help-33

[still temple]

sqrt(171)/2

[still temple]

its sqrt(171)/2 😭

[alpine gulch]

huh

[still temple]

but ya did it right

now we do final step

the rhombus area is already hinted by them

[alpine gulch]

so 1?2 into 6 into 8

/

this?

wait i still have to find area of triangle

[still temple]

yessir

[still temple]

two infact

[alpine gulch]

add them na?

like this?😭

[alpine gulch]

wait so i just add 16.25 + 16.25 right??? or what

[still temple]

yeh

now add them all together

[still temple]

yessss

[alpine gulch]

damn thank you soooo muchhh

i will forever remember you

[still temple]

no need to thank

[alpine gulch]

yes there iss

ghanta no need

[marsh citrus]

@alpine gulch Has your question been resolved?

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[sacred ermine]

can someone help

[still temple]

,rotate

[still temple]

what question is it?

[sacred ermine]

number 8

[still temple]

alrighty

let x be the lenght of the depth of watsr

[sacred ermine]

yep

[still temple]

so from here the top lenght is 10-x

from here we can use pythagoreom theorem

and see that $(10-x)^2+7^2=10^2$

[elfin berry]

Project_Nova

[still temple]

it should be easy from here

i think its called polynomials

[sacred ermine]

so then it would be one hundred minus x squared plus 49=100 which will equal 149-x squared =100 which that will equal -x squared equals to -49

[amber birch]

Nope

$(a + b)^2 \ne a^2 + b^2$

[elfin berry]

south

[sacred ermine]

what

i haven't learnt this deep into it

[still temple]

well

$(a+b)^2=a^2+2ab+b^2$

[sacred ermine]

hold on so what does this mean

[elfin berry]

Project_Nova

[sacred ermine]

yeah

[still temple]

this is what it actually equal to

[amber birch]

So here a = 10 and b = -x

[still temple]

yep

[still temple]

so its $10^2-20x+x^2+7^2=10^2$ for ya

[elfin berry]

Project_Nova

[sacred ermine]

i have no idea what to do with these 😭

worded problems are just difficult idk why

[still temple]

wheh?

[still temple]

do you know what next to do here?

do you know the quadratic formula

i believe we have forgotten one step also

a condition

that $x>=10$

[elfin berry]

Project_Nova

[still temple]

[lone hare]

youre overcomplicating it

[still temple]

do tell how

[still temple]

how we finna find x then 😭

do educate me

[lone hare]

(10-x)^2 + 49 = 100
(10-x)^2 = 51
since 10-x is positive,
10-x=sqrt(51)
x=10-sqrt(51)

[still temple]

wohh

i see, thanks sir

also your answer is here john

honestly this the first time ive seen it solved this way

[lone hare]

letting x to be the required length was more confusing

you can see it more easily if you considered x^2+7^2=10^2

[still temple]

x has to be positive indeed

or that lenght

[sacred ermine]

oh okay thanks

i kind of get it

those questions make absolutely no sense

!close

.close

[marsh citrus]

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[marsh citrus]

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[meager glacier]

Can anyone help me see what I did wrong here? (Markscheme says the answer is C

[meager glacier]

Please tell me if my handwriting or any of my steps are unclear

The blue paper is my working

[still temple]

it is A

how would it be C?

[meager glacier]

idk

[still temple]

😭

[meager glacier]

Maybe a mistake from the markscheme

but i also got A

[still temple]

what is markscheme

[meager glacier]

answer sheet

[still temple]

are you sure that it says C?

[meager glacier]

damn

been reading the wrong section

ty for the help lol

[still temple]

😂

all good lol

happens

[meager glacier]

it was A so we good

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[ebon depot]

I'm given to planes s_1 and s_2
"Determine the tangentline T to the curve γ in the point A

[ebon depot]

How do I solve questions like these?

Since i want the tangentline which i assume is :
y = kx + m
I'm therefor missing k and m.

So i input A:s cordinates into r(t) ?

@glass silo Are you awake? ;-;

Or should it be :
y = f(a) + f'(a)(x-a)

[hexed junco]

You have to find the tangent vector using differentiation.

[ebon depot]

tangentvector?

But it says tangentline

[hexed junco]

what is r(t)?

[ebon depot]

But if i were to find the tangent vector, would it be by using this: ?

[ebon depot]

r(t) is a vector

Weird of him to write tangentline then

If it's a vector that i want to find

I would assume it's any of these formulas

[hexed junco]

vector equation of line, also you don't need partial derivatives as r(t) is a function of t only

not that it really matters

[hexed junco]

Don't need a formula just find dr(t) / dt

[ebon depot]

I'm sorry i don't think i understand

I already figured out r'(t) in a previous step a)

do i just plug in A:s cordinates into the vector?

