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[pearl pond]

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[sharp smelt]

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[still dirge]

Please explain why is the drawing the way it is

[lucid anvil]

it might help to understand what the drawing is actually meant to show

here the drawing is just the region S that you're integrating over

which corresponds to [0, pi/2]x[0, pi/2]

the drawing doesn't convey any info about the integrand

[still dirge]

Thanks, I do know this.

I meant this: "it might help to understand what the drawing is actually meant to show
here the drawing is just the region S that you're integrating over
which corresponds to [0, pi/2]x[0, pi/2]
"

However, what I don't understand is:
why is pi/2 where it is.

this is not a coordinate system, is it

[lucid anvil]

we're just viewing S as a subset of the xy plane

so like the x coordinate is in [0,pi/2]

as is the y coordinate

[still dirge]

but there's only numbers on xy plane, not angles

pi/2 is 90*

[lucid anvil]

pi/2 is a number

[lucid anvil]

we're not considering pi/2 as an angle here

it is just a number

[still dirge]

okay

thanks for clarifying

.solved

[pearl pond]

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[sharp smelt]

Hi,how'd I find the mean here? Can I just ingtegrate from theta t infty

that feels sus

I'm getting a mean of 1?

[past perch]

could you show your work i only managed to simplify and i think youd have to work with limits

[sharp smelt]

$\alpha \theta^{\alpha} \int_{\theta}^{\infty} \frac{1}{x^{\alpha}}dx$

[past perch]

the mean is xf(x)

[sharp smelt]

oh shoot

I messed uo there

[past perch]

yeah you wont get a numeric value i think

[jolly parrot]

Wai

[past perch]

yeah now just power rule

provided a neq 1

[sharp smelt]

yea, gives us , uh

[past perch]

a > 1

[sharp smelt]

$\frac{\alpha \theta^{\alpha}}{(1-\alpha) \theta^{1- \alpha}}$

[jolly parrot]

Wai

[past perch]

really? afaik wouldnt we handle the infty lim separately

[sharp smelt]

well, that goes to zero

[past perch]

as x->infty the integral diverges

oh wait that's for a<1

[sharp smelt]

now to find the MLE of mean using invariance

which is the part I don't gte

*get

[past perch]

wait this isnt checking out can you show the steps cos i think the theta term gets multiplied not divided

[sharp smelt]

oh you're right

yes

so we get $\frac{\alpha \theta}{1- \alpha}$

[jolly parrot]

Wai

[past perch]

$\alpha\theta^\alpha \left[\frac{x^{1-\alpha}}{1-\alpha}\right]_{\theta}^{\infty}$

[jolly parrot]

sam beam

[past perch]

alpha - 1 no?

0 -

so reverses

[sharp smelt]

right

[past perch]

||i have not done MLE 😭 ||

[modern talon]

so this is a pareto distribution

to find the mle of the mean using invariance, you must obtain values alpha hat and theta hat

i.e find mle's for alpha and theta

then sub those for alpha and theta in E(X) to find mu hat

[sharp smelt]

Ah, cool

I thought there's some way without that 😭

[modern talon]

unfortunately no

[modern talon]

invariance just means this: if you are trying to find an estimate for some parameter in terms of another parameter with a known MLE, then plugging the MLE into the function produces the MLE of the first parameter

i.e g(theta) is maximized when inputting theta hat

[past perch]

yeah i looked into it it had something with a gamma function(?) i'll check it out after my exams

[modern talon]

that, i am unsure of. gamma distribution is a prototypical test of making sure invariance works namely because you have two parameters, alpha beta, but thats as much as i can think of

pareto could be another choice but its a less clean dist than gamma is

[sharp smelt]

cool

thanks

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[torn lodge]

im not too sure how to do part d.
I've found the coordinates for R which was (28cos(theta)/25 , -21sin(theta)/25) and I tried setting the x coordinate to be a variable and the y coordinate to a variable and rearranged both so that i could sub into the identity of sin^2 x + cos^2 x = 1 but i wasnt too sure how that would prove that R has the same eccentricity as the ellipse

[torn lodge]

oh and the eccentricity from part a is (root7)/4

[heavy onyx]

what equation did u get when u subbed in sin^2(x) + cos^2(x) = 1?

[torn lodge]

second fraction is supposed to have 784 on the bottom*

[pearl pond]

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[torn lodge]

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[daring bay]

hello im stuck on the 1.5

[rough forge]

Was ist Material 1

[daring bay]

bro

im about to cry

[rough forge]

i dont have tissues but i can try to help

[daring bay]

so basically from -1,5 to 1,5 theres one solution?

i dont understand

yeah wtf man

[steep saddle]

maybe you should integrate it

[daring bay]

hmm

but what is meant by -1,5 < k < 1,5

[honest spear]

i mean you can use symmetry and the fact that the function is strictly increasing after a certain point

lets call area A(k)

thing about A(-1.5) compared to A(-0.5)

[daring bay]

hmm

its not the same

or wdym

should i just try integrating it

[honest spear]

i mean you can but that defeats the point fr

cuz then its trivial

its testing your intuitive understanding

of the intregral and area

as k increases

from -1.5 up until like 1 A(k) is descreasing

[daring bay]

yup yeah

but idk what is meant by solution

wait

[honest spear]

for how many unique k values is A(k)=4

[daring bay]

so the area from -1,5 to 1,5

is 4

[honest spear]

no

wait

no

its like 5.something

you are integrating from k to k+1

so the width is always 1

they are just saying there exists

a k

between -1.5<k<1.5

s.t

A(k)=4

[daring bay]

ohhh

how the hell do you do this with the graph omly

wtf

[honest spear]

do it with the calculus you know

ig

then try see how that relates to the graph

thing is you can see the area is descreasing then increasing

so it goes less than 4

then goes back up to 4

but the function is strictly increasing

so A(x) is strictly increasing

A(K)*

for K>1.5

so it'll reach 4

then just keep blowing up

so theres 1 solution

[daring bay]

by reach 4 do u mean reaching like

4 on the y axis

wait so you can count area like that?

