help-27

[spare crypt]

so like you're given 11 points of data

but for midpoint rule you're forced to pick the center of a delta_x, so the best you can do is 5 rectangles spacings instead of the usual 10

[celest sundial]

Oh ok, yeah since 5 data points are in the middle

So 15/5?

[spare crypt]

this sort of picture

yea 15/5

[celest sundial]

Ohh ok

How do I know next time? To do 15/5? Is there something special about the 15 in the question that indicates this?

[spare crypt]

it's just about the 11 data points they give you, so 10 intervals in general but with midpoint it's 5

better picture

[celest sundial]

Ah ok, thank you

That makes sense

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[timid peak]

I don’t get where they got the (2+2…+1)2 part

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[strange arch]

Find the intersection line via Gauß algorithm

or you can also use a position and a direction vector to describe the line

and then you also use the vector from that position to the origin

meaning you have a position vec & two direction vecs which form a plane

[atomic depot]

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[edgy roost]

I’m trying to find the coefficients for this power series by knowings its convergent sum.

I can find the radius of convergence easily, but for some reason the c values are wrong.

It seems like I correctly turned the fraction into a power series, I just don’t know why it’s wrong still.

[devout snow]

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[edgy roost]

<@&286206848099549185>

[acoustic leaf]

your power series has an x^{n+1} term in it, whereas the original was based on x^{n}

[edgy roost]

That’s because I distributed x into the power series making it x^{n+1}

I turned the second factor into a power series after factoring out x

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[silk bay]

am i in the right direction, kinda stuck

i got something like that, but the solution in the textbook is completely different

[hushed lance]

y^4+2y+1 is not (y^2+1)^2

[silk bay]

sorry its supposed to be y^4 + 2y^2 + 1

[hushed lance]

oh

lol

but uh

why would you use two constants after integrating

you can just use 1 since a constant-constant is just another constant

[fossil locust]

,w y * y' = x(y^4 + 2y^2 + 1), y(4) = 1

[hushed lance]

then you use the given condition of y=1, x=4 to solve for that constant

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[fossil locust]

Yeah you should just square root both sides so you have $y = \sqrt{-\frac{1}{x^2 + 2C_2 + C_1} - 1}$

[woven radish]

south

[fossil locust]

The -1 doesn't disappear

And also you can justify using the positive sign cause when x is positive, y must also be positive

[silk bay]

okay

thank you

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[uncut quarry]

hello

[neon folio]

👋

[uncut quarry]

I need help with this problem

find the volume of the solid generated by revolving the region bounded by the lines y= x+2 , y= x. y=3x. around x=4

im doing this by the cylindrical shell method

do i need to construct 2 cylinders or just 1

i think 1 and the hight will be the upper function minus the lower function

but i need someone to solve this so i dont get confused

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[uncut quarry]

anyone??

[devout snow]

@uncut quarry Has your question been resolved?

[sharp plume]

Personally I would split it up into two cylinders

@uncut quarry - So you know I am trying to help you

[uncut quarry]

ah ok i see

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[oblique torrent]

Hoping to get guidance here

[oblique torrent]

This is my work so far not sure what to do with x^3

[stable storm]

so we can divide the x^3 with 2x

since it’s in the denominator

[oblique torrent]

making it x^2 and cancelling an x?

[stable storm]

yep

any ideas on what to do with the x^2 now?

[oblique torrent]

Not overly I know I am trying to get U in there but theres no +1

[stable storm]

so we can rearrange u=1+x^2

so that x^2=u-1

and now we can plug that in for x^2

[oblique torrent]

how'd id do?

[stable storm]

maybe put some parenthesis around u-1

cuz it’s kinda looks like only the -1 is being multiplied by sqrt u

[oblique torrent]

true true

thank you thank you

[stable storm]

yw!

[oblique torrent]

4 of the 5 times I've used this server I've had you help me so i appreciate that lol

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[opaque garnet]

g is area under curve

[graceful geyser]

so how would u find absolute max

i tried 16 it was incorrect

[distant birch]

maybe 18

[opaque garnet]

by going further you can attain more area

[graceful geyser]

oh

[opaque garnet]

which means infty

[graceful geyser]

yeah i forgot 16 doesn't cover the whole areaa

[opaque garnet]

or 18

[graceful geyser]

yeah that was right

to find concave downward intervals can f' be used right

because f' is g''

or how do u find

[opaque garnet]

yes

[graceful geyser]

ok i got it thank you

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[loud kayak]

could someone do e), f) and g)

these are the answers

[fossil rampart]

I agree with e

f too

g too

@loud kayak

[vocal tartan]

e) convert sin^2(x) to 1 - cos^2(x) and factor
f) multiply top and bottom by cos(x) and apply L'Hopital
g) apply L'Hopital

[loud kayak]

teacher didnt allow me to use L'Hopital

[fossil rampart]

Can you use equivalent ?