[hexed junco]

to find the tangent vector at that point, however u still need to go through the process of finding the equation of the line

[ebon depot]

Ideally i'd like to understand what you are talking about

He never mentions a gradient vector

or a vector

"Determine the tangentline T to the curve in the point A"

[hexed junco]

I can't read whatever language ur sheet is in and I can't see the full question, but what u essentially want to do is get an equation of the form X(t) = X0 - Vd

[ebon depot]

I'm supposed to determine the tangentline T to the curve in the point A.
So don't i just derive r(t). Input the x,y and z coordinates for A which results in a vector that is the line of the tangentline?

[hexed junco]

this is the form of equation you are looking for indeed. The difference is f(a) and f'(a) are vectors, and so y is a 'vector equation' of a line

last point the x will obviously change to t

[ebon depot]

$r(t) = \pmqty{e^t + e^{-t} \ e^t - e^{-t} \ e^{2t}-e^{-2t}}$

[elfin berry]

Merineth

[ebon depot]

So this is a curve in R^3 space, right?

[hexed junco]

what is the full question in English ?

[ebon depot]

"Determine the tangentline T to the curve in point A"

[hexed junco]

what is the point A or is it not given?

[ebon depot]

It is given

ah fuck

i missed to give

We are given A:s x, y and z coordinates

[hexed junco]

I have no idea what that says

[ebon depot]

I translated it for you?

"Determine the tangentline T to the curve in point A"

[hexed junco]

ok lol here. The tangent line : y(t) = A + t*r'(a) where a is the value of 't' that gives A [that is r(a) = A]. I mean to find the value of t you could try writing the expressions in r(t) as hyperbolic functions. I have somet to do tho so if u really can't get it after trying what I have just said ping helpers.

[ebon depot]

<@&286206848099549185>

[marsh citrus]

@ebon depot Has your question been resolved?

[marsh citrus]

@ebon depot Has your question been resolved?

[heady lagoon]

the tangent line would have to be perpendicular to the gradient vectors of both surfaces at the point and i think in the question it gives you the equations, gradient F dot product tangent vector

f1 f2 and r are given so i think u can find it this way (solve for t)

[ebon depot]

This is very hard for me to understand

I understand that i'm suppost to find a line

A line has the equations
L : (x,y,z) = A + t * v

Where:
A = The point A
v = directional vector

[heady lagoon]

yes

[ebon depot]

And i'm looking for a directional vector on the point A

[heady lagoon]

yes

[ebon depot]

If i input the point A:s x, y and z into the system of equation

Like so?

I attempted this but there is no t that solves all of these equations

No sorry that's r(t)

we want r'(t) ?

[heady lagoon]

yes

im not exactly sure what the question is asking for

[heady lagoon]

or if u are doing a or b

[ebon depot]

I'm doing b)

Trying to figure out the tangentline on the point A to the curve r(t)

[heady lagoon]

ok

[ebon depot]

My assumption was that

a tangentline can be written in the form

y = kx + m?

But that doesn't work right? For R^3

I went back to my onevariablecalculus and found that a tangentline to the curve y = f(x) can be written as $y = f(a) + f'(a)(x-a)$

[elfin berry]

Merineth

[ebon depot]

Some other dude said it's correct but f(a) and f'(a) are vectors instead?

But i'm not sure at all what i'm supposed to do

[heady lagoon]

i was thinking find t by getting dot product of gradient F and r'(t) = 0

[ebon depot]

Hm that does not seem right

If i understand it correctly, i have a curve in R^3 and it's asking me what the tangentline on the point A is

But what kind of line do they mean? y = kx + m, y=f(a) + f'(a)(x-a) or (x,y, z ) = A + t * v

why = 0?

[heady lagoon]

ok nvm i think you would also have to consider that the tangent vector is perpendicular to the other surface as well

i think someone else could probably explain this better/correctly

[ebon depot]

I severely doubt it, been waiting 3 hours for someone who might know. And for the past week none has been able to help

Seems like multivariablecalculus is just to advanced now

.close

[marsh citrus]

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[wispy niche]

What's the problem?

[ebon depot]

#help-48 Summed it up more clearly here

[marsh citrus]

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[robust heron]

Hi I am not able to find the mistake in my solution

[robust heron]

ill send the pic

if i understood the question correctly, the problem is with my dr integral

however wolfram seems to agree with my calculations

so the problem is probably with dA itself

idk whats wrong tho

wait nvm

my power is wrong, it should be r^26

.close

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[still temple]

need some help

[still temple]

how are they getting from the penultimate step to the last

same with this

for first one

idek

[normal sky]

integration by parts

[still temple]

yeah

so it's just integration by parts 2 times

?