[honest spear]

no like the area is 4

https://www.desmos.com/calculator/iq2yvmpcdx

heres something i threw up

if you press play on the k value

it'll show you how the area we are looking at changes as k increases or descreases

if i were to freeze k at the right moment then A(k)=4

[rough forge]

I think you paraemterized the black lines wrong

[honest spear]

yeah it goes further than it should for k<1.5 but its to help show how happens just as k increases

[rough forge]

It should just be x=k and x=k+1

[daring bay]

but how can you like imagine the area like that

how can u read the area from the graph

[honest spear]

pedantics

[honest spear]

its the area?

under the graph

[daring bay]

yo im stupid

[rough forge]

The integral equation means "Die Fläche von f(x) im Interval [k,k+1]"
-# Let's not talk about average value here

[daring bay]

yea

wait let me draw

[daring bay]

so this area equals 4?

[honest spear]

no

here me out

[daring bay]

wt fman

[honest spear]

confirm that yourself

you know $A(x)=(\frac{x^4}{12}-x^2+4x)_{k}^{k+1}$

[jolly parrot]

Nyxzore

[honest spear]

thats the integral of f(x)

[daring bay]

why /12

[honest spear]

4*3 is 12?

[daring bay]

oh

okay

yeah

[daring bay]

yes i got it

so?

[honest spear]

now to work out

this green area you have

i would need to integrate from -1.5 to 1.5

notice how thats 2 units long?

[daring bay]

yeah

wdym 2 units

[honest spear]

,w 1.5^4/12-1.5^2+4(1.5)-((-1.5)^4/12-(1.5)^2+4*(-1.5))

[steep saddle]

isn't that3

[honest spear]

:/

yeah

[jolly parrot]

Results provided by WolframAlpha

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

Requested by nyxzore

[steep saddle]

the answer is 1 k >= 1.5 because its strictly increasing

see for k = 1.5 A(x) < 4

since its less than 4 but the function strictly increases and therefore A(x) increases

there must exist a k such that A(x) = 4

and there can't be another one for the same reason

its strictly increasing so A(x) keeps growing to infinity

[daring bay]

i just cant seem to understnad what −1,5 ≤ k ≤ 1,5 means

i cant get it in my head

[steep saddle]

it means there is a k in that interval

such that A(x) = 4

[daring bay]

ohhh os

so

[honest spear]

.

[daring bay]

from -1,5 to 1,5 theres a starting point a and b for which the area is 4

basically

[honest spear]

well just a

b can be outside of that interval

a=k

b=k+1

[daring bay]

yes

but how does that HELP

for the graph thing

how did you guys figure it out in your head

[honest spear]

This is how

My function is decreasing starting from -1.5 and i eventually got an area of 4 so i must of started off with an area > 4 then descreased until i got area=4 at k

but its keeps decreasing so my area gets <4

but then it starts increasing again

[daring bay]

yeah

okay so

[honest spear]

for k>1.5

so lets say its stops decraesing at Area=2.71 - made up number

itll then start increasing again for k>1.5

[daring bay]

it stops decreasing at 1,5

so

but what does that mean

wtf

why is there only 1 solution then

[honest spear]

cuz its strictly increasing

the area starts at some number >4 right?

lets say 5.7

[daring bay]

yeah

you mean

yeah

[honest spear]

i go from 5.7 down because my function is decreasing my area must be decreasing

eventually by area gets to 4

thats the k they mentioned

now its keeps decreasing to some number

lets say 2.71

and finished decreasing

[daring bay]

yes

[honest spear]

then my function starts increasing

[daring bay]

then it has to increae

[honest spear]

at k>1.5

[daring bay]

yeah

[honest spear]

so my area go up from 2.71

[daring bay]

yep

[honest spear]

passes 4 again

[daring bay]

YES

[honest spear]

then blows up to infty

[daring bay]

yes this is just logic

wtf

[honest spear]

^

[daring bay]

obviously there can only be 1 solution right

yeah

i just gotta word it right

tysm

[honest spear]

<#

❤️

.done

!done

[pearl pond]

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[daring bay]

.close

[pearl pond]

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[daring bay]

hello, for the 1.2

[daring bay]

do i just calculate the inflection point like id do normally?

because i get f''(x) = 2x

[sweet junco]

yeah

[daring bay]

okay but 0 = 2x

wtf

[sweet junco]

you can divide both sides by 2

[daring bay]

so x is 0

yes but

[autumn trellis]

What is the issue here?

