help-33

[ember garden]

1+x^2 1 -x 2x

[hasty flint]

Yes no it wasn't I'm sorry for confusing you

[fringe nacelle]

parentheses missing, but it happened to not matter in this case

i thought thats what Jan was saying

[hasty flint]

Yeah it confused me

[ember garden]

Where should i put the parenthesis

[fringe nacelle]

its a mistake but etienne lucked out that it didn't matter

no derivatives in the denominator

(1+x²)1

[ember garden]

But isnt it the chain rule

[fringe nacelle]

thats if youre using the alternative form

D[fg^-1]=f'g^-1 + f d(g^-1)/dg dg/dx

you aren't

[ember garden]

I think im more confused than how i started

can u elaborate

[hasty flint]

This

[ember garden]

ah

ok so just simplify the deno

[hasty flint]

Look at the denom, no derivative there

[hasty flint]

Yes

[ember garden]

ok doki

[fringe nacelle]

D[fg^-1]=f'g^-1 + f d(g^-1)/dg dg/dx

this is for the other form, youre not using it here

[hasty flint]

The derivative of fx/gx is literally this. You don't have to take any extra derivatives. You can find this by using the multiplication rule for derivatives

[fringe nacelle]

and btw capital D if youre using Euler notation Dx[f], lowercase d if you're using Leibniz notation df/dx

looking good

[hasty flint]

Now simpify the top part

[ember garden]

ok ill proceed

ok

[hasty flint]

(What is the top part called in english?)

[ember garden]

numerator

[fringe nacelle]

in fact you probably could leave the denominator simply as (1+x²)² since expanding it isnt really a simplification, its just a choice of what you like better

[ember garden]

next time ill leave it that way maybe

[fringe nacelle]

and the middle line is called the vinculum but no one says that

what does the numerator of 6 simplify to?

[hasty flint]

So this wasnt how you simplify the numerator. Lets look at that:
x . 2x = 2x^2. So what tou get is 1 + x^2 - 2x^2

Yes agreed or not?

[ember garden]

hm

[ember garden]

sorry i do not understand...

[hasty flint]

So we agree that 2x . x = 2x^2 yes?

[fringe nacelle]

what's x2x my friend

[ember garden]

oh

yes!

[hasty flint]

So lets do that on the numerator

From 6

[fringe nacelle]

thats the part above the vinculum

[ember garden]

1+x^2-2x^2

[hasty flint]

Yes

Now simplify

[fringe nacelle]

same thing as 1+1x²-2x²

if that helps

[ember garden]

1-x^2

[hasty flint]

Yes, good!

[fringe nacelle]

[hasty flint]

Now, once you write the denominator, you are finished, I hope this wasn't too chaotic

[fringe nacelle]

the denominator i think you just made a typo for the mistake

[ember garden]

Thank you for helping me

[ember garden]

Huh?

[hasty flint]

Why does it say . in the denom? Is it a -?

[fringe nacelle]

how do you simplify 1 + x²+x²+x⁴?

[ember garden]

1+2x^2+x^4

[fringe nacelle]

yeah you wrote a dot

[ember garden]

no

its * and +

oh

sorry i think i reordered

[hasty flint]

Oh no, we are so close but the denom isnt right

[fringe nacelle]

reordering doesnt change + to ×

[ember garden]

Yea i made mistake

No worries i will recorrect

[hasty flint]

Exactly, the . Should be a +

[fringe nacelle]

you just need practice then youll do fine with the quotient rule

[hasty flint]

Yes

[hasty flint]

Yes!

[ember garden]

Yay

[fringe nacelle]

erase the part" <- chain rule "tho

[ember garden]

Ok

I will leave this channel open because i am working on a new one

[hasty flint]

Alright

I will try to help more clearly if you have another one

[ember garden]

ok ty

Hm

If i have -18xy^2, will i use product rule?

[hasty flint]

What is the full problem?

[ember garden]

Implicit diff

[fringe nacelle]

product and chain rule

d/dx[y²]=d(y²)/dy dy/dx

[hasty flint]

I do not know what those are lmao, guess I won't help

[ember garden]

Mmm

Im doing 2a

[fringe nacelle]

implicit diff is when your variable of interest,call it y, is writtzem in the form F(x,y)=blah blah,and in general y isnt solved for easily

otherwise its just chain rule

[ember garden]

Wait why we need chain rule

[fringe nacelle]

implicit differentiation uses the chain rule

[fringe nacelle]

^

[ember garden]

Hm ok

[fringe nacelle]

i should say that

here you assume y implicitly defines a differentiable function of x

so i should say you're given F(x,y(x))=0 and asked to find dy/dx

(you could always subtract one side from.the other to set it equal to 0)

[fringe nacelle]

not 40x⁴

[ember garden]

Oops

40x^3

[fringe nacelle]

how do you take the deriv of a product?