[normal sky]

they're just doing the integral of sec(x)

which is ln(sec+tan)

[still temple]

secx is not a standard function tho

[still temple]

how did you figure that out

[normal sky]

it's a standard integral

https://www.youtube.com/watch?v=CChsIOlNAB8&ab_channel=blackpenredpen

[blazing rampart]

nice bprp

you can use weierstrass substitution

I cant spell it

[still temple]

i thought only sec^2x was

[normal sky]

i mean "standard" is relative but intregral of sec is pretty well known

[still temple]

if i didnt know it

what would i do

[normal sky]

figure it out idk

[still temple]

?

[normal sky]

now you know it so

[normal sky]

idk how you want me to answer this question

[still temple]

ok nvm

what about

the second one

[normal sky]

it's another ibp

with integral of tan also being a well known result

[blazing rampart]

sec(x) has many methods

to integrate

[still temple]

OK

ok

such as

can you use double angle

or smth

[blazing rampart]

uh the method ik is weierstrass substitution

[still temple]

im sorry

but this is nonsense

they dont tell me anything about how to integrate the cotx

you're telling me i need to know all these standard integrals >?

[normal sky]

you don't need to memorize them

[still temple]

i thought if i get cotx

[normal sky]

but it's not difficult to do

[still temple]

it would just be -cosex^2x

[normal sky]

it's just a simple u sub

[still temple]

i dont get it tho

like

look what at what i said

[normal sky]

i am answering what you said

[still temple]

no

look

if cosec^2x integrated

gets you - cotx

wait i bugged out

yeah so anyway..

[normal sky]

.

[still temple]

what's the simple U sub in cotx

[normal sky]

u = sin

[still temple]

1/tanx

how do you identify which sub it is

[normal sky]

i mean, it's just experience

[winter smelt]

^

[normal sky]

if you're inexperienced, it doesn't make sense that you would "just know" which u to take

[still temple]

so we have

1/tanx

[normal sky]

but i don't think it's very difficult to see that in cos/sin

sin is a natrual choice

[still temple]

1/sin/cos

[winter smelt]

re-write that as cos/sin, and make the substitution u = sin(x) and try it out

[still temple]

.close

[marsh citrus]

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[still temple]

i got ti

[still temple]

yh yh ofc ofc

[winter smelt]

nice

[still temple]

but they write the answer directly

which makes me think they want us to just know it

lol

but it doesn't tell us these 'standard' integrals anywhere

[winter smelt]

also you could also notice that since cot = cos/sin, and cos is the derivative of sin, this intergrates to be some func with ln (natural log)

[normal sky]

like i said, "standard" is relative

[still temple]

yeah that's not rlly relevant tho

what you're saying

[normal sky]

i just meant that it's simple to integrate and the result is well known

[still temple]

how is standard integral defined then

like

it can't be that arbitrary

i thought there's a given list of integrals

if you want to change the description of it then that's on you

[normal sky]

sir i don't really wanna debate the semantics of this

[marsh citrus]

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[frank eagle]

Why is the integral $$\int_{-1}^1 \frac1{x} \ dx$$ not defined as a Lebesgue integral? What are the positive and negative parts of $1/x$ in $[-1,1]$?

[elfin berry]

Philip

[frank eagle]

.close

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[tardy idol]

im not sure conceptually how to go about this

at first i found her horizontal velocity... should i then find her vertical velocity and then find the angle between them?

[quick moth]

Find vert and Boris

Horiz

And then you can use pythag

To find the velocity of them combined

[tardy idol]

this is what I did but it says its wrong

[quick moth]

Sense check

How could the velocity be 71 ms

I see a mistake in your tantheta= o/a

[tardy idol]

nono the angle

oh woops

ok i corrected the arctan mistake and its still wrong

i got 18.8449 degrees

[quick moth]

Sry one sec

I’m about to get gone

Home

[tardy idol]

its ok! not time sensitive

i just have 2 attempts left so i wanna make sure its right so i dont lose points

[quick moth]

ok sry

im back

ok

so you have two componets

copx and copy

first off, copy is going to be easier

both jill and jack aren't heading the vertical direction so

idk what u did here tbh

[tardy idol]

oh

so true

wait no

[quick moth]

idk how u got this velocity

[tardy idol]

so in the beginning neither is oging in the y direciton

so 0

[quick moth]

you have the right from

[tardy idol]

and in the end jack has a vertical y component

[quick moth]

right here

[tardy idol]

did i just type it wrong maybe

[quick moth]

