help-38

[solid summit]

so ill do this last part

or will it also take 1992498 hours

[edgy cedar]

Well you can probably pit this into wolfram

Look up what Lagrange multiplier/the Lagrangian is

You need to find one with 2 constraints

[solid summit]

wtf

help

wtf

i dont have a g(x) and f(x)

or do i?

[edgy cedar]

https://youtu.be/5-CUqogfPLY

These are your constraints

[solid summit]

but i dont have x, y, z

[edgy cedar]

Your function is r(a,b)

[solid summit]

yeah

thats what i mean i dont have r(a, b, z)

[edgy cedar]

It’s still multi variable

[solid summit]

im so utterly lost you wouldnt believe it

so i cant just do the partial derivative and = 0?

[edgy cedar]

Yeah because you also have some constraints on what a and b could be

It’s still similar process but now you have to think about your constrains

[solid summit]

oh in that case i have a simpler expression for r

https://i.imgur.com/jyu7Oxg.png

does that make it less aids?

[edgy cedar]

What’s this

It’s totally different from what you had up there

[solid summit]

this is an expression without the constraints

[edgy cedar]

Well

Yeah you can probably very easily optimise this

[solid summit]

pogggg

even with those constraints?

[edgy cedar]

Well that’s a different story

[solid summit]

fuck

[edgy cedar]

It’s still a pain

[solid summit]

fuckk

ive never seen/ used partial derivatives and now im doing this

can u give me any hints?

[edgy cedar]

Unfortunately there’s no hint for you

[edgy cedar]

Look here

And do basically what he did

Put things into wolframalpha

Hope for the best

There’s no hint because I’m not doing this

[solid summit]

:(

if i buy u a coffee?

a big coffee?

[edgy cedar]

I don’t drink coffee :3

[solid summit]

a beer?

[edgy cedar]

No alcohol either lol

Good luck bro

I’m gone

[solid summit]

fffffffffffffffff

ty tho

[edgy cedar]

[solid summit]

@edgy cedar one thing, does this count as calc?

[edgy cedar]

Yeah multi variable calc

[solid summit]

cheers

[trim jolt]

@solid summit Has your question been resolved?

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@solid summit Has your question been resolved?

[solid summit]

yo @edgy cedar, im doing it now, and the video says he needs 5 eqn, so i need 4 (one less unknown, I only have a, b, lambda and u), what eqns do i choose?

[solid summit]

uh

this is impossible

[trim jolt]

@solid summit Has your question been resolved?

[edgy cedar]

Don’t you need all of them? You need to solve for a and b

Tbh I haven’t done one of these in a while so I may have forgotten something

[solid summit]

ah ok

well

whatever

[trim jolt]

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[dire kraken]

is it A?

[vagrant marsh]

Show work

I hate doing related rates

[dire kraken]

x = distance from lamp
y = shadow length

5 / (x+y) = 2/y
3y = 2x
3 * dy/dt = 2 * dx/dt
3 * dy/dt = 2 * 1.4
dy/dt = 14/15

my doubt is that my work didn't include the x = 10m

so I feel like this is wrong

or was that extra info

[paper pond]

can someone help me?

[delicate jasper]

Are ye serioues rn bro

[exotic pike]

@paper pond #❓how-to-get-help

[delicate jasper]

#❓how-to-get-help

[dire kraken]

diagram btw

[vagrant marsh]

I think it looks good

[dire kraken]

so was 10m extra info?

[vagrant marsh]

Uh yes

I believe so

[dire kraken]

interesting

[vagrant marsh]

In this case the similar triangles grow at constant rates

[dire kraken]

oh okay

thanks

.close

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[mortal sinew]

I genuinely have no idea where to start or how to do this question?

[mortal sinew]

Its part b which im struggling with

[sonic token]

i would think about the two perfect squares that 75 is inbetween

so like whats the nearest square below 75

[mortal sinew]

64 and 81?

[mortal sinew]

64

[sonic token]

yeah

so sqrt of 64 is 8 right

therefore n is 8

cuz an integer is just a non fraction or decimal whole number

u can verify this by using a calculator too

sqrt of 75 is around 8.66

and 8 <8.66< 9 is true

so its 8

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[mortal sinew]

Thanks @sonic token

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[pallid basin]

this is my work..idk what Im doing incorrectly

.close

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[pallid basin]

how do u simplify this

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[pallid basin]

hello

[pallid basin]

how u simplify this

it's for this problem

<@&286206848099549185>

[fringe minnow]

thats pretty simplified

not sure what else u want

[pallid basin]

it won't work when submitted

ill send example of a problem

[pallid basin]

this is what it looks like

[glacial karma]

aye aye

wassup

do u know

half angle

formula

[pallid basin]

ya

[pallid basin]

my only problem is simplifying radicals with a fraction

[glacial karma]

wait but

this doesn’t

seem quite right

to me

eh but I’m doing it

in my head

so I trust that’s correct

[pallid basin]

here ill send my work

[glacial karma]

yeye

[glacial karma]

where did

u get 16

from

[pallid basin]

ohhhhh

[glacial karma]

yeye

[pallid basin]

bro im so dumb

lol

[glacial karma]

nah it’s fine

happens to

The best of us

[pallid basin]

fs

[pallid basin]

for this problem I have to use the double angle formulas

However when u multiply across there I don't know how they get that final answer

[dense breach]

they multiplied the numerators to get the final numerator

and multiplied denominators to get the denominator

[pallid basin]