[ember garden]

Hm

Find the derv seperately

[fringe nacelle]

D[fg]=f'g+fg'

[ember garden]

Oh

[fringe nacelle]

and here i recommend using Leibniz notation

d[y²]/dx = d[y²]/dy dy/dx

[ember garden]

Sorry this one confuses me

[fringe nacelle]

would it help if i said

let h=y²

then dh/dx=dh/dy dy/dx

?

[ember garden]

Hm

Then d(y^2)/dx= d(y^2)/dy *dy/dx

Why would we do this

[fringe nacelle]

what do you think the chain rule is

[ember garden]

multiplying by the inside and outside

[fringe nacelle]

write out the form you know, ill show its equivalent

or i will

so you know this

[ember garden]

ok one sec

[fringe nacelle]

$$D[f \circ g] = (D[f] \circ g) D[g]$$

[elfin berry]

gfauxpas

[fringe nacelle]

yeah?

[ember garden]

f(g(x))'*g(x)'

[fringe nacelle]

write, but what does f o g mean?

it's the function that sends x to f(g(x)) right

[ember garden]

g inside of f

[fringe nacelle]

mm hmm

and when you write f(g(x))'

you mean f' o g

so

(f o g)' = (f' o g)g'

now watch how it's the same

you want to know

$$\dfrac {\mathrm d[f(g)]}{\mathrm dx}$$

[elfin berry]

gfauxpas

[fringe nacelle]

(f'og)=

$$\dfrac {\mathrm df}{\mathrm dg}$$

[elfin berry]

gfauxpas

[fringe nacelle]

and g' = dg/dx of course

[ember garden]

mmm

[fringe nacelle]

so you have

$$\dfrac{\mathrm df(g)}{\mathrm dx} = \dfrac{\mathrm df}{\mathrm dg} \dfrac{\mathrm dg}{\mathrm dx}$$

[elfin berry]

gfauxpas

[fringe nacelle]

so it's the same thing

[ember garden]

oh

[fringe nacelle]

cool, no?

[ember garden]

ye

[fringe nacelle]

so

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

[ember garden]

hm

ok

i think i need to take a break i amm getting dizzy

i will try again soon

[fringe nacelle]

sure, make sure yuou drink water and aren't too hot

[marsh citrus]

@ember garden Has your question been resolved?

[marsh citrus]

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[neat grove]

yo could someone label angles here

im a bit confused cuz i did it wrong but idk how

[indigo nest]

Well showing us what you did would be a start

[slim shard]

diagram is correct

[neat grove]

it isnt

because tsinø + 35sinø = mg is incorrect

[indigo nest]

Both of those angles won't be θ

[indigo nest]

One would be θ and the other would be 90 - θ no?

[neat grove]

ye but

im jss ayin its

[slim shard]

oh wait i didnt see that

[neat grove]

completely wrong what i did

[neat grove]

Like i tried doing the angle between the string and the vertical

but then idk what to do with that lol

[indigo nest]

Well that's one of the mistakes

See where fixing that gets you

[neat grove]

oh wait

i was correct

the first time lol

i shldve just kept it

between string and vertucak

Im done ty guys lol

i was js innatentive

.close

[marsh citrus]

Channel closed

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Use .reopen if this was a mistake.

[indigo nest]

Happens no worries

[marsh citrus]

Available help channel!

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[vivid beacon]

where did they get this 40 from

[cunning fiber]

40 miles from home - 0 miles from home = 40 miles traveled

[vivid beacon]

yh

[noble geyser]

no

[vivid beacon]

she goin from doc to home

[candid dune]

Can someone help me with my maths

[noble geyser]

40 miles / 1 hour

[noble geyser]

!occupied

[marsh citrus]

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).

[candid dune]

Okay

[noble geyser]

40 miles / 1 hour --> 20 miles / half hour

[vivid beacon]

yh i got the 20 miles bit

this 40

highlighted in green

[cunning fiber]

yes that's what I was answering

40 miles from home - 0 miles from home = 40 miles traveled

she started at 40 miles from home at the hospital

[noble geyser]

yea yea whats the problem

[cunning fiber]

and ended up at 0 miles from home when she actually got home

[vivid beacon]

home to hospital and hospital tp home is same distance

[cunning fiber]

also I got this, it's ok

I don't want a too many cooks in the kitchen situation

[cunning fiber]

mhm

[vivid beacon]

lol

[cunning fiber]

because you're going between the same two locations either way

[vivid beacon]

yh

[noble geyser]

kk bye then

[cunning fiber]

so you good now?