56(-5sin(34))/46

=-3.404

not -5.046

[tardy idol]

you're right i also got -3.4 when i typed it in

12.964 degrees

[quick moth]

ur x velocity

wrong too

p sure

56(-8)

= 56(-5cos(34))+46vjillx

[tardy idol]

46

[quick moth]

oh oops

jack mass 56 and jill 46

[tardy idol]

yes

[quick moth]

still

its wrong

should be around 4

u have it right

here

[tardy idol]

yeah it shouldnt be more than 5 huh

[quick moth]

56(-8)-56(-5cos(34))

=

-215.869

/46

=-4.693

[tardy idol]

why-5?

instead of 5?

[quick moth]

oops just 5

and 8 is pos

aswell

[tardy idol]

i thought 8 should be negative because its going in the opposite direction

[quick moth]

sliding due east

means hes going to the right

[tardy idol]

.....

awh man

i cant read

youre right

[quick moth]

tell me what u get

as ur new theta

make sure to carry down the decimals

[tardy idol]

should jills velocities still be negative then?

[quick moth]

doesnt matetr neg will cancel out

in the equation anyways

[tardy idol]

ok

[quick moth]

her horiz, no

vert, yes

but when you express an angle you wouldnt express it as a negative angle

you would BH

meaning below horizontal

[tardy idol]

54.046

[quick moth]

uh

did you do x/y

or y/x

[tardy idol]

x/y

[quick moth]

y/x

[tardy idol]

woops my bad i think im slow rn

[quick moth]

remeber from the unit circle that tangent is y/x

vertical velocity is y componet

[tardy idol]

yeah TOA

[quick moth]

and horizontal is x componet

[tardy idol]

35.954

[quick moth]

ye

idk how many decimals

they want but

[tardy idol]

rahhhh thanks for sticking with me

[quick moth]

np

[tardy idol]

decimal just 2

have a nice day

[quick moth]

u aswell

[tardy idol]

.close

[marsh citrus]

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[flint reef]

How do I expand and simplify

[flint reef]

First I put in -2 into the first bracket

[desert dirge]

you dont need to, but sure

[flint reef]

Wait hold on

Nvm

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[oblique musk]

how can i solve this

[oblique musk]

my first approach wil be break the whole thing into x^3-2x^2 and -4

so that it gets easy to solve then

am i right

?

[turbid mica]

yeah

[oblique musk]

cool
i reluctant a bit for that method cause i thought that this is rate of change of the function not the actually function

but i can use that manupilation

ok

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[topaz latch]

so im just not sure when applying the divergence theorem when I converted to spherical coordinates

I get the correct answer with dv = dz dr d theta

but not with dv = r dz dr d theta

oh sorry thats a bad picture actually

[fervent rampart]

i wonder if it's being integrated incorrectly?

[topaz latch]

since x = rcos theta

in the first image I made the mistake of doing x = cos theta

[topaz latch]

the textbook says this answer is correct for 620 pi

heres the problem again

[fervent rampart]

ok so our integrand is the divergence, which you found is: [ \nabla \cdot \vec{F} = 3x^2 + 3y^2 + 3z^2 ] What is that in polar (cylindrical) coordinates?

[elfin berry]

pnoןɔ

[topaz latch]

6cos^2 (theta) = 6sin^2(theta) + 3z^2

[topaz latch]

like I wrote above

[fervent rampart]

in polar coordinates we have: [ x = r \cos \theta, \quad y = r \sin \theta, \quad z = z ]

[elfin berry]

pnoןɔ

[fervent rampart]

r is a variable, not a constant, so we can't just substitute a value yet

[topaz latch]

ok

[topaz latch]

maybe the solution is incorrect?

[fervent rampart]

you can get the same number without necessarily having the right approach

[fervent rampart]

so returning to this, what is this expression in terms of r, theta, and z?

[topaz latch]

3r^2cos^2 theta, 3r^2 sin^2 theta and 3z^2

oh wait

is that my mistake

is that I substituted r for the constant value

[topaz latch]

Ok I think I get what you mean

but now im getting 120 pi

is there something wrong with my integration boundaries

nvm

I forgot to square r also

ok I got the correct solution the right way

@fervent rampart also is this a typo

is that supposed to say divF = 10

F = 10 means nothing to me

[fervent rampart]

i would assume so

[topaz latch]

Ok. cool thanks just making sure I didnt forget something

thanks for the help

[marsh citrus]

@topaz latch Has your question been resolved?

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[oblique holly]

This on least squares regression lines

Need help of any kind

[marsh citrus]

@oblique holly Has your question been resolved?