[daring bay]

thats just weird

it just appears unusual

and i thought there had to be a catch because it gives 5 full points

[sweet junco]

oh yeah you still have to find the value of a and the point

[daring bay]

why a

we only need x no??

i mean the y value

[sweet junco]

how do you find the y value

[daring bay]

wait

f(0) = 0

the fuck

[sweet junco]

is it though

[daring bay]

no

2a

my bad

[autumn trellis]

Any cubic of the form x^3+px+q will have an inflection point at x=0

[sweet junco]

yeah

[daring bay]

yeah so thats the coordinate?

[autumn trellis]

Yes, that's the inflection point.

[daring bay]

no way this gives 5 points

thats 5% wtf

[sweet junco]

I think it's 5 points because it's saying show it algebraically

[daring bay]

no i think that just translate dpoorly

let me check the solutions

yep wtf

damn

i thought id have to do some stuff with somelocus

when is that the case again?

[sweet junco]

locus points are relevant?

[daring bay]

wdym

for the exam?

[sweet junco]

in general

never seen them used

ever

[pearl pond]

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[daring bay]

so here

[daring bay]

can i just say (2k|0) and do 1/2k * x^2 = 0

[rough forge]

yes

[daring bay]

yo bro im actually starting to get a grip of sll this

okay but

1.2

do i just calculate the extremes then do the formula for distances

or

[sweet junco]

there's an easier way

I think

[daring bay]

hmm

let me think

yeha if

[sweet junco]

k is positive

[daring bay]

idk

[sweet junco]

so what can we say about the graph

it's always positive

[daring bay]

yeah

okay

[sweet junco]

except at two points

what are these two points

[daring bay]

uhh

[sweet junco]

hint, it's when Gk=0

[daring bay]

the zeropoints?

[sweet junco]

fk(x)=0*

[daring bay]

how do you know that

[sweet junco]

yeah

those are now the minimum values you have so what can you say about the x value of the maximum point of Gk between both minimum points

[daring bay]

wait but just because k is positive

wait

[honest spear]

[rough forge]

I guess you could try to show that the function is symmetric to x=k

[sweet junco]

if k is positive
then 1/2k is positive, x^2 is also always positive, and (x-2k)^2 is always positive

[daring bay]

ho yeah

okay so

[sweet junco]

so the lowest possible point is when the function = 0

[daring bay]

its between them

[sweet junco]

yep

[daring bay]

yeah but like

how does that help

[silver gulch]

Shouldn’t u move it to the right = 1.5 and the left 3.5 goes up

[sweet junco]

the distance of the maximum is between the two minimums and equal

that must mean the x value of the maximum is perfectly in between the x-value of the two minimums

[daring bay]

soo

we do the distance

from one zeropoint to the other

?

[sweet junco]

not yet

we're getting there

[daring bay]

okay

uhh

[sweet junco]

do this first

[daring bay]

yeah

wait but

a zeropoint isnt a minimum

[sweet junco]

what's a zero point

[smoky gull]

like f'(x) =0?

[sweet junco]

have I been misinterpreting y-intercepts with x-intercepts here

[daring bay]

f(x) =0

[sweet junco]

yeah

we just established that f(x)__>__0

so the minimums

[daring bay]

okay yes but

[sweet junco]

are f(x)=0

[daring bay]

okay

yeah

i guess

aso

[sweet junco]

what are you not getting about this explanation

[daring bay]

where were we

[daring bay]

like i thought f'(X) = 0 are minmums

maximums

[sweet junco]

those can indicate minimums or maximums

key word: can

even if you have f'(x)=0, you can't really tell if they're maximums or minimums from just the slope being 0, you also can't tell if cusps or similar are minimums or maximums

[daring bay]

yes but how r u so sure theres minimums at f(x) = 0

[rough forge]

The idea is that it's not a necessary tool here because f is always non-negative, but also has roots, so its minima lie on the x-axis

[sweet junco]

f(x)__>__0
what's the smallest possible value of f(x)

[daring bay]

0

yes

[sweet junco]

yes so what are the minimums?

[daring bay]

ohhhhhh

[daring bay]

but they still depend on k

but

yea

0

[sweet junco]

and

[daring bay]

0

?

[rough forge]

How many roots do you have

[daring bay]

wdym roost

roots

[rough forge]

Nullstellen

[daring bay]

2

[rough forge]

name them

[daring bay]

2k and the other one i didnt calculate

but

[rough forge]

u already said it

0

[daring bay]

should i do it rq

[rough forge]

so as nautilus said, f(x)=0 gives you here also the minima which are the "zero points" so at 0 and 2k f attains its minimum

[daring bay]

yeah the other one is ß

0

[daring bay]

oh shit

1/2k * x^2 = 0

that equals x = 0 right

[rough forge]

Yes

[daring bay]

ohhh

[sweet junco]

so now as we were talking about

[rough forge]

I would have gone here with the derivative, to conclude that the local max lies on x=||k|| and the prove f is symmetric to x=||k|| in order to conclude same distance from min to max

by symmetry

[daring bay]

how do u even derive that function

do you need the rule for that

[rough forge]

wait i realized you actually need the distance itself

[sweet junco]

it's between 0 and 2k

like

perfectly in between 0 and 2k

[daring bay]

yes

yep yea this makes sense

[sweet junco]

what value is directly in between 0 and 2k

[daring bay]

well

i mean

it could be anything

because of k

[sweet junco]

let k=1 what is the only number in between 0 and 2k

[daring bay]