[vivid beacon]

i dont understand the whole question

i only found out 20 bc tht was missing i dont even know wht im goin to do with the 20

[cunning fiber]

notice how the graph doesn't have a scale

we need to figure that scale out

the first part tells us that Ria stopped for a break

[vivid beacon]

yh

[cunning fiber]

stop for a break -> no moving -> distance from home doesn't change

[main parcel]

im fakelight

[vivid beacon]

waht

[main parcel]

so the graph starts out for constant speed, then it rests, then constant speed again

[cunning fiber]

based on this, we know the higlighted horizontal part is where she stopped moving

[main parcel]

easy peasy

lemon squeezy

[cunning fiber]

I'm already helping

it's fine

[main parcel]

okay bet

[cunning fiber]

@vivid beacon Do you agree so far?

[main parcel]

bye babe

good luck

[vivid beacon]

yh

[cunning fiber]

now you determined that she traveled 20 miles before taking a break, right?

[vivid beacon]

yh

[cunning fiber]

that tells us that this part of the distance axis is 20 miles

[vivid beacon]

thats 20

[cunning fiber]

correct

so at this point, you can do one of two things

you can either work out the scale explicitly (4 boxes = 20 miles)

or you can notice that each of these are the same height

and hence both of them represent 20 miles

does that make sense?

[vivid beacon]

yh

[cunning fiber]

yup that's exactly what I was thinking of here

[vivid beacon]

yh i followed that

[cunning fiber]

so we addressed what to do with the 20 miles

aka the info from this part

and that in turn tells us how to get the total distance of 40 miles from the hospital and the home

good?

[vivid beacon]

yh

[cunning fiber]

any other questions?

[vivid beacon]

yh

20 miles

[cunning fiber]

what about it

[vivid beacon]

no i undertand that its 20 miiles now

[cunning fiber]

alright

[cunning fiber]

now we just figured out the scale in the vertical direction along the distance axis

any idea of how to find the scale in the horizontal direction along the time axis

@vivid beacon Hello?

[vivid beacon]

hello

uhh no

[cunning fiber]

read the question

they tell you the relevant info

you just need to find it

[vivid beacon]

ik this

[cunning fiber]

that's what I was going for

except the 1:30 looks off to me

how did you get that

oh nvm you just converted between

mm

that could get confusing though

[vivid beacon]

same thing no?

[cunning fiber]

if you're switching between 24 hour time and 12 hour time

cause you might end up interpreting 1:30 as 1:30 am isntead of 1:30 pm

[vivid beacon]

k

[cunning fiber]

yeah I'd keep it consistent like that

[vivid beacon]

kk

[cunning fiber]

you can also do 1:00 and 1:30 if you wanted to

anyway

[cunning fiber]

we can now split all of this up into time intervals

that looks good

[cunning fiber]

^

[vivid beacon]

yh

[cunning fiber]

you've already figured out that this is 15:00, which is good

[vivid beacon]

i mean 3

[cunning fiber]

mhm

[cunning fiber]

now what does this time actually represent

(in the given context)

[vivid beacon]

that time she stoped for 30 mins?

[cunning fiber]

stopped means that your distance isn't changing

aka the graph is horizontal there

that was all the way back here (the entire horizontal part should be highlighted, my trackpad just went kaput)

[vivid beacon]

but wasnt she at the hospital for 30 mins aswell

[cunning fiber]

mhm

if that's what you meant, you should've said something like "when she arrived at the hospital"

your original wording matches the wording of the question for something different

which would probably be marked incorrect

[cunning fiber]

is this what you meant though?

[vivid beacon]

yes thanks for telling tht sry

[cunning fiber]

ok

so she got to the hospital at 15:00

[vivid beacon]

yh

[cunning fiber]

and as you said

[vivid beacon]

i read the question

[cunning fiber]

she took 30 minutes there

so when did she actually leave the hospital

[cunning fiber]

good

[vivid beacon]

15:30

[cunning fiber]

now what happens after she leaves the hospital

[vivid beacon]

she goes home

[cunning fiber]

correct

and how long does it take for her to go home

or rather, what information do we have that can help us figure that out

[vivid beacon]

hmm

we have 32mph

we dont have the time

[cunning fiber]

mhm

we have a speed and want a time

[vivid beacon]

yh

[cunning fiber]

any formulas you can think of that involve speed and time?