[marsh citrus]

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[late salmon]

just wanted to check my answers on this

[late salmon]

for the one on the left i got -6 and the one on the right i got 3/2

.close

[marsh citrus]

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[waxen nest]

3. A 50.0 kg person on a trampoline bounces to a height of 5.0 m. On their second bounce they end up only 4.5 m high. On their third bounce they end up 6.0 m high.
   a. If they just bounced without using any effort, how much energy was lost in the second bounce to sound and heat?

[waxen nest]

Quick physics question

Extremely confused

Doesnt kinetic energy play a role

😢

[severe cypress]

@waxen nest try the physics server in #old-network

[worn plank]

you can say Energy at beginning = Energy after bouncing + Energy lost

ok know what don't trust me

[vast rivet]

hey friends

i was here, testing some calculus II questions, and im stopped in this:

the mission is integrate by substituition, but i failed, anyone can help?

[marsh citrus]

@waxen nest Has your question been resolved?

[marsh citrus]

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[reef hinge]

Is this correct or wrong

The reason I think my answer is wrong is because of the isomorphic invariant that includes the shades regions being different than eachother, therefore the graphs are different

G has 5 shaded parts and H has 2 does that mean this graph isn't isomorphic

[marsh citrus]

@reef hinge Has your question been resolved?

[marsh citrus]

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[devout oak]

I have to prove this equality

either by going from right to left or left to right

been trying the whole day

(calculus 3)

(its a long proof)

I can show my game plan if needed

[wispy niche]

Do you know how to use the chain rule?

[devout oak]

yep

I can show you my work

[wispy niche]

Ok

[devout oak]

okay so here is my game plan to proof this

my goal is to make a proof from left to right

but before that I did my tree to see what are the variables in question and what depends on what

[wispy niche]

Ok

[devout oak]

here is my tree thats how I view it

I know Z is my dependent variable

and he depends on both the function f(u) and g(v)

[wispy niche]

Yes

[devout oak]

and both of these functions depend of x and y

[wispy niche]

Indeed

[devout oak]

but we cant forget this little thing

since I did u=x-y and v=x+y I need to transform my 1/x to Us and Vs since I did a variable change

so I did the following

[wispy niche]

Ok

[devout oak]

so 1/x= 2/u+v

[wispy niche]

Seems legit

[devout oak]

because the ys cancels themselfs

so I end up with my z being:

[wispy niche]

Fair enough

[devout oak]

and I can start my proof from here

[wispy niche]

How did you proceed?

[devout oak]

my game plan is going from right to left

I started off like this

we know this:

so I have to find dz/dy

[wispy niche]

Ok

BUT

[devout oak]

yes

[wispy niche]

z is now in terms of u and v

[devout oak]

yup

I have to change that x2

[wispy niche]

So chain rule right?

[devout oak]

yep

I did this

u agree?

[wispy niche]

Yes

[devout oak]

ok buddy

then I calculated the du/dy and dv/dy

it gives 1

wich is like saying this:

[wispy niche]

Fair enough

[devout oak]

from this point things get heated

I have to find dz/dv and dz/du right

so I did this

[devout oak]

I replaced my z basically

thoughts?

my logic is I know Z is a multiplication

[wispy niche]

Ok, it looks complicated but good

[devout oak]

so I did the multiplication rule u'v*uv'

[wispy niche]

Good job there!

[devout oak]

wich led me to here

[wispy niche]

Ok

[devout oak]

the f(u) and g(v) become 0 because they are like constantes

so I remove those

[wispy niche]

But not the other ones that aren't constant

[devout oak]

yup

I end up with this

after doing the derivates

the law I used is this one:

thats how I did the 2/u+v

[wispy niche]

Ok, so far so good

[devout oak]

I used the red thingys so I can see my terms and not get mixed up

I end up with this and a bunch of things end up canceling themselfs due to the minus

[devout oak]

this is after calculating d/dy

[devout oak]

still used u'v+uv'

[wispy niche]

Give me a sec, I'm trying to check all of it

[devout oak]

no problem

I know there is a lot of ''writing'' mistakes so ill apologize for that

a bit messy

but the idea is there

[wispy niche]

Ok, my recommendation is to cancel those terms that occur with opposite signs to simplify the following equation

[devout oak]

yup

thats the purples Xs

[wispy niche]

But everything seems ok

[devout oak]

I end up with this

only 4 terms and 2 can get simplified to 0

from this point there might have been a mistake

do you think im allowed to say the first and third therm are equal to 0?