2 x 13 isn't 13

[dense breach]

nah

They divided both sides by 2 then

[pallid basin]

im confused rn

so they multiplied both fractions by 1/2?

[dense breach]

both the numerator and denominator

They had a factor of 2 on both sides that they cancelled out

[glacial karma]

aye aye

wassup

what’s the

confusion?

[pallid basin]

@glacial karma

[glacial karma]

I don’t understand

ur qn

multiply what across

[pallid basin]

oh so the denominator multiplied like 8 x 16

since the 2 reduced the first 16 into 8?

[dense breach]

Yeah

[pallid basin]

how would u do that with a odd number tho

[glacial karma]

4/11?

sqrt137?

[pallid basin]

it's for my problem

[glacial karma]

this one?

of a diff question

[dense breach]

Then you don't need to

[glacial karma]

ah alr

[pallid basin]

same type of question

just different figures

[pallid basin]

bruh then how is my shit wrong then

[dense breach]

Were you originally supposed to have 7√13?

instead of √137

Or did you directly get √137

[glacial karma]

how did u get

sqrt 137

[glacial karma]

,w 11^2 - 4^2

[solid kiln]

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[glacial karma]

it’s sqrt 105

Wait

lemme see ur work

[pallid basin]

@glacial karma

[glacial karma]

aye bro

not 137

[pallid basin]

the sin given gives u y and r

u need to get x

[glacial karma]

but adjacent is sqrt105

[pallid basin]

so u have to do the pythagorem theorem

idk

i remember doing it that way

[dense breach]

x^2+16=121

That part is right

What happened then

[pallid basin]

oml brooooo im dumb af

i added 16 instead of subtracting it

LOLLLLL

[glacial karma]

ah

[pallid basin]

bro what is it with me today

im making all the dumb mistakes

[dense breach]

it be like that sometimes

[pallid basin]

when I put this in it says it's wrong

[dense breach]

Lies in quadrant 2

That's relevant here

[pallid basin]

sin is positive

[dense breach]

indeed

[pallid basin]

oh but bc u multiply by cos which is negative then the answer becomes negative

[dense breach]

yeah

sin(theta) is positive but sin(2*theta) is negative

[pallid basin]

true\

bros carrying me lol

[dense breach]

lol

[pallid basin]

problem again lol

im on tan rn

[dense breach]

Wasn't this supposed to be negative

[pallid basin]

i just sent my first screen shot of the problem

[dense breach]

ah ok

[pallid basin]

the radical is throwing me off

[dense breach]

ah

denominator

$(\frac{-4}{\sqrt{105}})^2$

[pallid basin]

ohh wait

is the square root supposed to go away bc I squared it

[solid kiln]

monikanicity

[dense breach]

yeah

[pallid basin]

bruh

everything is comong to me after I ask u bruh

[dense breach]

It be like that sometimes

[pallid basin]

bruh what is this now

[dense breach]

Simplify?

Including noticing the numerator is a multiple of 105

so rationalize the denominator

[dense breach]

yeah rationalize denominator

[pallid basin]

so like do I do -840 - 89?

[dense breach]

nah

[pallid basin]

-840 + 89?

[dense breach]

no

[pallid basin]

ohhh

wait is it where I do the factor tree and factor out 105?

[dense breach]

I mean you can try that

Idk exactly what you mean but it might work

[pallid basin]

nvm it won't work

like lets say if it was square root of 50

u could simplify it to 5 square root of 2

that;s what I meant

[dense breach]

ah ok

[pallid basin]

but 105 has no pairs so it won't work

idk what u mean

[dense breach]

indeed

multiply both sides by √105

[pallid basin]

so this?

[dense breach]

wait what happened to the bottom

[pallid basin]

well when I multiplied the √ 105 by √ 105

cancelled out

and left me with 89

[dense breach]

They don't disappear

[pallid basin]

but it cancelled?

u can't leave √ 105 at the bottom

[dense breach]

√105 * √105?

[pallid basin]

oh so it's just 05

105

bc 105 x 105= 11025

√ 11025 = 105

[dense breach]

well that's not how I would do it but yes

[pallid basin]

so do u add the 105 with the 89?