[loud kayak]

whats that?

[fossil rampart]

Like sin(x) ~ x when x tend to zero

[loud kayak]

nope

these are my teachers solutions

for the others

he didnt do those 3

idk y

[fossil rampart]

Or in other word that sin(x)/x tend to 1

[loud kayak]

oh yea we did that

[fossil rampart]

Okay so you can do g

[vocal tartan]

hmm for f) you can multiply the top and bottom by 1 + cos(x)

[fossil rampart]

Just try to use the rule your teacher applied

For g

[loud kayak]

i see

[fossil rampart]

So what would you do ?

[loud kayak]

i got it

its the right answer

i understand how to do e) as well

what about h)??

[vocal tartan]

Same idea as g, notice that the cos(x) doesn't actually affect anything since it goes to 1

[loud kayak]

okok

ty

i understnad now

thanks @vocal tartan @fossil rampart

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[lucid mason]

I am sooo confused 😭 this is like one of those tricky questions where u have to go backwards

the teacher said to do this i believe:

BUT i dont know what to do afterwards...

[dapper echo]

solve for a

a=-4(tan 111.8)

[lucid mason]

so would i do -4 (tan111.8) = a

[dapper echo]

yes

[lucid mason]

so 10 = a

woah that simple???

thats it???

[dapper echo]

yep

[lucid mason]

damnnn

thank u so much

😭

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[dim niche]

can someone help[ with d and e i laready have done a b and c

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[wide aurora]

can someone help me finish finding the power series representation of f(x)=3/(3-x), centered about x=2? I used the formula to find the constant of each nth term but im unsure how to create the general term with what I think is a fibonacci sequence

[stable storm]

notice how we can rewrite $\frac{3}{3-x}$ as $\frac{3}{1-(x-2)}$

[woven radish]

y0shi

[wide aurora]

okay

i see how you can do that

[stable storm]

do you remember what the power series representation of $\frac{1}{1-x}$ is?

[woven radish]

y0shi

[wide aurora]

not off the top of my head right now

its x^n

[stable storm]

yep

so we can essentially just plug (x-2) into that series

to get $\frac{1}{1-(x-2)}$

[woven radish]

y0shi

[stable storm]

and then we can multiply everything by 3 to get our power series

[wide aurora]

and that would give you the power series representation?

[stable storm]

yep

for 3/3-x

because we can manipulate our power series using substitution

addition subtraction multiplication and division

as well as differentiation and integration

[wide aurora]

can i show you what i did in pms?

easier if I just show you my work

[stable storm]

you can post it here if thats fine

,w power series representation for \frac{3}{3-x} centered at x=2

[woven radish]

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[wide aurora]

yeah I got 3 as the constant

[stable storm]

@wide aurora so i dont believe the answer at the bottom there was correct

since you had a n! at the bottom

[wide aurora]

so the series representation wouldn't be like a taylor series?

I worked out my answer to get that series expansion

[stable storm]

well it would be

its just that the n! should cancel out

when you take the derivative for the constant

so can you not use substitution and manipulation to the 1/1-x series?

also from your work, it seems like you divided by n! twice

since you evaluated the derivative, divided by n! and then when you plugged it into the taylor series formula

you divided by n! again

[wide aurora]

where did i divide by n! twice?

and I supposed i could, i'm just not very good with this

[stable storm]

so you evaluated the derivative at the point

and then divided it by n!

and used that as your constant

but instead of just putting that constant in front of the term, you divided it by n! again

[wide aurora]

i think i just left it out

in the denominator for that part of my work

[woven radish]

y0shi

y0shi

[stable storm]

and we evaluate this at x=2

so we would just get -3

then when we plug it back into the taylor series formula

[wide aurora]

okay

[woven radish]

y0shi

[stable storm]

f'(2) would be -3

so the second term there would be -3(x-2)

[woven radish]

y0shi

[stable storm]

and then we plug in 2

we'll get 6

see how it keeps dividing out to 3 or -3

when we do 6/2!

for the constant in front

[wide aurora]

okay give me a minute to understand im monkey brain rn

[stable storm]

also i might need to go now, you can ask other helpers for help if you still have questions

[wide aurora]

thanks

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[wide aurora]

i see what you mean now

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[craggy field]

why is it wrong

where did i go wrong about it

[eternal tapir]

It's centered at 1/3

I didn't verify the radius but assuming it's 1/12

You need to take 1/3 +- 1/12

And check endpoints

[craggy field]

what do you mean by that

[eternal tapir]

1/3 + 1/12 and 1/3 - 1/12

[craggy field]

OOOh it is 1/6 instead of 1/3. I wrote a wrong number in my steps

[craggy field]

what would adding radius with endpoints achieve me?