[wispy niche]

No, cuz 'u' and 'v' depend on 'y'

[devout oak]

ya I was thinking that now that I read it again

so I guess im left off here

[devout oak]

if everything before is correct, it means that this expression (ignore the 0s) is equal to this

do you have in mind a way to transform that to this?

or do we restart

[wispy niche]

Wait no, srry my mistake, forgot that $x=(u+v)/2$, so $1/x=2/(u+v)$

[elfin berry]

bondalton

[wispy niche]

So the y-derivatives are zero indeed

[devout oak]

ohh yeahh thats the logic I used

mb im awake for like 18 hours

because there is no y so its a constante

and equals to zero

[wispy niche]

Yep

[devout oak]

so after that I did this

[wispy niche]

Sorry for the confusion

[devout oak]

all good

I end up with this and this is where iv been stuck for the past 5 hours

idk how to do this

the one in red

I still havent transformed my x^2 we can do it if you want

[wispy niche]

That's just the chain rule again

For instance

[\frac{\partial}{\partial y}\left(\frac{\partial g}{\partial v}\right)=\frac{\partial}{\partial v}\left(\frac{\partial g}{\partial v}\right)\cdot\frac{\partial v}{\partial y}+\frac{\partial}{\partial u}\left(\frac{\partial g}{\partial v}\right)\cdot\frac{\partial u}{\partial y}]

[devout oak]

it says failed to render

[wispy niche]

Omg, give me a sec haha

[devout oak]

no problem man

[elfin berry]

bondalton

[devout oak]

hmm

[wispy niche]

It's just the chain rule applied to $\frac{\partial g}{\partial v}$

[elfin berry]

bondalton

[devout oak]

I cant manage to see it

on the tree

[wispy niche]

So, you see how like $z$ depends on two variables and you used the chain rule here in a similar fashion?

[elfin berry]

bondalton

[devout oak]

yup

[wispy niche]

Same thing with $g$, it depends on two variables and it's partial derivative depends on those two variables as well

[elfin berry]

bondalton

[devout oak]

thats where you lose me because g isnt a variable its a function

I cant grasp my head around that

[wispy niche]

The thing is that $z$ is also a function! Indeed, $z=z(f(u),g(v))$, so $z$ is a function of $u$ and $v$

[elfin berry]

bondalton

[devout oak]

true I see ur point

give me 2 mins im thinking about it

[wispy niche]

Ok

[devout oak]

but I know the behavior of the function Z tho

[devout oak]

how would you write g(u)=

sry f(u)

in order to derivate it

all we know is the value of u

[wispy niche]

Well, $f(u)=f(x-y)$

[elfin berry]

bondalton

[devout oak]

yup

[wispy niche]

And that's as much as we know about $f$ and $g$

[elfin berry]

bondalton

[devout oak]

I agree

[wispy niche]

BUT we are assuming that $u$ and $v$ are functions of $x$ and $y$

[elfin berry]

bondalton

[wispy niche]

Indeed, writing $u=x-y$ for example

[devout oak]

yes because of the variable change we did at the begening

[elfin berry]

bondalton

[wispy niche]

So taking the $y$-derivative of $u$ makes sense right?

[elfin berry]

bondalton

[devout oak]

if I continue down this path do you think at one point its going to prove the equality?

[wispy niche]

Totally!

[devout oak]

would you have chosen a different game plan that is shorter?

I know someone told me to evalute both sides of the equation and see if we end up with the same value but he didnt try it

[wispy niche]

I would've tackled the problem in a straight forward manner

Pretty much all you did, but without the $u$'s and $v$'s

[elfin berry]

bondalton

[devout oak]

wym without the u's and v's?

what would your tree look like

[wispy niche]

That's the thing, I wouldn't use a tree

But that's just me

Try to use what I told you, give it a shot 🙂

[devout oak]

ill try it

do you think after I use it ill be close to it?

we still havent touched the x^2

also where are the d/dx gonna come from?

I have trouble viewing how we can end up with the left expression ngl

[wispy niche]

Yup, it's just tons of work and chain rules

[devout oak]

can I show you what my teacher did?

[wispy niche]

Sure!

[devout oak]

because idk why but I feel im doing it 4 times longer than her

Her game plan was the same but instead of going from left to right , she did right to left

This is what she wrote on the board

I got inspired by her to do similar

[wispy niche]

Yes! She's doing the same thing I'm telling you, applying the chain rule once more to the partial derivatives

[devout oak]

okay tysm ill give it a try

thank you for ur time 🙂

[wispy niche]

No problem! 🙂

[devout oak]

is there a way to give a tip or its against the rules?

since u took a lot of time

its my first time using this server

also what are your credentials if I may ask just curious

[wispy niche]

No need for that, it's fun helping others 🙂

[wispy niche]

Credentials? Like what am I studying?