[dense breach]

√x * √x = x always

[pallid basin]

ya

[dense breach]

nah

find a common factor in the bottom and top

[pallid basin]

what's the bottom number

bc u have the 89 too

so when we multiplied the √ 105 with √ 105

we were left with 105

i mean 105

but u have that 89 left over

[dense breach]

it's just multiplied by 105

[pallid basin]

so it's -8√ 105/89?

[dense breach]

yeah

[pallid basin]

alr thx sm bro

[dense breach]

you're welcome

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[ornate ridge]

How do I solve x^3 + 4x^2 - 7x + 11 = 0?

[lament jewel]

guess

[trim lichen]

throw the rational root theorem at it

see if anything comes of that

[lament jewel]

whats that

[trim lichen]

if it doesn't then youre kinda screwed

[ornate ridge]

And what's that?

[glacial karma]

yeee

the roots

aren’t nice on this one

[wispy palm]

cubic formula

[trim lichen]

if a polynomial with integer coefficients has any rational roots, then those roots are all of the form ±(divisor of constant term)/(divisor of leading term)

[wispy palm]

;D

[lament jewel]

good ol cubic formula

[ornate ridge]

Ooh, thx

[glacial karma]

BRUH

HAHAHA

[wispy palm]

have fun

[trim lichen]

@lament jewel don't tell me you don't know what RRT is given that youve been studying complex analysis and the likes

[slim scroll]

LMAOOOO

[lament jewel]

idk ive only seen remainder theorem n quad formula

hahaha

[amber python]

@lament jewel is special

[trim lichen]

[lament jewel]

thanks

[ornate ridge]

I'm tired . close

[glacial karma]

this is Ann’s fav

[lament jewel]

.close

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[lament jewel]

no way

this looks so alien

[amber python]

https://en.wikipedia.org/wiki/Rational_root_theorem

for your learning experience

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[echo dove]

I'm stuck on a calculus problem: f(x) = 7e^x sin x, [0, 𝜋]
I got 7e^x (sin x + cos x) but I need to get coordinates in the parenthesis instead of the derivative

[echo dove]

.close

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[stray spindle]

is this an error in the book? or

[echo dove]

can I get someone to help me with my math problem in my channel or is there a queue that I need to wait for?

[night zodiac]

1.2/6

[stray spindle]

ohh missed that... thank you

[night zodiac]

You just wait

[echo dove]

got it

[stray spindle]

.close

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[midnight jackal]

Im not sure I'm understand what an element pf R^w is

is the sequence endless?

also is poposition 2.4.10 is Let dim(V) = dim(W). A linear transformation T: V -> W is injective if and only if it is surjective

from my limited understanding of this question I think dim R^w = R^w but R is not a linear transformation.

which is why it doesnt follow 2.4.10

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[harsh plover]

Why does an object accelerate if it is changing direction

[harsh plover]

Can someone explain to me pls

[silent canyon]

Velocity is a vector quantity that's why

I suggest you go to the sister physics server

[harsh plover]

haha doesnt really answer what im looking for

why is does it being a vectory quantity mean it accelerates

anyway have oyu got an invite to the phsyicc one

[night zodiac]

Velocity is changing

If velocity is changing, there is an acceleration involved

[harsh plover]

ahh

is there a way to differentiate the two accelerations

like acceleartion due to change of direction and acceleartion due to change of pseed

speed

it seems quite confusing

[night zodiac]

Changing direction relates to changing speed

[harsh plover]

hmm

dont worry man

im going to do some more work

.close

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[wraith hinge]

Can anyone help me with this question i just can’t figure out where to start

[wraith hinge]

waltuh, what question waltuh

[torpid depot]

What is initial velocity? Final velocity and distance covered?

So you have u,v,s. To find a you can apply one of the three equations of motion.

[wraith hinge]

Alright thank you

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[bronze canopy]

let f(x)=ln(x)/2x, which value gives the maximum value of f(x)
a.)1
b.)sqrt(e)
c.)e
d.)e^2

[bronze canopy]

for this question?

do i just differentiate f(x)

until i get the simplest expression

so if i do f''(x) ln(x)/2x

o ye

[past comet]

,w d/dx log(x)/(2x)

[solid kiln]

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[bronze canopy]

uh

its ln(x)

not log

wait theres no way they equal the same thing

[past comet]

log=ln

[bronze canopy]

,w d/dx' log(x)/(2x)

[solid kiln]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by 745367241428#9376

[bronze canopy]

,w f''(log(x)/(2x))

[solid kiln]

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[past comet]

Log is used as ln everywhere past highschool

[bronze canopy]

o

uh

uh

,w d/dx -(log(x)-1)/(2x^2)

[solid kiln]

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[bronze canopy]

,w d/dx 2(log(x)-3)/(2x^3)

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[past comet]

Can I ask what you are doing?

[bronze canopy]

im just differentiating

till the simplest term

[past comet]

What?

