when you say radicand?

I'm honestly confused at how you bypassed chainrule

yeah

ok

but you still need to multiply the others via chain

[cunning raven]

Sorry my keyboard is going nuts again

ok it's back

thanks for the hard work trying to get that damn bot to compile

[cunning raven]

do you just ignore the square root at the very outside?

[cunning raven]

what about here then

it's missing the square root at the very outside?

so we're ignoring the first part of the chain basically for now

im noob at math

OHHHHHHH

I get it now

[cunning raven]

I only get it up to this point

then after I'm confuse

[cunning raven]

i don't know what you mean by denominator

is this part of the chain ?

[cunning raven]

I'm honestly questioning if the problem intended the derivative to be this cancer

[cunning raven]

yeah

but again im questioning if the problem intended on the derivative to be this complicated

yeah i see that

are you sure that this is the only way to solve this question?

or if the approach is even correct to begin with?

yeah i see this way

I'm confused why they put something like this in Calc1

is there a more easy/simple way

wo

w

how do I test the crit number after this?

yeah

[cunning raven]

how did you go from this

[cunning raven]

to this?

the other one is the simplified ?

I see

the numerator is also unpleasant

[cunning raven]

I honestly just

yeah I see

yeah, if you don't mind, can you help me with this one? I'm on my last submission attempt on my webassign so if I get this question wrong 1 more time I lose permanent marks.

no

webassign is hw

I have to use the
dy/dx = (dy/dt)/(dx/dt) thing

I know I got (dx/dy) correct

but (d^2y/dx^2) is

aafdasdf

I already submitted wrong (d^2y/dx^2) 4 times

I'm on my last submission attempt

[cunning raven]

wait but isn't it

(d^2y/dx^2) = ((d/dt)(dy/dx))/(dx/dt)

[cunning raven]

uh

this is also complicated method

yeah

yeah but how do i know that it's the actual answer

im trying it right now

My keyboard is problematic again

f(x)=lnx(1/x)

dy/dx = (1/x^2)(1-lnx)

aight

is this the correct answer 100%?

if you don't mind can you help me check for this one?

i can't hand it in wrong

aight ty

hello?

which two?

aight

THanks a lot. Do you mind answering one last quick question?

for the same question

this was the line of thinking i was going through

but idk about the interval part

how do i determine that? set it equal to 0?

I see, and how do I pinpoint that?

I'm talking abotu like systemic and structured approach

ah right thanks for jogging my memory

so the deciding point is whether t is greater than or smaller than 3/2

so is the answer (3/2, infinity)?

so it's better to transform the derivative to be in terms of x then solve?

i see.

Aight I gtg now. Thanks a lot for the help.

[lone heart]

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@thin osprey Has your question been resolved?

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[devout summit]

Modifying it to A(x-h)+B(y-k)+C=0

[devout summit]

Changing h shifts the line left or right. While k allows up and down movement

I understand

Suppose you wanted to shift the line Ax+By+C=0 left "h" units without changing slope, the new equation would be
A(x+h)+By+C=0

Umm you mean changing C to C+Ah?

Yes that would shift your line left by "h" units

Yes sure. Negative values of h in
A(x+h)+By+C=0 shifts to h units right

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[zinc hamlet]

Hello

[zinc hamlet]

I need help with going to polar coordinates. I have x^2 + y^2 = a^2

But i want to use sqrt(x) = r cos (phi) and the similar to y

And my question would be about radiuss, if I use this substitution, then radiuss = sqrt(a) or i have to calculate putting everything in x^2 + y^2 = a^2 ?

I tried putting x = r^2 cos^2 (phi) and y = r^2 sin^2 (phi) in x^2 + y^2 = a^2 and tried to plot it, but it did not look similar to a circle

For more context, I need to integrate region bounded by z^2 = x*y and x^2 + y^2 = a^2. Just using x = r cos phi and y = r sin (phi) gives me an awful integral I dont know how to integrate, therefore I am using different substitution, but now I cant figure out the upper limit for radius

<@&286206848099549185>

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@zinc hamlet Has your question been resolved?

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.close

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@desert zephyr Has your question been resolved?

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@desert zephyr Has your question been resolved?

[vivid crater]

#help-8, ty

[late bridge]

what's stoping you from arguing that since $\langle w , g\circ f (u)\rangle = 0 $ it is the case that $\langle f^{\star}\circ g^{\star}(w) , u\rangle = 0$

[ocean seal]

Brun。

[late bridge]

genuine question, idk if that does it

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[pliant sonnet]

someone can help me to understand linear functions

[icy trail]

hmm, what specifically

[alpine sable]

@pliant sonnet

[pliant sonnet]

gimme a sec

This

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[wide plaza]

I need to prove triangle LJM congruent to triangle LNK using one of the triangle congruency theorems. (SSS, ASA, SAS, AAS, or HL). I can't figure out how though...

[devout summit]

EFC CDF? Is it a different question?

[wide plaza]

oh sorry

@devout summit

look now again

thats the correct thing

my bad

[devout summit]

Better

[wide plaza]

So.... What do I do, precisely?

[devout summit]

Notice LJM and LNK

LJ=LN, LM=LK

[wide plaza]

yes

I see that

[devout summit]

There is also an angle between the sides which is common

Namely, angle JLM and NLK

[wide plaza]

yes

i see that

njk does not exist btw

[devout summit]

Right. I edited it to NLK

Now that a pair of sides and an angle between them are shown to have equal measures, which test should we use? @wide plaza

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@wide plaza Has your question been resolved?

[wide plaza]

omg im so sorry

euclid

im back now

[wide plaza]

wdym by test?

@devout summit

[devout summit]

Congruency theorems

[wide plaza]

so the pair of sides are lm and lk right?

[devout summit]

Pair of sides congruent are just the ones already given (in the problem)

LM, LK and LN,LJ

[wide plaza]

hmm

[wide plaza]

ok yea i get that

[devout summit]

So which theorem?

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[dire crown]

Can someone give me a quick explanation of what indefinite integrals are? I under stand how normal integrals work.

[pine kettle]

I have not learned integrals from a teacher yet but isn't the indefinite integral where you add C to your antiderivative because