1

[sweet junco]

now what is 1

[daring bay]

the highpoit

maximum

[sweet junco]

1 is k

don't overthink it

[daring bay]

okay

[sweet junco]

what is perfectly in between 0 and 2k is (0+2k)/2=k

[daring bay]

i mean but

[sweet junco]

so we know x=k

happy 4th of july

[rough forge]

just product rule and then you can easily factorize

[sweet junco]

oh wait it's may

[daring bay]

wait

what

what did u do there

[rough forge]

this assumes that we know the function is symmetric at x=k

[sweet junco]

we're essentially told that

[rough forge]

oh i cant read mb

i'll leave lmao

[daring bay]

no

[sweet junco]

nah u stay I gotta sleep

[daring bay]

no wait

here

[sweet junco]

and I fear my explanations would confuse them more

[daring bay]

why did you do (0+2k)/2k =k

[rough forge]

Mittelwert

Both 0 and 2k are equally far away from the maximum, so you just compute the Mittelwert

[sweet junco]

the simplest way I can say this is midpoint formula

except you remove the y value

[rough forge]

denom is only 2

(a+b)/2

[daring bay]

holy shit

so you basically sum them up and divide by 2

is that the distance then too?

[rough forge]

no

[daring bay]

thats the x value

?

of maxim

[rough forge]

it only tells you at what x the max lies

[rough forge]

yes

[daring bay]

im so smart man

[rough forge]

now you can draw a triangle and use pythagoras

[daring bay]

so then y value

wait so then

y value of middle value

and then ^2-^2

in root

x, y

[rough forge]

what

your max lies on (k,f(k))

[daring bay]

like this

[rough forge]

Distance between two points

[daring bay]

yea

[rough forge]

but that's just the same

it's more intuitive to come up using Pythagoras by urself

[daring bay]

okay but

[sweet junco]

if you randomly know about arc length formula you can decompose it into this

[daring bay]

i do 100%

i memorized it from vectos

but

i can use the (0|0) as one point too?

[sweet junco]

sqrt(1+(dy/dx)^2)dx=sqrt(dx^2+dy^2)=sqrt((x2-x1)^2+(y2-y1)^2)

[sweet junco]

yeah

that makes it easier

[daring bay]

yeah so

[sweet junco]

this is a crime against math tho

[daring bay]

the problem is

i think i know how to solve problems, but

there could so much variation

[sweet junco]

learn adapt overcome

you do not win unless you lose first

[daring bay]

yea but my finals in like 2 days basically

but ty

[sweet junco]

I'm in a similar situation

I've been procrastinating studying multivariate calculus and also forgot the past lessons for the past 12 weeks

idk tho it is possible ¯_(ツ)_/¯

[honest spear]

...?

[daring bay]

bro we are fucked 😭

[honest spear]

isnt adonis helping

plus how are you asking more than a question a day 💀

1.1 or 1.2

[sweet junco]

can't people ask infinite questions

[daring bay]

no i understood it

[daring bay]

last exam i got like 70% because i was asking hella questions

people here r really crazy

[honest spear]

is chrue but if you are your time is better spent reading and understanding the topic

[daring bay]

they help so much

[sweet junco]

I don't think I am my country has laughable education standards

[honest spear]

same same

[sweet junco]

reading doesn't always help + they were curious

[rough forge]

no

[sweet junco]

this is reminding me we have technology and we can use calculators to integrate now

[daring bay]

waizt

im trying to derive the function

just for practicing

[pearl pond]

@daring bay Has your question been resolved?

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[daring bay]

yo so how do i sketch the graph with like 0 info

[rough forge]

did u read the text

[daring bay]

yes

u mean the z - a part

?

[rough forge]

yea

You got pretty much everything in Material 1

z>a for interval

then z=a at some point (notice how m behaves there too)

then a>z

[daring bay]

yeah but how does that help

[rough forge]

draw some lines

in the z>a the difference that remains purple, after you subtract the green is what you get for m'

Which makes sense in the interval where z>a <=> m'>0 means m is increasing but the slope flattens out as you can see at z=a

which means m'=0 so it touches the x-axis

then m decrease which means m' will be below x-axis

but notice that as m decreases it seems to flatten out again which would imply something like m' -> 0

[daring bay]

hmm

yeah i guess but what does that line mean

for our m‘

[rough forge]

z-a

you are just subtracting function values

[daring bay]

yeah

[rough forge]

if you do some of these you could get a rough idea

ig something like

,w plot D(x/(x^2+1)) on Plotrange -> {{0, 5}, {-.5, 1}}

stupid wolfram

[daring bay]

soo

what does the purple line

what is it in m'

[rough forge]

the difference

[jolly parrot]

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[rough forge]

You have two distances

z distance and a distance now you do z-a, if z is greater then the purple part denotes what's left when we take away the green part (distance of a)

Which is the distance from the value of m' at that point

[daring bay]

hmm

wait i just realiezd

i shouldnt even be doing this exercise

[rough forge]

[pearl pond]

@daring bay Has your question been resolved?

[pearl pond]

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[haughty cosmos]

I am stuck in last part

I found the potential line that is tangent to line and it goes through origin

I think I have to find the gradient of line

But how do I find the gradient of the line

[pearl pond]

@haughty cosmos Has your question been resolved?

[pearl pond]

@haughty cosmos Has your question been resolved?