[bronze canopy]

uh

nvm

continue on how you would do it

[past comet]

You differentiate once

Set it equal to 0

Those solutions are critical points

You can then verify what kind by for example differentiating again

[bronze canopy]

uh

ok

[shut prism]

yo

[bronze canopy]

hi

[shut prism]

im learning like horizontal transformations and vertical transformations

and im really confused

[bronze canopy]

o

[shut prism]

.close

[bronze canopy]

.close

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[zinc ibex]

I am a beginner at math so I have trouble solving the problem.

I wanted to know what kind of theory I should study upon or any advice on solving my problem.

My problem is as followed.

A , B = Non zero positive integer (not a floating point)

I have a finite list of A-B pair , For example List = [{A1,B1} , {A2,B2} , {A3,B3} , --- {An,Bn} ]

I wanted to find how many pair is within a single pair (I wanted to group them together)?

My sample example data is as follow.

+----+-----+
| A | B |
+----+-----+
| 30 | 201 |
| 31 | 179 |
| 31 | 178 |
| 30 | 148 |
| 31 | 117 |
| 30 | 87 |
| 31 | 165 |
+----+-----+

{A1,B1} to {A6,B6} is in a single group and {A7,B7} is in another group.

{A1,B1} to {A6,B6} is in a single group because A1+B1 = 202 and
{A1+A2+A3+A4+A5+A6} = {30 + 31 + 31 + 30 + 31 + 30 } = 184 (which is less than B1 [201] )

{A7,B7} is another group because 184+A7 = 184 +31 = 215 (which is greater than B1).

After the end of the group {A6,B6} , I wanted to reset the process and start it from the start of next group {A7,B7} until I run out of the number of pairs in my list.

I wanted to find out the pair which is the start of the group {A1,B1} , {A7,B7}. Or is it at least possible to find out how many group will there will in my finite set of list.

Please tell me if you wanted me to clarify the detail of my problem or wanted me to rephrase it in another way.

[marble wharf]

well just add the numbers An until you are too big and then start new

[zinc ibex]

Could you please tell me if there is set-based logic of adding a number until they are too big and start new?

[marble wharf]

not really

what do you actually want this for

[zinc ibex]

I am actually performing a date calculation. The "A" column is actually a "date" and I wanted go find out how many date to get

[marble wharf]

do you want to do this in some kind of software? in your head? on paper?

[zinc ibex]

I have written it in software and I wanted to verify if I am right

[zinc ibex]

For example

+-------------+-----+
|    Date     | Day |
+-------------+-----+
| 2021-07-24  | 201 |
| 2021-08-24  | 179 |
| 2021-09-24  | 178 |
| 2021-10-24  | 148 |
| 2021-11-24  | 117 |
| 2021-12-24  |  87 |
| 2022-01-24  | 165 |
+-------------+-----+

In the above example (same as previous data) , I wanted to put "2021-07-24" to "2021-12-24" into a single group and so on

I am curious in whether I can mathematically determine how many group I should have in a list in order to verify the output of my program

[marble wharf]

well to mathematically determine that you would just add up the numbers. which is probably the same as what your program does

[zinc ibex]

My program actually does use other way to find the group so is adding the numbers a good way to verify it?

I am not sure if there is actually edge case which invalid it (adding the number up). I can only guarantee that A and B are non zero positive integer

[marble wharf]

what does your program use

but sure you can add them to verify I guess

[zinc ibex]

My program actually needed to find a start of the group. and then it perform other operation on the start of the group

I know my program is having a problem on some edge case so I thought I can verify it with math

[marble wharf]

just a simple

sum = 0
i=1
while sum <= B1
  sum = sum+A_i
  i++
end
return i``` or something should work

(I hope I don't have an "off-by-one" error in there)

[zinc ibex]

I am using it in database so set-based logic is better for me

and am trying to find it more math-like way instead of procedural like

logic

[marble wharf]

algorithms are math...

can you calculate sum(A1, ..., An) with your system?

[zinc ibex]

I can calculate but the problem is I needed to know "n" value

[marble wharf]

can you do something likefor each n calculate sum(A1, ..., An) ?

I know that loops in databases are bad tho

unsafe and all that

[zinc ibex]

that what I am concern about

My database loop take too long to run but this is a math channel so

I thought somone know a way to optimize it

by using math logic

[marble wharf]

well then the question you should have asked is "hey can someone in here optimize database code for me"

you didnt even mention that you are working in a database for 30 minutes

[zinc ibex]

well that was my mistake.

[marble wharf]

I only have very basic knowledge about databases, definitely no clue how to optimize queries or stuff like that, sorry. maybe try it in #computing-software ? not sure. or maybe the CS server under #old-network

[zinc ibex]

well I can optimize query but I thought there is similar problem in math (finding a group based on a certain condition) instead of adding up a number

[marble wharf]

well the condition here is based on adding numbers

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[orchid beacon]

A metal brace sits at 55 degrees to the horizontal and reaches 4.2m
up a wall.