I'm a grad math student

[devout oak]

alr again thank you for your time

[devout oak]

ah okay sounds cool im a first year engenieering student

[wispy niche]

Hey, good luck with the maths!

[devout oak]

For the next one I’m assuming it’s the same thing right?

[wispy niche]

Yep, same trick, lots of work 😦

[devout oak]

Alr ya it does take time

Again thanks a lot friend

[wispy niche]

No problem!

[devout oak]

.close

[marsh citrus]

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[marsh citrus]

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[main mica]

Equalavent pentagons neighboring sides are MN and NQ in that pentagon there is a point P so it makes MPN equalaveng triangle. Need to find angle NQP

[main mica]

<@&286206848099549185>

[calm harbor]

Equivalent pentagon?

[main mica]

MNP is equal all sides are equal

So far ik that pentegons each angle is 108

Equal triangles every angle is 60

[calm harbor]

Are you sure?

[main mica]

Wdym?

[calm harbor]

Wait, is it what the question said?

[main mica]

If a triangles every side is equal all angles are 60 wdym

[calm harbor]

It’s not equal

[main mica]

My guy the problem says it is

MNP is equilateral

[calm harbor]

Oh ==

Nvm, let’s move forward

[main mica]

So i need to find angle NQP

[calm harbor]

So the question says MN=NP=MP

And that’s the only given info

[main mica]

Yes

We know that pentagons each angle is 108 degrees

Have any ideas?

[calm harbor]

It’s loading

[main mica]

Oh i see

What dose that tell me tho?

[calm harbor]

NP=NQ

[main mica]

66 is the answer

Thansk

[calm harbor]

!done

[marsh citrus]

If you are done with this channel, please mark your problem as solved by typing .close

[main mica]

.close

[marsh citrus]

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[marsh citrus]

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[cloud token]

hey i need a quick help with my activity just one here it is

[shadow nest]

what have you tried

rather what are you confused on

[cloud token]

i quite lack of knowledge at the moment and that is why I'm confused at 3 and 4 (reason side)

[shadow nest]

ah

so what do you notice with angle 2 and 3

the form two intersecting lines right?

meaning you can apply this

when you have opposite angles in an intersection like that

you can say that those opposite angles are congruent

for example, the blue ones have the same measurement with each other

while the yellow ones have the same measurement with each other

to be specific, these pair of angles are called "Vertical Angles"

so the reason will be

Vertical angles are ???

[marsh citrus]

@cloud token Has your question been resolved?

[marsh citrus]

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[main mica]

Need to find the black parts area if one legs middle point is connected to the other legs corners.

[marsh citrus]

@main mica Has your question been resolved?

[main mica]

<@&286206848099549185>

[lofty gyro]

what's the original question?
there isn't enough info here

[main mica]

Oh yea

I forgot

So the full traps area is 20

Need to find that triangles area

[lofty gyro]

is it a trapezium?

any more information?

[main mica]

Trapezoid

[lofty gyro]

then you can split it into two parts

[lofty gyro]

with 3 lines parallel to each other

then you can
let l be lower base
let u be upper base
let h be height
and try to calculate the areas

[main mica]

The blu one is the middlelin right?

[lofty gyro]

yes

note that the length middle line within the shape is (l+u)/2

[main mica]

That *h=20

What next tho?

[lofty gyro]

, rotate

[main mica]

Why/4??

[lofty gyro]

hmmm

[lofty gyro]

note that the length middle line within the shape is (l+u)/2
hint

[main mica]

Mhh do i need a system?

Btw i think this can be done simpler were not using the legs middle line fact

Middle point

[lofty gyro]

well, you'll just have to write and out and spot a pattern instead of solving for the variables

[lofty gyro]

do you recall anything from here? @main mica

[main mica]

Mhh yes

So is the triangles area 10?

I see

I still don’t get the :2 part well

[lofty gyro]

oh

it's just area of triangle

base times height ÷2

[main mica]

.close

[marsh citrus]

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[marsh citrus]

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[limpid heart]

Help me if my answers correct

[marsh citrus]

@limpid heart Has your question been resolved?

[limpid heart]

<@&286206848099549185>

<@&286206848099549185>

<@&286206848099549185>

[marsh citrus]

@limpid heart Has your question been resolved?

[limpid heart]

.close

[marsh citrus]

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[potent rain]

How do I show that this diverges?