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[floral sail]

do I need trigonometry to solve this integral?
$\int \frac{4t^3 - t^2 + 16t}{t^2 + 4}$

[woven radish]

jean

[sonic smelt]

Yes

[floral sail]

aight

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[floral sail]

is $ (\sqrt{x})^3 = \sqrt{x^3} $ ?

[floral sail]

$(\sqrt{x})^3 = \sqrt{x^3}$

[woven radish]

jean

[floral sail]

they are the same?

[strange arch]

ys exponent rule

(a^b)^c = a^(bc) = (a^c)^b

in R+

[floral sail]

$(a^b)^c = a^{bc} = (a^c)^b$

[versed juniper]

is it still the same if it was 2 instead of 3 and x was negative?

[woven radish]

jean

[floral sail]

aight

wait

it means that

[strange arch]

No, it's true for a e R+, b,c e R

[floral sail]

$x^{\frac{3}{2}} = x^{\frac{2}{3}}$?

[woven radish]

jean

[strange arch]

no wait

what

why 3/2 to 2/3

[versed juniper]

whats R+

[floral sail]

because

[strange arch]

Real & ≥0

[versed juniper]

oh ok then yeah

[floral sail]

$\sqrt{x^3} = x^{\frac{3}{2}$

[woven radish]

jean
Compile Error! Click the reaction for more information.
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[strange arch]

in C the exponent rules are more well applied imo

[strange arch]

Yes that's true

But why would it equal ^2/3

[floral sail]

because

$(\sqrt{x})^3$

[woven radish]

jean

[strange arch]

= (x^1/2)^3

= x^(1/2*3)

[versed juniper]

rewrite sqrt as half power to make it clearer

[strange arch]

= x^3/2

[floral sail]

thanks lunatic and latte

[strange arch]

np

[floral sail]

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[devout snow]

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[winter torrent]

those being elements of Z and not N feels very strange

[loud aspen]

indeed

[winter torrent]

anyway try induction

[loud aspen]

i have expressed that gcd(a,b) > 1 => ax+by > 1

[loud aspen]

what would be the base case? c = 0?

[winter torrent]

sure

wait with c=0 it's false

right?

[loud aspen]

yeah

true

[winter torrent]

i think Z was a typo

either just show it's false or prove it for N

[loud aspen]

alright, i will try this idea, thanks!

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[bitter halo]

how do I find f'(0)?

[loud aspen]

you do the derivative of the branch for f(x), when x=0. in this case, 0'= 0

[devout snow]

@bitter halo Has your question been resolved?

[bitter halo]

but we should know first if the function is differentiable or not

a function is differentiable iff its differentiable in all the points of their domain, shouldnt we know if the function is continuous as well

?

[heady tulip]

ok so first you need to show the limit as x approaches 0 of that thing is zero (but x is not 0)

[bitter halo]

ok

does that include both 0^- and 0^+

how do I do that?

[jaunty mantle]

well you just do it from definition

[bitter halo]

if it isnt continuous it isnt differentiable

[jaunty mantle]

$\lim_{x\to 0^+}f(x)$

[woven radish]

Frosst

[jaunty mantle]

and then you do it for 0^-

[heady tulip]

$f'(0)=\lim_{x\to 0} \frac{f(x)-f(0)}{x-0}$

dam

[woven radish]

svc

[heady tulip]

you know f(0)=0 from the second part of the piecewise function

so you have to show the limit exists

[bitter halo]

okay

,, f'(0) = \lim_{x \to 0} \frac{f(x)}{x}

[woven radish]

milanesa de pollo

[bitter halo]

mmm

but f(x) is this

with the x in denominator or without?

[heady tulip]

with

is f(x)/x

[bitter halo]

okay

how do I get rid of x^2 in denominator

[heady tulip]

no idea

[bitter halo]

currently I have this

$$
\lim_{x \to 0} \frac{ln(2x+1) + \cos(x^2) -2x -1}{x^2}
$$

[woven radish]

milanesa de pollo

[bitter halo]

should I try lhopi or is there other way out?

[devout snow]

@bitter halo Has your question been resolved?

[open current]

you can use taylor's expansion for ln(1+x) and for cosx this will give you the solution faster

[bitter halo]

taylor or maclaurin expansion

I will try

[heady tulip]

idk but ln(2x+1) and -(2x+1) looks suspicious

i mean, looks good para hacer algun truco o algo asi no se

sabes q olvidalo

no se intenta cualquier cosa xd

wolframalpha lo hace asi sin usar l'hopital

[open current]

sorry I only understand English.. but if you expand ln(2x+1) you get 2x-(2x)^2/2! + (2x)^3/3!.. and so on 2x and -2x will cancel out each other and if you divide the rest by x^2 you will get -4/2 and other terms will have some positive power of x which you can put as 0

similarly expand cos(x^2) as 1-(x^2)^2/2!.. and so on 1 and -1 will cancel out and divide the rest by x^2

[open current]

you obviously can use l'hopital this is just another method

[heady tulip]

I meant to say that's how wolframalpha evaluates the limit without using L'hopital

but yea ur right, any method should give the same result

[bitter halo]

oh yeah by taylor expansion of ln(2x +1) it seems way nicer

[bitter halo]

so did you got -2 aswell?