[orchid beacon]

can someone help draw this out?

i struggle at making these type of diagrams and i think thats giving me the incorrect answer

[grim saffron]

In general in math, a drawing is only supposed to illustrate. If the accuracy of drawing is somehow crucial to get a correct answer, you're doing it wrong

[orchid beacon]

the question is a metal brace sits at 55 degrees to the horizontal and reaches 4.2m up a wall. The distance between the base of the wall and the base of the brace is:

i know its either Sin or Cos but i can't tell what to use so i wanted to illustrate it to find it out

[grim saffron]

By "4.2m up a wall", do you mean:
a. The distance between the base of the wall to the point where the brace and the wall meet, or
b. the length of the brace itself?

[orchid beacon]

the questions just says '4.2m up a wall' and i assume it means the brace is acting as the hypotenuse

[grim saffron]

Yeah, the brace is acting as hypotenuse and the wall forms 90 degrees angle to the ground.

[orchid beacon]

right

[grim saffron]

But I think the phrase "reaches 4.2m up a wall" is ambiguous. Is the question already in English?

[orchid beacon]

yeah it is its just how it is in the textbox

whats throwing me off is the degrees

[grim saffron]

Can you show the photo of your textbook?

[orchid beacon]

yes lemme just take a photo

[grim saffron]

,rccw

[orchid beacon]

the answer i got was 2.41m, but the textbox says its E

i did 4.2*cos(55)

[lilac wind]

it reaches 4.2 meters up the wall, meaning that its the opposite which is 4.2 meters. knowing this, should you be using cos(55) or something else?

[orchid beacon]

oh

sin

right !

ok i appreciate it, these questions get around ma head sometimes despite not really being that hard

but thank you !!!

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[tough wadi]

How is this reflexive?

[tough wadi]

Shouldn't it be mm = mm ?

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[dire oyster]

How many real subsets of the three-element set {2; 3; 4}?

[dire oyster]

the answer is 6 although I do not know how to solve it

[silent canyon]

for example {2,3} is a subset

[dire oyster]

its a subset of {2,3,4}?

[silent canyon]

yeah

[dire oyster]

then

{3,4} too

[sturdy leaf]

yes

[dire oyster]

and {2,4}

thats three

[sturdy leaf]

though i would have said 7 is the answer. wouldnt {} also be a real subset?

that sais i assume real means proper

[dire oyster]

ye

probably {} is not a proper subset in this case

what is the other 3?

[silent canyon]

{2}, {3}, {4}, {2,3,4}

technically I would have included {}

[dire oyster]

i see

[silent canyon]

i would have said 8 lol

[dire oyster]

these are the answer choices

[silent canyon]

interesting

so they dont consider the empty and the whole set itself

[dire oyster]

wait a second

i have the solve key

maybe its not 6

it is 6

[silent canyon]

ok they asked for "real" subsets

i dont know the difference

[dire oyster]

sadly me neither

okay

thanks for the help

have a nice day

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[river badger]

find a collection of points that meet the requirement:
|z-i| < 4

[civic bloom]

a disk with centre at i and radius 4, like (x)^2 +(y-1)^2 <16

[civic bloom]

method is , put z= x+iy and solve further

[river badger]

thnkas

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[velvet bolt]

f^2=4f^2 *cos alpha

cos alpha=f^2 / 4f^2

cos alpha =1/4

did i do this right?

[west sphinx]

How did you arrive at this step?

[velvet bolt]

i added

2f^2 and 2f^2

[west sphinx]

Is 2 + 2cosx = 4 cosx?

[velvet bolt]

ohh

ok

i see

thanks

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[night zodiac]

Do you know what $A \cap B$ means?

[solid kiln]

dldh06

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[static pewter]

yes

[static pewter]

yes

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[cerulean hearth]

i don't get the u-sub
i mean why sqrt(3) at the top instead of 1/sqrt(3)

[marble wharf]

where at the top

du = dx/sqrt(3)

do you mean that?

[stark bison]

Cuz dx = sqrt3 du

[cerulean hearth]

numerator

[cerulean hearth]

ah ok
i didn't see

thx

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[cerulean hearth]

.reopen

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✅

[cerulean hearth]

at the last step, i don't get why the constant they put before is 1/sqrt(3)

i would put sqrt(3)/3

[marble wharf]

well thats the same thing

$\frac{\sqrt 3}{3} = \frac{\sqrt{3}}{\sqrt 3 \cdot \sqrt 3} = \frac{1}{\sqrt 3}$

[solid kiln]

Denascite

[cerulean hearth]

hum ok i see

thx

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[wraith hinge]

How do I draw a line on the horizontal axis

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[wraith hinge]

<@&286206848099549185>

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[void oak]

how to prove that the derivative of x^2 is 2x without power rule

[void oak]

also im a bit confused on why it wouldnt be x

somebody already helped me iwth this, but im confused where he got the equation from

please help

[glacial cove]

Use $\frac{f(x+h)-f(x)}{h}$

[solid kiln]

Mezza

[glacial cove]

This is differentiation by first principles

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[wraith hinge]

.open

hey..!