[dense jasper]

tbf if you let $a=|a|e^{i \theta}$, then
$$az^{\ast}+a^{\ast} z=0 \implies x \cos \theta+y \sin \theta=0,$$
which is a line through the origin with normal angle $\theta$. For this line to be tangent to the circle, the perpendicular distance from the centre must equal the radius. This gives you $\cos \theta$, and so $\tan \theta$ immediately follows.

[jolly parrot]

Civil Service Pigeon

[pearl pond]

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[livid goblet]

<@&268886789983436800>

[pearl pond]

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[junior elm]

Hello i'm trying to solve this equation and after double checking everything the answers dont match, i dont exactly know what im doing wrong either

[livid goblet]

thats right

what was the right answer?

[junior elm]

it said that the answer was 9

[torn marsh]

it's not lol

[junior elm]

no???

[dense jasper]

,w integrate 4x+3x^2 dx from 1 to 2

[jolly parrot]

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[dense jasper]

you can check like this yourself in #bots

[junior elm]

whaat

[warped violet]

By the way

The notation is a little funky

Writing 2x^2 + x^3 = ... = 13

Is incorrect

[junior elm]

are all the answers wrong in my practice sheet then 😭

[warped violet]

Well we can't know that 🙂

[junior elm]

i checked this too

i keep getting 5

[warped violet]

Well that is incorrect

[livid goblet]

,w integrate 6x+4x^3 dx from 0 to 1

[warped violet]

Lol

[jolly parrot]

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[livid goblet]

can you show working

[junior elm]

it said 4 in the sheet too

[junior elm]

hold on

[warped violet]

So all the answers are in fact not wrong

[livid goblet]

🥀

[junior elm]

do the zeros make 1 and then it becomes 5-1?

wrong reply oops

[livid goblet]

4x^3 integrate to x^4

not 2x^4

[junior elm]

oh

oops

wait it WOULD be 2x though

4/2=2

[livid goblet]

why would it be 2x

??

[junior elm]

just like with 6

let me show

[livid goblet]

ye but x in 4 is cube

[warped violet]

But again

I strongly suggest working on notation

This is an integral sign

An integral sign from 0 to 1 means nothing

Like writing 7x +

[junior elm]

Wouldnt it be that

[junior elm]

i watched vid of organic chem tut lol thats where im writing all these from

[livid goblet]

yes

[warped violet]

Either use brackets around [] the expression

Or a straight line |

But not the integral sign

[warped violet]

I assume they wrote a straight line |

[junior elm]

oh yeah

straight line i just noticed

[warped violet]

But yes

if we disregard notation again 😭

The first integral sign looks like an opening bracket (

[junior elm]

oh NOW i got the answer

[warped violet]

I do suggest working on your notation though. Not just because I like cleanly written math, which I do, but also because it will benefit you. Either on tests (though I don't know how strict your professor is), or on the exit exam (or whatever you have, where it is probably pretty strict)

[junior elm]

yeahh i'll do that

just practicing for now

i do have another question

[warped violet]

Yeah of course

Ask away

[junior elm]

i solved this but i dont understand what happened to the 5 in the middle

[warped violet]

Though I am going to sleep now, so the person with an extremely long, unique username will probably help you

[warped violet]

Best of luck though 🙂

[junior elm]

thanks a lot

[livid goblet]

you should decode it before you go to sleep

[livid goblet]

chain rule

[warped violet]

Oof, I fear my life will end before I get to sleep if I do that

[junior elm]

could you explain that?

[livid goblet]

trust its simple

[livid goblet]

okay

so

whats dy/dx of cos(5x)

[junior elm]

minus 5?sinus?

[livid goblet]

?

[junior elm]

i dont exactly know

[livid goblet]

its equal to -5sin(5x)
basically we treat cos(5x) as 2 functions together

lets say y = cos(u) and u = 5x

so that y = cos(5x)

[junior elm]

oooooh

[livid goblet]

taking derivative of y = cos(u) gives dy/du = -sin(u)
derivative of u = 5x give du/dx = 5

now to get dy/dx
you multiply dy/du by du/dx

du cancels out to give dy/dx
hence dy/dx = 5 × (-sin(5x))

this is chain rule

[junior elm]

ooooooh i see

[livid goblet]

basically as you know integration is opposite of differentiation
so you reverse that to find integral of -5sin(5x)

[junior elm]

yeah

i think i get it now

thank you

[livid goblet]

np

[junior elm]

.close

[pearl pond]

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[pearl pond]

Please don't occupy multiple help channels.

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[fading nexus]

I was able to mostly do these problems but I don’t know I am supposed to know what x and y represent after I find the max or min using my graphing calc

[wicked edge]

For which question, 1 or 2?

[merry depot]

2

[wicked edge]

The question is for OP.

[fading nexus]

Both and just in general

[wicked edge]

For the first one, you got $V=\frac{90x^2-x^4}{4x}$

[jolly parrot]

Mercury.

[wicked edge]

Then, if you chuck in the calculator, V should be y, and x still be x.

It just the change of variable's name.

[fading nexus]

Yea I just don’t understand how I would know on like a test the y value is the volume

[wicked edge]

Depends on what the situation the question is asking about.

[merry depot]

mercury uer not being very helpful tbh

[fading nexus]

I just don’t know how to figure it out like on a test I would have no idea

[wicked edge]

Okay? Are you contributing any?