I dont get it am kinda confused

what a mess my handwriting sorry

[bitter halo]

you also speak spanish nice

[bitter halo]

thats the answer? -2

[heady tulip]

es xq hablo español

si

[bitter halo]

nice

so f'(0) = -2?

[heady tulip]

si

[devout snow]

@bitter halo Has your question been resolved?

[open current]

yeah i believe -2 is the answer

[devout snow]

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[fresh gyro]

This is much I solved

Which is correct option?

[fresh gyro]

From this

Question from Linear Programming

<@&286206848099549185>

Anybody?

[devout snow]

@fresh gyro Has your question been resolved?

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[rustic jetty]

so we can simplify that to $\cot(\theta) = \frac{1}{\sqrt{3}}$

[woven radish]

Dork9399

[rustic jetty]

or $\tan(\theta) = \sqrt{3}$

[woven radish]

Dork9399

[rustic jetty]

do you know any angles for which $\tan(\theta) = \sqrt{3}$?

[woven radish]

Dork9399

[late wyvern]

no I don't

[rustic jetty]

what is $\tan(\frac{\pi}{6})$

[woven radish]

Dork9399

[late wyvern]

sqrt3/3?

[rustic jetty]

what is $\tan(\frac{\pi}{3})$

[woven radish]

Dork9399

[late wyvern]

just root 3

[rustic jetty]

so can you answer this now?

[late wyvern]

sqrt3/3 and sqrt3

[rustic jetty]

so we know that $\tan(\frac{\pi}{3}) = \sqrt{3}$

[woven radish]

Dork9399

[rustic jetty]

so we know that $\cot(\frac{\pi}{3}) = \frac{1}{\sqrt{3}}$

[woven radish]

Dork9399

[late wyvern]

thats about right I guess

[rustic jetty]

so thats one answer

[late wyvern]

so is it pi/3,pi/6?

yo

<@&286206848099549185>

[devout snow]

@late wyvern Has your question been resolved?

[devout snow]

@late wyvern Has your question been resolved?

[last tartan]

pi/3 + pi(n)

[devout snow]

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[random vine]

How taking alpha = i did the author conclude that x and y are orthogonal?

[random vine]

This is problem 7

[trail eagle]

It would be nice to have the question itself, but to me it looks like a typo? Like a = 1+i instead of something

[random vine]

I can send now 2 secs

Question 7 here

I’m more so concerned with the right to left direction of the iff statement

[devout snow]

@random vine Has your question been resolved?

[random vine]

<@&286206848099549185>

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[devout snow]

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[devout snow]

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[vivid hill]

what do i do here

[wispy eagle]

are you sure your first step is correct

[sweet elk]

How did an e appear

[vivid hill]

i meant to erase

[vivid hill]

no

that’s why i’m asking

idk how to do it

[vivid hill]

i thought u log them each then do e but idk ..

[wispy eagle]

revisit log rules

[vivid hill]

uhm

i don’t have my notebook can u just tell me what it transforms to

i’ll do the rest

then u can check

[vivid hill]

i

is this one right ,

?

[wispy eagle]

have you subbed them back in and tried to check yourself?

[vivid hill]

bro plz just give me the answer

i’m abt to cry

[wispy eagle]

same here

[vagrant cedar]

hey bro

just give her the answer alright

[vivid hill]

thanks 😊

<@&286206848099549185>

[wispy eagle]

We are not here to do your homework for you, but we are here to help you learn how to do your homework #rules

[vivid hill]

i know

[sweet elk]

Did you factor correctly?

[vivid hill]

let me know what i did wrong

what should i do instead

[wispy eagle]

.

[sweet elk]

How u get x-2 x+1

[wispy eagle]

.

[vivid hill]

that’s not helping bc it doesn’t work even if i sub back in

log rules i said idk

i tried

[sweet elk]

That expands to x^2 -x -2

[vivid hill]

oh

oh

i

did

wait

bc i favored

factored

[sweet elk]

You factored out a -1?

[vivid hill]

no

the 2x - x^2 + 6 - 3x = 4

simplified to 0= -x^2 -x +2

[sweet elk]

Yea

[vivid hill]

then i favored and got (x-2)(x+1)

FACTORD+

[sweet elk]

But if you expand it

It doesn’t match

[vivid hill]

yes it does

[sweet elk]

No it was -x^2

[vivid hill]

oh

[sweet elk]

If you expand u have x^2

[vivid hill]

oh wait

let me retry

-(x+2)(x-1)

[sweet elk]

-1(x^2+x-2)?