[wraith hinge]

help

pls

help

[unreal loom]

💀

just sub in 2x

and ur done

[misty cosmos]

lels

[wraith hinge]

🫤

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[wraith hinge]

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[nova fractal]

I don’t know how to do this please help also I’m pretty sure your supposed to use Desmos to find the answer

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[nova fractal]

<@&286206848099549185>

This is what I got idk if it’s right

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[lone egret]

need help with 7 and 9 please 🙂

[lone egret]

i know it’s simple just wanna make sure i’m doing it correctly

[civic bloom]

Conside like 16/5+4/1

[civic bloom]

Can you solve now

[rugged latch]

i think its division

[civic bloom]

Damm

[lone egret]

yes it’s division

[civic bloom]

My eyes hurts lol

[lone egret]

sorry it’s not that clear 😭

[civic bloom]

Flip it 16/5 x 1/4

[lone egret]

why don’t i flip the 16/5

you know?

[civic bloom]

Flip after division sign

Not before division sign

Its like 16/5/4 = 16/4*5

[lone egret]

thank you!!!

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[light scroll]

can someone help me how to find A

[grim saffron]

Pertama-tama turunkan dulu f

Saya mau lihat bagamana Anda menurunkannya

[light scroll]

this is as far my calc went

this f’

[grim saffron]

Tinggal masukkan x=1

[light scroll]

udh tp mencari A nya bingung

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[solid kiln]

BlokirKominfo

[grim saffron]

Saya tidak baca aturan bahwa A harus bilangan bulat.

[grim saffron]

Kalau begini, tinggal bagi kasus untuk A bilangan ganjil atau A bilangan bulat

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[stable jacinth]

This is a problem for Surfaces Defined Parametrically and Surface Area. I need to set up a double integral but im not sure how to do it and really confused with the fractions. Help please?

[cyan zinc]

solve for z in the eqn for the plane

@stable jacinth

[stable jacinth]

Z = 5-2x-2y

dz/dx = -2

dz/dy = -2

[cyan zinc]

ok now let's set this up as a double integral in x and y

we have

$$\iint_{C} (5 - 2x - 2y) dx dy$$

where C is the ellipse

[stable jacinth]

According to my notes it becomes int sqrt(1+ dx^2 + dy^2) dxdy

And i need to set the limits for the double integration

Is this correct?

[cyan zinc]

yeah that's right, i made an error there

[stable jacinth]

Its fine. I just dont know how to find the limits. Can i get some help with that?

[cyan zinc]

yeah sure

observe this

x^2/36 + y^2/4
x^2/6^2 + y^2/2^2
(x/6)^2 + (y/2)^2 = 1

let x/6 = cos(t) and y/2 = sin(t)

[stable jacinth]

(cost)^2 +(sint)^2 = 1?

[cyan zinc]

yes

but we need to plug that into our integral and compute the jacobian

[stable jacinth]

Is that the dy component?

[cyan zinc]

let me write it out

we start with

$$\iint_{C} \sqrt{1+f_x^2+f_y^2} dx dy$$

[solid kiln]

Mr. Gamer

[cyan zinc]

we found that f_x = f_y = -2, so squaring that gives 4.
we get sqrt(1+4+4) = sqrt(9) = 3

$$\iint_{C} 3 dx dy$$

[solid kiln]

Mr. Gamer

[stable jacinth]

Okay got it so far

[cyan zinc]

now we have to change coordinates from:
(x,y) -> (r,t)
(using t for theta here, don't feel like typing greek rn)

are you familiar with the jacobian?

[stable jacinth]

No im not. Never heard of it sorry

[cyan zinc]

ok so what we have to do basically is find what our dA element becomes under a change of coordinate system

we have the equations:
x = 6rcos(t)
y = 2rsin(t)
for:
0<=t<=2pi
0<=r<=1

so basically we have to take the determinant to change:
dx dy -> f(r,t) dr dt

let me show you how that's done, latex for determinants is kind of painful so i'll write it out

[stable jacinth]

Thank you

[cyan zinc]

dx/dr = 6cos(t), dx/dt = -6rsin(t)
dy/dr = 2sin(t), dy/dt = 2rcos(t)

now all we have to do is take the determinant of this matrix:

| 6cos(t) -6rsin(t) |
| 2sin(t) 2rcos(t) |

what do you get

[stable jacinth]

Matrix? is this linear algebra?

[cyan zinc]

it's calc 3 so yes

there's no other way to get the jacobian factor haha

[stable jacinth]

Oh I actually didnt take linear yet. Ill try it tho?

I know this is cross product but never did something so complicated. Only numbers

[cyan zinc]

$\begin{vmatrix}a&b\c&d\end{vmatrix} = ad - bc$

[solid kiln]

Mr. Gamer

[stable jacinth]

ad is 12rcos^2(t)

bc is -12rsin^2(t)

[cyan zinc]

yeah you're right

[stable jacinth]

Is it just 12r?