[vivid beacon]

S=D/T

[cunning fiber]

perfect

we just don't have that distance

[vivid beacon]

yh

[cunning fiber]

well at least it's not stated in the question

but we can figure that out

I think we did that earlier

what's the distance

[vivid beacon]

20 miles

[cunning fiber]

no

that was the distance before she stopped for a break

your axes for the graph here is perfect

use them

[vivid beacon]

is 20 no?

[cunning fiber]

but she's not going home from where she took a break

she's going home from the hospital

[vivid beacon]

yh

[cunning fiber]

it never said that she was taking a break on the way home

look at the second section specifically

[vivid beacon]

speed of 32mph

[cunning fiber]

it tells us was at the hospital and then goes home

that implies that nothing else happens between those two things

[vivid beacon]

yh

[cunning fiber]

so ... how long was she traveling at the speed of 32 mph?

[vivid beacon]

30

[cunning fiber]

what

[vivid beacon]

im not sure

[cunning fiber]

we talked about this earlier

[cunning fiber]

^

[vivid beacon]

20 miles

• 20 miles

[cunning fiber]

yes

[vivid beacon]

there 2

hospital to break then break to home

[cunning fiber]

you could've also just read 40 miles off directly

[cunning fiber]

since you already did the work of labeling

[vivid beacon]

i dont have a ruler

[cunning fiber]

doesn't rlly matter tbh

[vivid beacon]

yes

[cunning fiber]

^

so can you figure out how long it takes to travel those 40 miles at 32 mph

[vivid beacon]

32=40/t

40/32?

[cunning fiber]

which is?

[vivid beacon]

5/4

[cunning fiber]

write it in hour:minute since all our times here are in that format

[vivid beacon]

60*5/4

what

am i doing

[cunning fiber]

that could work tbh

[vivid beacon]

sorry i hv been doong math for way too long now my brain hurts

[cunning fiber]

there's multiple ways to convert 5/4 hr to hr:min

you could convert it all to minutes and take out the hours

or you could write it as a mixed number and (kind of) read it off that way

[vivid beacon]

1 and 1/6

hours

[cunning fiber]

typo?

[vivid beacon]

this wht i mean

[cunning fiber]

your denominator changed from 4 in 5/4 to 6 in 1 1/6

[vivid beacon]

wht

[cunning fiber]

nvm

How did you get $\frac{5}{4}=1 \frac{1}{6}$

[elfin berry]

Civil Service Pigeon

[vivid beacon]

no i did

5/4*60

which is 70/60

[cunning fiber]

@dapper canopy I'm already helping, it's ok - I don't want a too many cooks in the kitchen situation since we're already doing quite well right now.

[vivid beacon]

aurelis pls dont interept there is too many chefs in the kitchen

[cunning fiber]

recheck that

[dapper canopy]

I am not talking, chill out

[cunning fiber]

remember that $\frac{5}{4} \times 60=\frac{5}{4} \times \frac{60}{1}=\frac{5 \times 60}{4 \times 1}$

[elfin berry]

Civil Service Pigeon

[vivid beacon]

oh is 75

75/60

[cunning fiber]

good

[vivid beacon]

1.25

[cunning fiber]

actually you could convert this to a mixed number

or you could say that 1.25 hours = 1 hour + 0.25 hours

they're functionally the same thing tbh

ig it depends of if you can convert instinctually or not in terms of marginal time saved

[cunning fiber]

with the non-whole number of hours (aka the 0.25 hours) being what you'd convert to minutes

[vivid beacon]

3:30+1.25 hours

[cunning fiber]

typo?

where is # coming from lol

[vivid beacon]

sry

yes

[cunning fiber]

all g

1.5 hours though?

you just found that it's 1.25 hours

[vivid beacon]

typo

[cunning fiber]

^

good

how do we add those times

[vivid beacon]

4:45

[cunning fiber]

good

actually

since the question is asked in 24 hour time

[vivid beacon]

16:25

[cunning fiber]

maybe I should've said leaving it in 24 hour time made more sense

[vivid beacon]

cook

[cunning fiber]

typo?

[vivid beacon]

yes

16:45

sry

[cunning fiber]

yup

[vivid beacon]

that it?

[cunning fiber]

say that's after 16:30 ig

[vivid beacon]

kk

you solved that so easily why cant i do that when do i unlock the power?