Yea

[vivid hill]

ok then

i get

x= -2, x=1

[devout snow]

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[modern hatch]

all i know is that it cant be B or D because it would make sense for those to be the bounds of the integral

so its either A or C but idk how to show the work

[eager fox]

Idek what the question is asking lol

[versed juniper]

am i stupid or smth how are you supposed to integrate the equation of circle

id integrate the semicircle and multiply by 2

[eager fox]

They want volume

Not area

[versed juniper]

yeah same idea

oh nevermind it isnt

[modern hatch]

dam mocha so studious helping everytime

[modern hatch]

they want to find the volume of a circle with a diameter of 6

with the cross sections of volume being

equilateral triangles

[eager fox]

Can u show me a picture?

[modern hatch]

of?

[eager fox]

Of the figure

[modern hatch]

theres no specific figure

[eager fox]

I dont understand what equilateral triangle

[modern hatch]

thats the cross section, or the shape thats being used to calculate volume

[eager fox]

Like this?

[modern hatch]

yes

assuming those 3 lines represent xyz axis

then yes

[versed juniper]

i got C

[modern hatch]

its A

[versed juniper]

uh avtually nvm

yeah

[modern hatch]

i got C too

why did u change ur mind

how do i get A

[versed juniper]

integrating the top half of the circle is half the thingy

but since its area of crosssedtion you need to multiply by 4 to make up for the halved side

[modern hatch]

half of the thingy

😭

[versed juniper]

thj gu

[modern hatch]

what do u mean

by this

[versed juniper]

u know how doubling the side length of a square quadruples the area

same idra

[modern hatch]

no

i dont know how

ok wdym to make up for the halved side

[versed juniper]

draw it out

[modern hatch]

how is it halved

[versed juniper]

cuz ur integrating half the circle only

[modern hatch]

how do u know

[versed juniper]

sqrt

only positive

[modern hatch]

so wouldnt you

multiply by 2

not 4

[versed juniper]

no because the area is quadruped when side length is doubled

[modern hatch]

what is the side length]

[versed juniper]

cross section area

[modern hatch]

whats a cross section area

[modern hatch]

i was in a state competition during the day we covered cross sections

😭

[modern hatch]

what am i looking at

[versed juniper]

math at 1am

[modern hatch]

tthe latest it could be 4 u is 11

if ur in the us at least

but anyways

[modern hatch]

how did ur 36-x^2

reduce to 9-x^2

[versed juniper]

idk

idk anymore

[modern hatch]

stop give up.

this unit test is my last chance to get an A

[devout snow]

@modern hatch Has your question been resolved?

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[verbal pagoda]

did i do this right

[verbal pagoda]

can someone check my work pls

[restive river]

,rccw

[restive river]

wait what is rhe question asking of your even

convert to polar?

[verbal pagoda]

yes

[restive river]

ok 1 nad 2 r fine

[verbal pagoda]

what abt 3/4

[restive river]

3 is fine

something went wrong in 4

r*r = r is what u said p mucj

[verbal pagoda]

?

[restive river]

,rccw

[restive river]

u don't need to solve for r or whatever

just keeping it in the form [
4r^2\6\cos\theta\6\sin\theta = 7
]
Is ok

[verbal pagoda]

we don’t need to solve for r?

[restive river]

no

i mean like they didnt solve for y or x either in their equations

[verbal pagoda]

how would u change polar to cartesian equations

[restive river]

reverse the process u just did

rcos(theta) = x
rsin(theta) = y

for the first one [
r =12\6\sec\theta
]
try to get a $r\6\cos\theta$

[verbal pagoda]

divide by r?

[restive river]

no

i mean you can

but multiply both sides by cos(theta) seems wiser

[verbal pagoda]

so you’d get x = 12?

[restive river]

yuh

[verbal pagoda]

how would u do r = 7sintheta

[restive river]

multiply both sides by r

[verbal pagoda]

then what do you do with the r^2?

[restive river]

,, r^2 = x^2 + y^2

[verbal pagoda]

i’m confused

[restive river]

like

its just Pythagorean theorem i guess?

uh

[verbal pagoda]

so

x^2 + y^2 = 7y ?

[restive river]

yeah that works im pretty sure

[verbal pagoda]

can you help on r^2sin(2theta) = 16

[restive river]

yeee

so like

sin(2theta) has a trig identity

do yk it

[verbal pagoda]

2sin x cos x ?