[fading nexus]

Wait on the second one I am confused why are we inputting the x value into the r value to find the height

[limber nimbus]

y would be whatever you're finding the min or max of, and x is the value you input to achieve that min or max

[fading nexus]

I put in the height value in for h tho

[limber nimbus]

i dont think there is an x in the second problem

[fading nexus]

So why is the radius the x value

[limber nimbus]

i think it would be good to avoid terminology like "x value" and "y value" tbh

there's different variables here, and some are written to depend on others

[fading nexus]

Not in the problem but it’s like the first value of the max value

Like my calc gives the x and y we are finding the volume so the y would be the volume

But how would I know the x is the radius

[limber nimbus]

you're basically asking "how do i know if i should solve for V in terms of r or in terms of h before maximizing"

the answer is it doesn't matter, either is correct

but one will often be easier than the other

[fading nexus]

I just don’t know how I will solve these problems because without having the answer key I would have no idea what to do next because I would not know the radius is the x value so I would not be able to find the height because I would have nothing to input

[limber nimbus]

again, you don't have to use radius as the x value

[fading nexus]

I’m able to do everything find up until I get the value of the max

[fading nexus]

I just don’t know how I am supposed to know that the x value is the radius

[limber nimbus]

im not sure if youre listening to what im saying

[fading nexus]

idk I’m trying I’m just really confused

[limber nimbus]

okay so lets solve using h instead \

we get the same equation of $3 - r^2 = 2rh$ \

rearranging, we get $r^2 + 2rh - 3 = 0$ \

using the quadratic formula, we get $r = h \pm \sqrt{h^2 + 3}$ \

we have to use $r = h + \sqrt{h^2 + 3}$, because $h - \sqrt{h^2 + 3}$ is negative \

so now we can plug in $r = h + \sqrt{h^2 + 3}$ into $V = \pi r^2 h$, and now we have an expression for $V$ in terms of $h$. We can go to our calculator, use $h$ as the x and $V$ as the y, and maximize

[jolly parrot]

snowflake

[merry depot]

ty for leaving

[limber nimbus]

it's more work this way, but it's still a totally valid method

[fading nexus]

I understand that

but that’s not the part I’m confused on

[wicked edge]

<@&268886789983436800> irrelevant information and keep not respecting other helper. I was contributing help and this guy jumped in.

[fading nexus]

like I am a me to do everything I was able to get the max value but I am just not able to identify why the x and y represent

[dim linden]

if you have nothing constructive to add, dont talk in the channel.

[limber nimbus]

the point of this is to show that there are multiple valid things to pick for x

there isnt a special unique x for any given problem

[fading nexus]

Yes I know that but I have no idea how to figure out what it represents

[limber nimbus]

what do you mean by represent

[fading nexus]

Like the x represents the radius but I only knew that because I looked at the answer key

[limber nimbus]

because they decided that x should be the radius

[fading nexus]

But there’s a right and wrong answer? How would I know how to continue the problem I alrwdy found the value of height i need to input the radius

[limber nimbus]

they asked for both a height and a radius to maximize the volume

you can either

1. solve for height in terms of radius and plug in

or 2) solve for radius in terms of height and plug in

if you do 1), you will get a height h and a volume V. then you can solve for r after

if you do 2), you will get a radius r and a volume V. then you can solve for h after

either way, you will get the same exact answers for r, h, and V

[fading nexus]

Like in this problem it turns out both the height and radius equal 1 but not all spinels are like that

[limber nimbus]

thats completely irrelevant

once you find radius = 1, you plug that into your equations to solve for height, and then you get that height = 1

[fading nexus]

I’m just very confused

[limber nimbus]

it again seems like you're looking for a unique way to solve the problem, and there isnt one

[fading nexus]

But the thing is I don’t know that I found out the radius equals 1 without looking at the answer key

[limber nimbus]

okay are you saying you dont know how to find the radius after finding the height and volume

[fading nexus]

Like I have (1,pi) but I don’t know how to know that the pi is volume and fhe 1 is the value of the radius

[limber nimbus]

what did you input into your calculator

[fading nexus]

like I am worried on the test I will think the volume is 1 and the radius is pi

[fading nexus]

I put the v= equation it’s hard to type out

And then I graphed it and found the maximum volume

[limber nimbus]

okay so you inputted the exact equation you wrote down with V and r, except on your calculator you replaced r with x and V with y

[fading nexus]

Oh wait that kinda makes more sense

I never thought about it that way 😭 like it seems obvious now

Sorry I feel like I was overthinking it to much

[limber nimbus]

its fine i think just next time be more direct about your question, it was very unclear that you were confused about the actual calculator output

[fading nexus]

Ya sorry my bad

I’m stuck on what to do next for this problem

[limber nimbus]

,ccw

,rccw

[limber nimbus]

okay so you have an equation with 1 variable, which means you can solve it for a solution

you would use a root finder or equation solver on your calculator

[fading nexus]

should I try to isolate w? That’s the type of thing I did in my previous problems

[limber nimbus]

you could, it's difficult here because you have a cubic equation

you could also try expanding everything and factor to find roots, but that can also be difficult with cubics

have you seen factoring for cubics and higher in class before?

[fading nexus]

I don’t think so

[limber nimbus]

or factor theorem or anything like that

[fading nexus]

could I find the solution by finding the max or min?