[cunning fiber]

I'm probably older than you

also I probably have a different

how do I word this

I have a different experience with math due to my interests

[vivid beacon]

yh

[cunning fiber]

I was interested in math so I pursued a lot more of that in my secondary school years

(i just started uni for context)

[vivid beacon]

nice

what country education?

[cunning fiber]

also there might have bit of "sink of swim" in my education lol

that made me adapt

[elfin cairn]

you have a lot to learn xD

[cunning fiber]

oh don't worry ik

[elfin cairn]

have fun!

[cunning fiber]

I got a 30/60 which I'm happy with

[vivid beacon]

what does that mean and you know better english then me i live in the uk

[cunning fiber]

sink or swim = figure out how to succeed or you fail

[elfin cairn]

you said "of"

[cunning fiber]

swim being succeed (since you make it through the metaphorical water, representing challenges)

sink = you die

oh

oops

Also I grew up in the united states lol

[vivid beacon]

this is too complicated ty for the help

hope to be gd at maths 1 day

.close

[marsh citrus]

Channel closed

Closed by @vivid beacon

Use .reopen if this was a mistake.

[cunning fiber]

[marsh citrus]

Available help channel!

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• Show your work, and if possible, explain where you are stuck.
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[prisma iris]

trying to use gauss jordan elimination but i keep getting fractions, got the last two wrong due to the same issue. what mistake am i making?

[prisma iris]

<@&286206848099549185>

[small berry]

!15m

[marsh citrus]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[small berry]

From line 2 to 3. R3 = 4R1 + R3. The -4 in the bottom right became -20. Should've been 12

[marsh citrus]

@prisma iris Has your question been resolved?

[prisma iris]

i did get a different answer with this correction, but it's still wrong

<@&286206848099549185>

[cunning fiber]

Oh

[cunning fiber]

Part 1

Part 2

Correct?

[prisma iris]

yes

[cunning fiber]

To get to the red row, were you doing 12(green row) + (blue row)?

[prisma iris]

no i did 4

so i could get a 0 in the blue row

[cunning fiber]

Oh so you’re writing the row operation for line n

In line n-1

ok

Sign error

[small berry]

honestly for these types of problems the best approach is to just toss your paper and restart from scratch

[prisma iris]

yeah probably but i've spent 4 hours on this assignment already and im not even halfway done

[small berry]

like you're just making us play where's waldo with your mistake... by the time we find it you could've probably re-done your problem

anways looks like pigeon got it

[prisma iris]

it's not as much about the exact problem but i've gotten the wrong answer everytime i've done gauss jordan so im making a consistent error and i can't find it

[cunning fiber]

Tried to do a binary search by rref at the start except it was like 1/6 of the way in

Then realised it was a multi post

[small berry]

quicksort the mistake

[cunning fiber]

basically

[cunning fiber]

@prisma iris hello?

[prisma iris]

i don't know im getting -20 now when i redo it

[cunning fiber]

no

What is 4(4)-4

[prisma iris]

okay

[cunning fiber]

You good now?

[prisma iris]

i'm redoing it from there right now, i think this is right

it was right

thank you for your help and sorry again for posting it twice

[marsh citrus]

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

[prisma iris]

.close

[marsh citrus]

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• Show your work, and if possible, explain where you are stuck.
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[glad bear]

Need help with question 20

[glad bear]

Help

[vale iris]

What are you thinking?

[glad bear]

Idk how to find the y intercept

[vale iris]

Okay

So the y intercept is the y value when x is 0

To find it, you can just substitute the value 0 for x

[glad bear]

Can u explain more

No we’re not doing that

We’re doing how to find the slope

Idk how to

[vale iris]

But you just asked about y intercept?

[glad bear]

Idk bro I’m failing math

I need help

Hello

[vale iris]

1 sec

Alright here's a drawing of a graph

This is the point where it crosses the y axis

So that's the y intercept

Make sense?

[glad bear]

Ohhh ok

So how would I find the slope of the equation or is it alr provided

[fresh oyster]

in general for anything of the form y=mx+b, m is the slope, b is the y-intercept

[vale iris]

The slope of the graph is how steeply it is angled upwards, if that makes sense

[glad bear]

Oh ok

[vale iris]

If the line goes up 2 for every 1 it goes right, then the slope is 2

That's the intuition you should have

[glad bear]

Ohh thanks I get it now

[vale iris]

And yeah if the equation is given to yuo then this is how you find it

But slope and y intercept are very different so don't get them confused

[glad bear]

OKK

[fresh oyster]

so, lets make sure
whats the slope and y-intercept?

[glad bear]

The slope is 2 and the y intercept is 4

[fresh oyster]

for 20?