[restive river]

yuh

so

,, r^2\6\sin{2\theta} = 2r^2\6\sin\theta\6\cos\theta = 16

[restive river]

now uh

do you see what to do

[verbal pagoda]

is it 2xy

[restive river]

yes

u can simplify

say divide by 2

and solve for y

[verbal pagoda]

ohh ok ty

[restive river]

,rccw

[verbal pagoda]

i’m not sure what function to put first in the integral

[restive river]

i dont think it matters in this case, since you are looking for the geometrical area presumably

so whatever order u choose will come down to a difference of a - factor

and the geometrical area will always be positive

[verbal pagoda]

then what do i put in the integral?

[restive river]

,, A = \412\int_\alpha^\beta r_1^2 - r_2^2\dd\theta

[restive river]

alpha and beta are the points of intersection between the two

and r_1 and r_2 are just ur equations

[verbal pagoda]

what abt the next one

pi/2 to 3pi/2 is when it makes the half of the thing

so multiply by 2 and then i subtracted the area from the first provlem

[restive river]

,rccw

[restive river]

one sec

[restive river]

ok i dont think this is right

[verbal pagoda]

why 😭

[restive river]

well what is 17.586 on the right?

first of all

[verbal pagoda]

area i got from first problem

[restive river]

oh

okay then yeah thats correct

[verbal pagoda]

it wants outside of 6cos0

[restive river]

assuming that area is correct

yeah

ur work is good although are u sure the area is 17.586? seems to be approximating something irrational

[verbal pagoda]

that’s what i got on the calc

[restive river]

hopefully u didnt like erase a pi somewhere

[verbal pagoda]

i’m not sure how to do the last one

[restive river]

yeah so

u need to differentiate first of all

,, \dv[y]x = \4{\dv*[y]\theta}{\dv*[x]\theta} = \4{r\6\cos\theta + r'\6\sin\theta}{r'\6\cos\theta -r\6\sin\theta}

[verbal pagoda]

ughh

[restive river]

think this is it from memory

[verbal pagoda]

is it when dydx is 0?

[restive river]

nah

like

the pole is when r = 0

so solve [
5\6\cos{2\theta} = 0
]
first of all

[verbal pagoda]

pi 3pi ?

wait idk

pi/4

[restive river]

like

um

its gonna be a general solution

its gonna repeat

[verbal pagoda]

i’m confused

why did i need to find the dydx

[restive river]

it is 0 at pi/4 but cos is a periodic function

it repeats

so u need to find the general form of the solution

[restive river]

bc u r trying to find the tangent lines, to recover the slope u need to compute the derivative

[verbal pagoda]

so what do i do after i get pi/4

[restive river]

.

when is r 0 again after pi/4?

[verbal pagoda]

3pi/4v

?

[restive river]

ye

[verbal pagoda]

so it’d just be the pi/4 before 2po

pi

[restive river]

so theta = pi/4 + kpi/2

for any integer k

[restive river]

wdym?

oh

yes

[verbal pagoda]

well it js needs 0 to 2pi

[restive river]

yuh so thats fine

anyways u got ue equations now

time to compute the derivative

[verbal pagoda]

so i js plug in the pi/4 into the derivative

[restive river]

u need to use thie

[restive river]

yuh

that Will get u the slope

[verbal pagoda]

and then i still have to find the x y value

[restive river]

what do you mean exactly

[verbal pagoda]

well doesn’t the tangent line require the y and x value at the angle

[restive river]

ye

y - y_1 = m(x-x_1)

sorry wasn't sure of what u meant there lol

but yeah just compute the derivative and plug in pi/4 and 3pi/4

[verbal pagoda]

don’t i also have to do 5pi/4

[restive river]

yeah

[verbal pagoda]

bruh 😭

[restive river]

although i believe it will be ok since the thing will repeat to pi/4

u will see

[verbal pagoda]

i’ll check

hm

[restive river]

hmmmmm

[verbal pagoda]

i got 1 and -1 for the slopes

[restive river]

i cant verify ur algebra but yeah

there will be different lines

[verbal pagoda]

but i don’t think im doing the x and y stuff right

[restive river]

show me ig

[verbal pagoda]

this is what i’m putting in

[restive river]

oh boy ok lets aee

i mean

yeah

thats right

[verbal pagoda]

aren’t they just 0

[restive river]

x = rcos(theta) and y = rsin(theta)

yeah

[verbal pagoda]

but doesn’t that just give y = x

[restive river]

if u think about it it does make sense because like Ur thing is a bunch of petals

,w plot r = 5cos(2theta)

[woven radish]

Results provided by WolframAlpha

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

Requested by qylooo

[verbal pagoda]

i did all that for y = x

[restive river]

😭

its ok u done now tho :D

also dont forget y = -x

it can be either

[verbal pagoda]

yes

tysm for helping

[restive river]

nw

damn its been 2 hours i didnt even realise

[verbal pagoda]

no way 😭😭

[restive river]

😭

anyways good luck !! u can close this now if u dont have anything else ig

[verbal pagoda]

ty again!!