[turbid mica]

dont you need a limit

[shadow warren]

compare it to $\frac{\frac{-\pi}{2}}{x}$

[elfin berry]

y0shi

[shadow warren]

the lowest value arctan can give is -pi/2

[potent rain]

ah, right, so it's always lower than that, and that diverges. I tried to compare it to $\frac{\frac{\pi}{2}{x}$, which obviously didn't work

[elfin berry]

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

[potent rain]

Hm

[potent rain]

Don't think so no

but alright then I get it lol, thanks

.clos

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[marsh citrus]

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[tame ivy]

how does it go from there

[late geode]

did you crop the image

[late geode]

you conveniently cropped out the important presence of the fraction and 3 in the denominator

sec^2(t)(1 + tan^2(t)) expands to
sec^2(t) + sec^2(t)tan^2(t)
you shouldn't know the integral of the first term
you could do a sub like u=tan(t) for the integral of the second term

[tame ivy]

ah so i go from this

sec^2(t) + sec^2(t)tan^2(t)

ok its not a one step solution right

ok ty

.close

[marsh citrus]

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[proper lintel]

What am i doing wrong here????

[proper lintel]

How did my notes get those answers

[proper lintel]

0 + 2/5 √2 |v| - 2/5 √2 |z|= 0
5 √38/38|u| - 3 √38/38|v| - 3 √38/38 |z| = 0
5 √38/38 |u| + 5 √38/38 |v|+ 5 √38/38 |z|= 300

I plugged systems of equations into calculator… and it was onviously wrong

Answwre key says

[marsh citrus]

@proper lintel Has your question been resolved?

[proper lintel]

<@&286206848099549185>

[marsh citrus]

@proper lintel Has your question been resolved?

[proper lintel]

😦

I am cooked

Noone can solve this

PLEAAASEEE ::(((

It’s over

It is so over

[proper lintel]

I just did the quiz

I losed

🪦

I dont need help anymore

[marsh citrus]

@proper lintel Has your question been resolved?

[proper lintel]

No, but I don’t need it to be resolved anymore

[marsh citrus]

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[red mason]

This is like 9th grade math but I haven't learned everything yet, so I am I need of help!

The figure shows two squares (black lines) with parallel sides. The big square
has side length 3 and the small one has side length 2. What is the area of ​​the blue shaded quadrilateral?

[marsh citrus]

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[smoky plover]

the area is the sum of 4 triangles + the square

we know the square's area

so we need to find the sum of the 4 triangles' area

the area of a traingle is base*height/2

use that to solve the problem

[red mason]

I got about all the info that is the triangles bases are 2 but I can't find the heights

[smoky plover]

you dont need to find the heights individually

find the sum of the heights

[red mason]

The height of like the whole thing?

[smoky plover]

the sum of the heights of the triangles

[turbid mica]

all the heights added together

[red mason]

how do you see that?

[smoky plover]

see what

[red mason]

The heights

[smoky plover]

the height is the perendicular distance from the opposite vertex to the base

[marsh citrus]

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[red mason]

I haven't learned that and I can't understand that sentence

[red mason]

lemme translate

[red mason]

Yeah, I got all that but all the heights like different

[smoky plover]

but the sum?

[red mason]

How do you find the sum?

[smoky plover]

well look at the sum of the heightns of triangles on opposite sides

what do you notice?

[red mason]

They're bigger

[smoky plover]

no

[red mason]

and 2 of them are the same?

[smoky plover]

no

[red mason]

what????1!! im so dumb

[smoky plover]

hint: 3-2

[red mason]

1?

[smoky plover]

what does 3-2 represent here

[red mason]

the height between them? like its fully in the middle?

[smoky plover]

wdym

[red mason]

like 0.5, 0.5

the bottom one and the over

[smoky plover]

ok i'll just give you the hint

[red mason]

yeah ty

[smoky plover]

the sum of opposite heights is 1

let the first height be x, and the second be y

then we have x+y+2=3

because the sidelength fo the big square is 3

and the side length of the small sqyare is 2

the 2 heights and the side length fo small square comprise of the side length of the big square

[red mason]

are we talking about the heights of the bottom one and the one that is over?

or the sideways one?

right to left

[smoky plover]

either one

[red mason]

Okay can you just give the answer for now and I'll figure it out

[smoky plover]

sorry we cant do that

[red mason]

I don't udnerstand math in english

oh okay

[smoky plover]

translate

[turbid mica]

is it this

[smoky plover]

not really

ig ill drwa it out

[turbid mica]

dang

oh it's 1/2 height

[red mason]

Okay but, my school we do geometry last semester so I haven't learnt everything

So this is like unknown

And I still don't understand in my native language

[red mason]

Oh okay

So its 0.5? Or the heights doesnt matter?

[smoky plover]

the sum of the opposite heights is 1