[fresh oyster]

the slope isnt 2 and the y-intercept isnt 4

[vale iris]

For my example it is

[glad bear]

Oh mb

[fresh oyster]

y=(10/7) x?

[glad bear]

OKK

[fresh oyster]

wait, is this the one were talking about?

[glad bear]

Yes

[fresh oyster]

oh i understand, they were using an example

[glad bear]

I think the slope is 10x/7 but idk

[fresh oyster]

almost

x shouldnt be in the slope

or else its not linear :P

[glad bear]

Oh so this isn’t linear

[fresh oyster]

it is

so we must not report x in the slope

[glad bear]

So what would be the slope

[fresh oyster]

so, y=mx+b, we want m specifically

[glad bear]

Yes

[fresh oyster]

so we have y = (10/7) x + 0

so its just m = 10/7

[glad bear]

OHHH OKK THXXX

I GETVIT NOW

[fresh oyster]

make sure to close the channel if youre done

or ask more questions if not

[glad bear]

How do I do that

???

[fresh oyster]

.close

[marsh citrus]

Channel closed

Closed by @fresh oyster

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[marsh citrus]

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[buoyant jetty]

i need help with A

[static quarry]

could work directly with the definition

[buoyant jetty]

of directional derivative

[static quarry]

wait maybe i mistranslated

lemme read again for a moment

[wary lichen]

Im not precisely sure but i think that this is Directional Derivative Work
Basically:

∇f ⋅ v = Directional Derivative when v is a versor.

[wary lichen]

from there you basically get a function based on x and y, but we know that they are 0,0

[buoyant jetty]

we dont know v

[wary lichen]

Your function stops being from x,y of f(x,y) and becomes from (sin theta,cos theta)

[static quarry]

don't you already need to know that f is differentiable before you can use the gradient to compute directional derivatives?

[wary lichen]

you have to "check all" v

[buoyant jetty]

True

[static quarry]

and part (b) is where you show that it's differentiable

probably they want you to just compute the directional derivative directly

using the limit definition

[wary lichen]

I read the = and =/= in reverse 🥀

[buoyant jetty]

also this shit is a piecewise

[buoyant jetty]

for sure

[wary lichen]

But, yeah, anyways youll have to deal with a limit

[buoyant jetty]

anyone have the dir der as limit def

[static quarry]

probably want something like g(t) = f(tx, ty), where (x,y) is a fixed unit vector

then differentiate g with respect to t and evaluate the result at t=0

[buoyant jetty]

wdym

[static quarry]

yea, my (tx, ty) is your hv

and a = 0

[buoyant jetty]

i see

can i get some help?

v = (v1,v2)

[wary lichen]

Cant you just show that the upperpart lim(x,y)->(0,0) = 0?

[buoyant jetty]

that shows that f is continuous

i think

you guys want to do b)

?

i thought maybe starting with a)

[static quarry]

^ use this for (a), did you try it?

[buoyant jetty]

on it

[wary lichen]

-> By extension b)
and probably shows you can work with the partials

idk

[fresh oyster]

use $$\nabla_vf(\vec a)=\nabla f\cdot v$$
where $\nabla_v$ is the directional derivative for f in the direction of v

[static quarry]

your numerator is just f(hx, hy), because f(0,0) = 0

[elfin berry]

Cycadellic

[buoyant jetty]

we need f to be differentiable for that

[fresh oyster]

oop

right

[static quarry]

we don't know that f is differentiable yet (that's part b)

[buoyant jetty]

yeah

[static quarry]

you can't use the gradient to get the general directional derivative unless you know f is differentiable

for all we know it could have completely random derivatives in every direction

(alternatively you could do part (b) first and then use the gradient, your choice really)

[buoyant jetty]

i think doing a then b was the intended thingy

[static quarry]

yea i presume so

[buoyant jetty]

help dude

this limit is nasty

[static quarry]

well you can factor out h^2 in num and denom and cancel it

then you also know something about v1^2 + v2^2, because (v1,v2) is a unit vector

[buoyant jetty]

im hardstuck

[static quarry]

did you lose the 1/h in front?

[buoyant jetty]

my bad

[static quarry]

aside from that it looks ok so far

[still temple]

Yo guys do yall know lim?

In math

I get often confused

[amber birch]

as we said already, get your own channel

[still temple]

Like how bruh

[amber birch]

go to #help-30

[still temple]

Ok ty

[buoyant jetty]

Help

[static quarry]

yea this all looks fine

i think

you know one more thing

[buoyant jetty]

?