.close

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[devout snow]

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[marsh thistle]

How do I make a or c with my calculator using normalcdf/pdf and invnormal?

[devout snow]

@marsh thistle Has your question been resolved?

[marsh thistle]

<@&286206848099549185>

[fossil locust]

For a note that 80 is the mean

So the area to the left of X = 80 is just 0.5

0.5 + 0.25 = 0.75

So you need invnormal(0.25, 80, 10)

For c, the area to the left of a must therefore be 0.05

For a, b, c, d you only need to use invnormal

[marsh thistle]

thanks

I'll try and close it if it works 🙂

it works! ❤️

.close

[devout snow]

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[marsh thistle]

Hmmm how does this one work? (b, c, e, f)

.reopen

[devout snow]

✅

[devout snow]

@marsh thistle Has your question been resolved?

[devout snow]

@marsh thistle Has your question been resolved?

[devout snow]

@marsh thistle Has your question been resolved?

[devout snow]

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[wind karma]

.reopen

oh

[wind karma]

i'm doing a paper on something related to probability and statistics, and i searched up for resources on irwin-hall distribution which i need a lot. considering i'm supposed to derive some stuff, i wanted to derive the distribution function from the probability density function and i wanted to cite a proof, however i am not really well versed in advanced calculus (i'm taking ib) so i have no idea how to just get this from integrating

and i also don't have huge knowledge in probability and stats (i know the basics) and i'm not supposed to fully understand the concept, however i need some steps

[eternal tapir]

all the stuff not containing (x-k)^n are just constants

so just use chain rule to differentiate (x-k)^ n

and you get n (x-k)^(n-1)

[wind karma]

oh that's actually true lmao i forgot it's just dx

[eternal tapir]

then n/n! = 1/(n-1)!

[wind karma]

yea i get it tysm!!!

[eternal tapir]

np

[wind karma]

.close

[devout snow]

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[restive river]

Hello. I need help with a problem for stats!

[restive river]

I was wondering where my teacher got 0.78. (: Sending pic!

[devout snow]

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).

[restive river]

this is my channel?

[devout snow]

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).

[restive river]

,rccw

[restive river]

q = 1 - p pretty much

like 1 - 0.22 = 0.78

Thank you!

.close

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[restive river]

https://media.discordapp.net/attachments/1230158901555757149/1230186452717928488/DocScanner_17-Apr-2024_9-28_pm_1.jpg?ex=6632676a&is=661ff26a&hm=7726b66cd7b0da66ebe2d6549021b92be2c01e9a618cde79cd1d591a40f23ec7&=&format=webp&width=298&height=415

[devout snow]

Please don't occupy multiple help channels.

[restive river]

.close

[devout snow]

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[restive river]

Hello

[restive river]

So,

I get all the words they are saying, I hear them clearly, I have read it multiple times, but I just do not get it

[lime harbor]

"A is the original amount of a sample"
"h is the half-life of the substance"
then they say
"the half life of carbon 14 is 5740 years"
"carbon-14 in a sample is 200 g"

is it clear to see which variables take which values?

[restive river]

Im just so very confu

Confused, idk were to even start

But, I do understand the variables, or what they mean at least

Except for half life

Idk if it means half the life of a substance or something else

[lime harbor]

half life is just the amount of time it takes for a substance to reduce by half

[restive river]

Then why does there need to be 1/2 there?

[lime harbor]

imagine t = h
we know that P(t) = A(1/2)^(t/h)
so P(t) = A(1/2)^(h/h)
P(t) = A(1/2)^1
P(t) = A(1/2)
in other words, when the time passed is equal to h, the inital amount is equal to half of itself

the 1/2 being there would ensure that A becomes half of itself once h time passes

or once t=h said differently

[restive river]

So, h equals the time it takes for a substance to half itself, and t equals time passed, so when t gets to the time it takes for a substances to half itself, it then equals h?

[lime harbor]

yeah

[restive river]

So,

This helps, but I still dont understand why you need to divide t by h, or have (t/h) as a exponent of 1/2

Wait so I think im getting it a little, its an exponent of 1/2 because thats how many times its going to half itself?

[lime harbor]

yes, the base is 1/2, and the exponent is by how many times it would half itself

because 1/2 would get multipled exponent times

[restive river]

Ok, but I still dont get by you would need to divide t by h

[lime harbor]

ok ignore h for now
if i had just P(t) = A(1/2)^t, that would mean that when t=1, it would half itself
but what if something takes 5 units of time to half itself
well you'd have to compensate by dividing t by 5; that's how i see it

so h would be how long it takes to half itself

[restive river]

So, h is the number t would need to be divided by to get a certain number of t?

And then t would be the number of times something half’s itself?