Since the sin and cos cancel out

[cyan zinc]

so then when we perform the subtraction we get:

12r exactly

now we can finally set up the integral

$$\iint_{C} 3 dx dy$$

[solid kiln]

Mr. Gamer

[cyan zinc]

becomes

$$\int_{0}^{2\pi} \int_{0}^1 (36r) dr dt$$

[solid kiln]

Mr. Gamer

[stable jacinth]

Okay ill try solving it out

Got 36pi

Worked. Thank you!

[stable jacinth]

Is it possible to ask one more? if you have time.

[cyan zinc]

ok sure

[stable jacinth]

This is the question

And this is my work done so far

Im pretty sure i made a error somewhere but unsure whats wrong with it

[cyan zinc]

alright so the bounds on your integrand look good

not too sure about the jacobian factor or the radicand however

let's double check that

can you see how we'd get that:
2x = rcos(t)
2y = rsin(t)

[stable jacinth]

Im pretty sure its the 7 that im missing something

Yes. I understand that

[cyan zinc]

wait i see what it is

our radicand should be:
1+(2x)^2+(2y)^2
(Let x = rcos(t), y = rsin(t))
1+(2(rcos(t))^2+(2(rsin(t))^2
1 + 4r^2(cos^2(t)) + 4r^2(sin^2(t))
1 + 4r^2

[stable jacinth]

Ahh calculation mistake

Okay ill try again.

[cyan zinc]

everything else looks good

just change the radicand into 1+4r^2

[stable jacinth]

It doesnt seem right. I got 157.53473

[cyan zinc]

why are you getting a decimal instead of an exact answer

[stable jacinth]

the exact answer is (pi(135sqrt(5)-1)/6

[cyan zinc]

yeah that checks out

[stable jacinth]

Says that it is wrong.

[cyan zinc]

i think i see your error now

when z = 7, we have x^2 + y^2 = 0, so our lower bound for r is 0

when z = 11, we have x^2+y^2=4, so our upper bound for r is 2

@stable jacinth

[stable jacinth]

Oh

Can you clarify please?

[cyan zinc]

sure

let's consider our original eqn

z = 7 + x^2 + y^2

it just gives us a single bound on z (z<=11), so we have to find the minimum value that z can attain

x^2+y^2 will always be nonnegative, so the minimum value for z is 7. when z = 7, we get:
7 = 7 + x^2 + y^2
0 = x^2+y^2
0 = r^2
0 = r

Our upper bound for z is given as z = 11, when that occurs:
11 = 7 + x^2 + y^2
4 = x^2 + y^2
4 = r^2
2 = r

[stable jacinth]

Ohh okay i understand

Got it. Thank you!

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[civic fox]

if x = $\sin \left( \frac{2\pi}{7}
\right) + \sin \left( \frac{4\pi}{7}
\right) + \sin \left( \frac{8\pi}{7}
\right)$ and y = $\cos \left( \frac{2\pi}{7}
\right) + \cos \left( \frac{4\pi}{7}
\right) + \cos \left( \frac{8\pi}{7}
\right)$ then find x^2 + y^2

[solid kiln]

Chochan
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[civic fox]

@split chasm

[strong breach]

dude

cant you like open pi?

oh wait nvm

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@civic fox Has your question been resolved?

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[full dock]

I’m not sure how to do these

[full dock]

How does question 2 work?

[trim lichen]

question 2? do you mean part b?

[full dock]

oh yeah

i dont quite get it

am i supposed to plug in the values of g into x or something

[split chasm]

$\boxed{?}$ just seems to be another function

[solid kiln]

ℝamonov

[full dock]

well a and b

i dont know how to do

[split chasm]

for g(-1) substitute all xs in
g(x) = x^2 + 2x
with (-1)
and simplify

[full dock]

so then -1^2 + 2(-1)?

[split chasm]

not quite

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

[full dock]

ohh

but what does it mean evaluate g(-1) and g(2)

how come there is 2 g functions

[split chasm]

those are two separate problems

[full dock]

ohhh

okay

i thought you had to do it in the same equation

so after simplifying the equation it would be

g(x) = (insert equation)?

[split chasm]

no

[full dock]

wha

[full dock]

how would i write it then?

[split chasm]

for g(-1) substitute all xs in
g(x) = x^2 + 2x
with (-1)

[full dock]

what about the ?(x)

what letter do i put

[split chasm]

wdym

[full dock]

you know how theres a letter before (x)

like f(x)

g(x)

things like that

it says ?(x)

[split chasm]

yeh...

then write a boxed ?