[limber nimbus]

okay then they probably want you to use your calculator to solve this

[limber nimbus]

in a sense yes but that would probably be more confusing and you shouldnt do it like that

we arent really trying to maximize or minimize anything here. we have 1 equation with 1 variable, and we just want to figure out what that variable equals

in the previous examples we had equations with 2 variables, and we wanted to maximize one variable by varying the other one

here we just have 1 variable in the equation, so there's nothing to vary; there's only going to be a few possible solutions, and most likely only 1 valid one

[fading nexus]

I’m just stuck because I don’t really know what u mean by a equation solver

[limber nimbus]

like, (2w + 1)(w)(2/3 w) = 312 is an equation

you can't pick just anything for w, most choices for w would not satisfy this equation

we want to find a w that does satisfy this equation

a simpler version of this is like seeing the equation 3x + 7 = 12, or x(2x + 5) = 13. you have an equation with 1 variable, and you want to solve for what the variable should equal to make the equation true

with linear and quadratic equations you can do this directly, but with cubic ones you often will want to use a calculator, unless the equation is easy to factor

do you have a ti84?

[fading nexus]

Yes

[limber nimbus]

if you do 2nd trace, then you should see a zero command

[fading nexus]

Is there another way to do this? It’s just our teacher never showed us this before so I’m worried she intends for us to do it a different way

Cus she does not like it when people overuse their calculator

[limber nimbus]

mm okay, it's kind of hard if you haven't seen something like factor theorem

but maybe you have and just dont recognize the name

first, expand the equation so that you have a polynomial that = 0

then multiply the equation by constants to get rid of any fractions

you then basically just guess roots

but if you want rational roots, you only have to check factors of the constant term in the polynomial, divided by factors of the highest-power term

once you find one that works, you can factor it out, and you get a smaller polynomial

with cubics, once you find one, you're left with a quadratic, and you can just apply the quadratic formula or try to factor directly

this method doesn't work very often though, since lots of polynomials have all irrational roots, so you might not even be able to find that first root

here though i checked and there is a rational root so you should be able to find it

[fading nexus]

I looked at the way my teacher did it and for some reason she said find the intersection of the 2w+1 😭

So now I’m even more confused

[limber nimbus]

like she wants you to use your calculator?

[limber nimbus]

right okay she does want you to use your calculator

intersection and zero do very similar things

for intersection, you want to find a w so that the equations

y = 312
and
y = (2w+1)(w)(2/3 w)

intersect. if they intersect, that means that the y-values are the same so 312 = (2w+1)(w)(2/3 w) and you have a valid solution for w

zero basically checks for intersections between an equation and the x-axis, so you make one side 0 and plug the other side in

both of these commands are in 2nd trace

[fading nexus]

Oh wait we are putting both y=312 and the y = the long one in?

That makes more sense

[limber nimbus]

right, intersection needs 2 equations

[fading nexus]

I got confused I thought she meant just out in the 2w+1

And I was panicked

And when we get the interaction the x value would be the width and the y value would be the volume?

[limber nimbus]

the x value is W and the y value is irrevelant here

[fading nexus]

Ah ok

[limber nimbus]

y is just a dummy variable we use to set the two sides equal

[fading nexus]

And then I would just put in the with into the h equation

[limber nimbus]

youre gonna see y = 312 here because the LHS is just 312

[fading nexus]

What’s LHS?

[limber nimbus]

left hand side

[limber nimbus]

more generally y will just be whatever thing you were setting equal

here y will be the volume since you made the equation by writing volume 2 different ways, but we also already know the volume is 312 so y doesn't really tell us anything here

[pearl pond]

@fading nexus Has your question been resolved?

[pearl pond]

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[turbid lark]

i understand how to get the theta values. how am i uspposed to get these r values?

thx

[turbid lark]

. . .

neverm...ind...

.close

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[timber cape]

oof

[pearl pond]

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[wicked edge]

!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

[lost fiber]

1

[pearl pond]

@lost fiber Has your question been resolved?

[split void]

as a side note, if you do not find any helper helping after a long time, you may consider posting this is an advanced channel.

#advanced-algebra

may be suitable; but I may be wrong; so please find a suitable channel

[pearl pond]

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[polar yew]

Could you guys do it on a piece of paper please? I’ve been stuck on this one since yesterday and I’d understand it a lot better visually

[light helm]

did you read what i said yesterday

note how you applied power rule to the outer cube
then multiplied by the derivative of the inner (2x+1)
similar idea to what you have here
(sec(x))^3
out is the same, the cube
and inner is the sec(x)
so applying chain rule would result in
3sec^2(x) * d/dx sec(x)
and you can use your table for the derivative of sec(x)

[pearl pond]

@polar yew Has your question been resolved?

[pearl pond]

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[swift spindle]

Hi is this right

[lament dawn]

there doesnt seem to be an error

i think its right

[sweet junco]

should be

[swift spindle]

answers do this, so not quite sure

[cursive wraith]

well the final answer is the same isn't it

[lament dawn]

this solution is wrong

i think

[cursive wraith]

oh wait

[swift spindle]

yes but idk where they went wrong

[lament dawn]

du/dx should be -4cos(2x)sin(2x)

[cursive wraith]

yeah they're missing a 2 in the derivative of cos^2(2x)

[lament dawn]

they forgot to apply the chain rule to the innermost function

[swift spindle]

oh wait yeah. true. thanks guys

.close

[pearl pond]

Channel closed

Closed by @swift spindle

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
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• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[low kestrel]

hi

[wicked edge]

Hi.