[lime harbor]

honestly i'm having trouble understanding this 😅

[lime harbor]

as for this, when there is no h (same as h=1), then yes, t would be the number of times it halfs itself

if we look at the equation without h, t represents the amount of times you multiply by one half right
P(t) = A(1/2)^t

(without h)

so suppose i divide t by 2 or something

P(t) = A(1/2)^(t/2)

that means to divide A by half, I need a t of 2 this time, not 1

because once t=2, the exponent = 1, so then you multiply by 1/2 once

[restive river]

Oh, because 1/2 is the base of the exponent, so that means if t is divided by any number, it would still be multiplying itself 1/2 times?

[restive river]

I think I just confused myself typing this

[lime harbor]

it would be multiplying itself by 1/2 times only once t = the divided number

cause when t = the divided number, or t=h, then the exponent = 1

[restive river]

Because that means t would equal one?

Yea

[lime harbor]

ye, so then when the exponent is one, (1/2) gets multiplied just once

so A becomes half of itself

[restive river]

Okay, somI now understand it, but

One more question,

If I wanted something to be halfed twice, so basically t=2, does that mean h would have to equal half of t?

[lime harbor]

if you wanted something to be halfed twice, that would not mean t=2, but rather the exponent =2 (t/h=2)

unless by t you meant exponent

then yeah that sounds right

because if h is half of t, then t/h = t / (0.5t) = 2

so (1/2) has an exponent of 2

[restive river]

So, t/h would have to equal 2 correct?

[lime harbor]

yes

[restive river]

Ok, thx, I got a grasp of it now

[lime harbor]

but it's kinda confusing to come up with an h relative to t

but yeah gl

[devout snow]

@restive river Has your question been resolved?

Channel closed

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Use .reopen if this was a mistake.

[devout snow]

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[smoky hamlet]

The binomial series for ( 1 + x ) n converges if, and only if, − 1 < x ≤ 1 .

[smoky hamlet]

can som1 explain what converges means?

[opaque nova]

?? It's like if a limit converges, same sorta idea with an infinite series

[smoky hamlet]

i dont know what limits are mb

but a binomial series is already infinite

[opaque nova]

Honestly then, why u doing this question then?

Sorry that sounded mean

Didn't mean to, just genuinely curious

[smoky hamlet]

it's in my curriculum

limits arent

[opaque nova]

Fr? R u limits aren't in you curriculum but convergent series are??

[smoky hamlet]

yea?

bra please explain it without limits

[opaque nova]

I don't really think there's anyway to solve this without some knowledge of how limits work. Lemme think about it for a sec

[smoky hamlet]

ok just go ahead ill try to understand

@opaque nova where'd ya go

[opaque nova]

So what have you learnt about infinite series?

Like technique wise

[smoky hamlet]

that they never end

[smoky hamlet]

and they become tiny and tinier after every term

[opaque nova]

Okay so that's not really true, only in some case.

Now with the question, it's an iff right so u have to go in both directions.

I'll help with the -> direction, the other should be straight forward by contraposition

But so, proving that a series diverges is easier than convergence.

[smoky hamlet]

i dont need to rpove it

i just need to understand the meaninmeaning of convergence

[opaque nova]

Oh shit my bad, that's way easier then.

Honestly your best bet is youtube. I'll try my best to explain tho.

[smoky hamlet]

i couldn't find it on google

[opaque nova]

Basically imagine your taking a bunch of numbers that keep getting closer to 0 fast enough that when you add them together you get a fixed number

[smoky hamlet]

can you please find a video for me

that explains convergence

[upper schooner]

Basically you can take it to mean "gets you a finite number"

For example, if you did 1 + 1 + 1 + ... (and didn't stop at any point) that just gets larger and larger and goes to infinity

There are other series where you can keep adding terms together and it will give you a finite number - are you familiar with geometric series?

[smoky hamlet]

yup

multiply to get new term

there is a coomon ratio

common

[upper schooner]

Thought so, then when your common ratio has absolute value less than 1, the infinite series converges

[fervent ledge]

lol havent learnt that yet

maybe ill come back after i learn all these things

[upper schooner]

In that, for example, something like the series 1 + 1/2 + 1/4 + ... will reach a finite number, which is 2

The whole a/(1 - r) thing, if you've seen that

[fervent ledge]

nah

[upper schooner]

Whereas if the common ratio does not have an absolute value less than 1, the infinite series would not reach a finite number

[fervent ledge]

i remeber putting the choice Both A and B on an mcq in my exam. it asked that if r is less than one, then the series
A- converges
b- diverges
c- both a and b

[upper schooner]

Ouch

[fervent ledge]

lol i was wrong haha

ill learn tis after i do the geometric one

.close

thank you btw

.close

woops

[upper schooner]

wait you're not OP, are you?

[fervent ledge]

im on my alt