[full dock]

do i just keep it as ?(x)

[split chasm]

$\boxed{?}$

[solid kiln]

ℝamonov

[full dock]

and is the ?(x) = 2x^3-x^2+6 part of the g(x) = x^2 + 2x

or is that its separate thing

[split chasm]

two different functions

[full dock]

oh okay

so do i follow the same value input for part A with the equation on the right

[split chasm]

wdym

[full dock]

in A) it says g(1), g(2)

do i use that for the one on the right as well

[split chasm]

it says g(-1) and g(2)

[full dock]

oh okay my bad

do i use those for the one on the right

[split chasm]

i still don't get what you mean

[full dock]

do i input these values onto the equation on the right equation

in the image

[split chasm]

you use the g(x) function when finding g(whatever)
you use the ?(x) function when finding ?(whatever)

[full dock]

the right box

do i use g(x) values in part A with this as well

[split chasm]

no?

[full dock]

so what values do i use

[split chasm]

whatever it tells you to use

[full dock]

ohh

these ones right

[split chasm]

yes...

[full dock]

sorry for being dumb

im new to this

[split chasm]

for g(-1) substitute all xs in
g(x) = x^2 + 2x
with (-1)
for g(this) replace all xs in g(x) = x^2 + 2x with (this)

g(🍎) = 🍎^2 + 2🍎

[full dock]

and i apply that logic for the rest right

[split chasm]

for g(-1), performing the described action gets you
g(-1) = (-1)^2 + 2(-1)

and simplify to get the value they wanted

same idea for the rest yes

[full dock]

and what does part C mean

[split chasm]

same idea

[full dock]

express in terms of x?

[split chasm]

for g(this) replace all xs in g(x) = x^2 + 2x with (this)

your result will have x in it

[full dock]

ohhh okay

that makes sense now

thank you a lot

when i substitute value with x like

x-5

would it look like

x(x-5)^2 + 2?

[split chasm]

no

keep looking back at

for g(this) replace all xs in g(x) = x^2 + 2x with (this)

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[wary rivet]

Find the largest volume of the rectangular parallelepiped with the maximum surface area among the rectangular parallelepiped that can be enclosed in a sphere. ( r = 1)

[wary rivet]

i set f(x,y,z) = 2xy+2yz+2xz subject to x^2+y^2+z^2=4

but i can't solve this

.close

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[undone knoll]

just as a quick question if i have -x+1 and i insert for e.g. -3 than it would be -(-3)+1 and not -3+1 right? :0

[turbid spruce]

yep

[long basin]

Yes

Poyo

[undone knoll]

thanks! just wanted to make sure because i have like
f(k) : k + 1, if k ≥ 0
-k +1, if k < 0

so yea thank you <3

.close

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[wraith hinge]

.

hi

[wraith hinge]

I was wondring if I could remove the x-1 with 1-x

x-1/2(1-x)

[hardy zodiac]

Is this $\frac{x-1}{2(1-x)}$

[solid kiln]

Sup?

[wraith hinge]

yes its like this

[hardy zodiac]

Well u cant cancel right away, you need to make the x-1 into 1-x or vice versa

[wraith hinge]

how do I do that

[hardy zodiac]

recall that (a-b) = -(b-a)

[wraith hinge]

so it becomes 2-(x-1)

[slate cloud]

Implicit multiplication usually has precedence over division, however, since that is convention, I would recommend putting parentheses around it just to make sure it's unambiguous

[wraith hinge]

aight

[slate cloud]

right now you're saying 2 minus (x-1)

(which isn't equal to 2(1-x))

[wraith hinge]

I see

I was trying to simplify it and thought I could simplify it even more

[slate cloud]

Well you can. This is true.

[wraith hinge]

I do not know how to do it tho

[slate cloud]

what is 2*(-x) equal to?

[wraith hinge]

-2x?

[slate cloud]

yup

not 2-x

[slate cloud]

So, therefore, can you now transform 2(1-x) with this?

[wraith hinge]

it should be
2-2x

[slate cloud]

well, this is true, but we want to get a factor of (x-1) here

to simplify the fraction

[wraith hinge]

sorry I do not understand what you mean

[slate cloud]

we know that (a-b) = -(b-a)

and we want to get a factor of (x-1) in 2(1-x)

[slate cloud]

so just use this in the expression

(but correctly)

[wraith hinge]

so x?

[slate cloud]

2(-(x-1))

[wraith hinge]

why the added parentheses

[slate cloud]

Because in the original, it's two times 1 - x

So after applying the property, we get two times -(x-1)

not two minus x-1

[wraith hinge]

ohhh

I see

cuz its two

thanks for the help

imma close this one now

.close

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[wraith hinge]

it's given that AE = BE, AF = 3DF, and we need to prove that BD = 2DC, I am super frustrated trying to solve this thing

[trim jolt]

@wraith hinge Has your question been resolved?

[torpid depot]

wait let me try

@wraith hinge

do you still need help?

[wraith hinge]

Yes

[torpid depot]

a hint would be to draw DM || EC where M is aa point on AB