[low kestrel]

are you good with statistics?

[wicked edge]

Any questions?

[wicked edge]

Depends on the question level.

[low kestrel]

l need some help with trigonometry as well

[lament dawn]

hey, dont ask to ask, just ask!

!da2a

[pearl pond]

Asking the actual question right away is more likely to get responses.

Asking "Can I ask...?" or "Does anyone know about...?" doesn't give people enough information to decide whether they can help, and answering can feel like a promise to help with the actual question, which they might find themselves unable to.

[low kestrel]

alright!

[wraith jacinth]

ask ahead!

[low kestrel]

eliminate m from the following equations

x=1-sinm,y=1+cosm

[wicked edge]

Ask away!

[fleet ridge]

What are you stuck on?

!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

[low kestrel]

1

[fleet ridge]

I see

Recall that $sin^2x+cos^2x=1$

[jolly parrot]

-TimeLord-

[fleet ridge]

Try using this

[lament dawn]

do you know of an identity involving sine and cosine that does not result in any trigonometric function?

ah

well

that

[fleet ridge]

Oops

Probably should have waited

[lament dawn]

no \sin and \cos is slightly triggering sorry LOL

[sterile python]

[fleet ridge]

Thats a thing?

$\sin{x}$

[lament dawn]

yeah, $\sin^2 x + \cos^2 x = 1$

[wicked edge]

do \sin{}

[jolly parrot]

wjs

-TimeLord-

[fleet ridge]

Oh snap

[lament dawn]

more you know!

[fleet ridge]

Yes. I'll probably forget it tomorrow

[wicked edge]

[low kestrel]

prove that (sec x +tan x)(sec x - tan x)= 1 can you also help me prove that equation please?

[fleet ridge]

Did you solve your original problem though?

[sterile python]

wait, stick on each question please.

[fleet ridge]

I would prefer to look at that first

[sterile python]

Do not jump from this to other question in a short time!

[lament dawn]

what the others said!!

ill assume youre working from first principles since this is another well known pythagorean identity, you can always rewrite, say, trigonometric functions as a ratio of sides, and use the idea that O^2 + A^2 = H^2 and so on

otherwise, you can also derive this from the original pythagorean identity sin^2(x) + cos^2(x) = 1 by rewriting sec(x) and tan(x) in terms of sine and cosine

[lament dawn]

why did hutao become nefer what happened

[sterile python]

Many waifus I can not afford...

[wicked edge]

What in the actual.

[low kestrel]

thanks so much reze

[lament dawn]

no problem!

[low kestrel]

timelord, l didn't solve my first problem though

[sterile python]

How about me

[fleet ridge]

Ok. Did you try using this?

[sterile python]

Layla blud and that freaky emoji...

[lament dawn]

thanks nefer for keepin jahoda employed

[low kestrel]

but y'all been so kind and helpful to me..Thanks a million

[proper nova]

what the fuck

[cinder flower]

freoji

[proper nova]

ts a help channel 😭

[sterile python]

You mean in the game right

[low kestrel]

timelord l already tried mathgpt and discussing with my friends at school

[pearl pond]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[sterile python]

https://tenor.com/view/reze-tongue-out-reze-chainsaw-man-moving-her-head-reze-head-moving-reze-chainsaw-man-gif-15937294161332759512

[fleet ridge]

What the hell is mathgpt

[low kestrel]

l'll try out that app though

[fleet ridge]

Who made ts

[prisma kernel]

wtb is happening here

[proper nova]

!redir you people

[low kestrel]

look it up

[pearl pond]

This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.

[cinder flower]

wtb

[honest oyster]

We lost the plot somewhere

[lament dawn]

well, rearrange x = 1 - sin m, y = 1 + cos m in terms of sine and cosine

i.e.

make sin m and cos m the subject

and then employ the identity (holy crap my keyboard is lagging i dont even know how) sin^2 x + cos^2 x = 1

[wicked edge]

Pure chaos here.

[lament dawn]

@low kestrel bump

[lament dawn]

||my favourite green eyed goblin||

[sterile python]

Wut 😭 who is that

[fleet ridge]

Its Reze from Chainsaw man

[low kestrel]

reze what do i do next?

[wicked edge]

Guys redir

[low kestrel]

l have cosm = y-i,sin m = x+1

am l on the right direction?

[lament dawn]

right! now you square both sides

[fleet ridge]

Wait y-i?

[lament dawn]

y - 1*

[low kestrel]

wait, is sin m supposed to be x + 1 or x -1

l feel like l've messed up somewhere

[lament dawn]

right, well youve given us x = 1 - sin m
so, sin m should be 1 - x by adding sim m on both sides and subtracting x

[low kestrel]

but sin will be negative right?

is that ok?

[lament dawn]

hm? what do you mean

sin m + x = 1 - sin m + sin m
sin m + x = 1
sin m = 1 - x

[cinder flower]

@lament dawn can i send you a friend request?

[lament dawn]

okie!

[low kestrel]

x = 1-sin m x-1 = -sin m

[sterile python]

How about me

[cinder flower]

no

[lament dawn]

sure! this works too, you could multiply both sides by -1

[sterile python]

🥺

[lament dawn]

LOL

[low kestrel]

thanks

you're a blessing reze