[static quarry]

use v1^2 = 1 - v2^2

to get it all in terms of v2

and then you just want to find what values of v2 make this equal to 3

[buoyant jetty]

how

this shit is a cubic

[static quarry]

yea cubics are nasty

i just wolfram alpha'd it

one real root

[buoyant jetty]

we can use rational root theorem

[static quarry]

sadly i don't think the root is rational

[buoyant jetty]

, w x^3 -x + 1 = 0

[elfin berry]

Results provided by WolframAlpha

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

Requested by renatoxbz

[buoyant jetty]

dude are you for real

[static quarry]

oh wait, i put x^3 - x^2 + 1 instead of x^3 - x + 1

[buoyant jetty]

still

this solution we are getting

is just nasty

[static quarry]

well that's not good, it has abs value bigger than 1

maybe an algebra mistake somewhere?

[buoyant jetty]

my friends were telling me there doesn't exist any v

[static quarry]

the only possible thing i can see is that we assumed y != 0 when we divided by it

but obviously if y is zero then the limit is zero

so that's not a solution either

maybe to confirm, assume f is differentiable (since you're gonna prove it anyway in b), find the gradient, and compute the directional derivative that way

see if you can verify there's no solution that way either

[buoyant jetty]

we need to

use the partial derivative as a limit aswell

[static quarry]

yea

just take (v1,v2) = (1,0) and (0,1)

in your formula

(this assumes the formula is right haha)

[buoyant jetty]

wdym?

what formula

[static quarry]

to find the partial derivatives you can use the formula you already have

[buoyant jetty]

here?

[buoyant jetty]

why

[static quarry]

well the partial derivative in the x direction is the same as the directional derivative in the (1,0) direction

and similarly for the other one

[buoyant jetty]

its equal to 0, so (1,0) and (0,1) for v1 and v2 doesn't work

in 3v2(1-v2^2) i mean

[buoyant jetty]

this is what my friend did for a)

[buoyant jetty]

it is right

[static quarry]

yea i get the same, 0 for both partial derivatives

but then how can it be differentiable?

if it is, then all directional derivatives would be zero

but we found that they are not in general

either we have a mistake somewhere or this question is just jacked up

[buoyant jetty]

it can be differentiable i don't see a problem with that?

[static quarry]

if it's differentiable then you can get the directional derivatives from the gradient

but your gradient is zero

and not all of the directional derivatives are zero

so something is wrong here

[buoyant jetty]

dude

when its close to zero we get the first case of the piecewise

what i mean is, when its close to zero but not zero we get the first case of the piecewise

[static quarry]

sure but we're calculating the derivatives at (0,0), not "close to 0"

[buoyant jetty]

say, this function is differentiable

at (0,0)

[static quarry]

how do you conclude this?

[buoyant jetty]

we haven't proved it yet

[static quarry]

i'm saying that based on what we found so far, it can't be differentiable at (0,0)

[buoyant jetty]

like, if it's differentiable at (0,0) then this limit is zero

[buoyant jetty]

i might be tripping

but why you say it's not dif at 0

because all the directional derivatives are 0?

[static quarry]

no,

[buoyant jetty]

exam is tomorrow dude I'm giga cooked

[static quarry]

because the directional derivatives are NOT all zero, but the partials are

if it were differentiable then the directional deriatives would all be of the form (f_x, f_y) . (u, v)

where f_x = f_y are the partials (zero)

so the dot product would be zero in any direction

but we found that the directional derivatives are not zero in general

[buoyant jetty]

how do you know the partials are zero

[static quarry]

we just found that no?

(v1,v2) = (1,0) gives you the x partial
(v1,v2) = (0,1) gives you the y partial

both result in 0

i call bs on this problem, unless there is an error in our work, but i have checked it twice and think it's ok

[buoyant jetty]

yeah

[buoyant jetty]

is from a 2024 exam dude

[static quarry]

harsh

do they have solutions for it?

[buoyant jetty]

i call bs too there must be a mistake

[buoyant jetty]

let me see

no there is no solution for this one specifically

i call bs aswell dude

[static quarry]

yea

[buoyant jetty]

that's a nice thing to know yeah because if you set v = (1,0) you get the first coordinate of the gradient

[static quarry]

yep

[buoyant jetty]

i see

[static quarry]

partials are just directional derivatives in the coordinate directions

[buoyant jetty]

giga cooked for tomorrow's exam, maybe it's just not differentiable?

[static quarry]

yea i think it is not

i wouldn't spend any more time on this tbh

i'd be pissed off if this was on an exam i had to take haha

although if instructors are decent they should give you extra credit if you find a mistake in the question

[buoyant jetty]

shit happens, maybe there is a mistake in the Q

[static quarry]

yea seems so

anyway gl on your exam, hope it goes well!

[buoyant jetty]

thanks bingo we will see

i appreciate the help

[static quarry]

sure, yw

[buoyant jetty]

. solve

.close

[marsh citrus]

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[marsh citrus]

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• Show your work, and if possible, explain where you are stuck.
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[paper stirrup]

How would you show that f(D)[e^kx V(x)]=e^kx f(D+k)[V(x)], where D is the differential operator?

I have proved it for f being a polynomial, but not sure how to extend it to rational functions

[wispy stump]

You know the exponential shift law?

[paper stirrup]

Nope. I didnt have a ‘rigorous’ definition of d operators other than it being the derivative

[tawdry rampart]

D + k [V(x)] = V`(x) + k•V(x)?

[paper stirrup]

Sure

That’s just straight from the definition of the d operator

How does that help?

[tawdry rampart]

I'm just confirming

f(D) = d/dx ?

That makes a lot of sense

[paper stirrup]

f(D) is any rational function in D

[stoic saddle]

D itself is d/dx

[wispy stump]

D itself is

[paper stirrup]

Does it help to show my proof for f a polynomial?

[stoic saddle]

sure

but you may probably want to write like, f(D) = q(D)^-1 p(D)

or something

or show us what your definition of f(D) is for a rational function f

[tawdry rampart]

$f(D)[e^{kx} V(x)] = \frac{d}{dx} e^{kx} V(x)$

[wispy stump]

Yeah assuming that q(D) is invertible on the function

[elfin berry]

Wumpus Man

[tawdry rampart]

Is this what you mean sunset?

[paper stirrup]

No

I’ll try to give an example

If f(D)=D+1/D

[wispy stump]

You need to conjugate

[paper stirrup]

Then finding f(D)[g(x)] means integrating g to get h, and then computing (D+1)g

[paper stirrup]

Could you elaborate

[tawdry rampart]

Ohkk

[paper stirrup]

I guess if f(D)=n(D)/d(D) then f(D)[q(x)] would involve solving the diff eq d(D)[y]=q(x) and then computing n(D)[y]

The particular integral of the diff eq to be precise i think

[wispy stump]

e^-kx q(D) e^kx = q(D+k)

Same for p(D)

p and q are polynomials

We need to assume that q(D) is invertible

On the function space

[paper stirrup]

Doesn’t tthat just equal q(k)?

Because q(D)e^kx=e^kx q(k)

[wispy stump]

q(D+k) is invertible too

But the first thing you gotta know is conjugation of D by multiplication of e^kx

[paper stirrup]

By invertible you mean the particular integral exists?

Whats conjugation?

This is all the definition ive got, and im not meant to try prove these but i wanted to

[wispy stump]

Pair of operations which are nearly identical

Could differ by signs

[paper stirrup]

Then i guess D^2(sinx) is conjugation?

I suppose it isn’t an operation

I got it i think. Your method of conjugation is more sophisticated than me expanding everything but in the end all i needed was to argue that i can extend my argument for polynomials to rational functions if the polynomial in the demoninator is invertible

Thanks

.close

[marsh citrus]

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[wispy stump]

Im sorry couldn’t explain it to you in detail cuz i went on a call

My bad

Lmk if you need the explanation

[paper stirrup]

On an intuitive level i understood now that I could treat q the same as p provided q(D+k) is invertible. I dont think i have necessarily the background to prove it formally so I’ll leave it at that. Thanks

[wispy stump]

Yeah also that way, it becomes easy

Also it should be an non zero invertible function

[paper stirrup]

Actually, I would want to understand how to actually define something like 1/(D^2+3). Thats because it’s the same(?) as solving the differential equation d^2y/dx^2+3y=whatever function you have, but theres usually infinite many solutions. In what way does the actual formal definition (which I dont know of) restrict the inverse of polynomials in D to spit out the particular integral?

That was the most stupid question ever on second thought

The infinitely many solutions came from solving D(something)=0

Forget it

[wispy stump]

No question is stupid and just wait for couple of mins, im solving a doubt

[wispy stump]

Yep

[wispy stump]

You still wanna understand this or you figured it out?

[paper stirrup]

A little bit, specifically how the particular integral is unique maybe

But i think that might be beyond my abilities

[blazing pulsar]

This isn't really my area of interest, but I'm not sure I understand why you would want to extend to the rationals as your next when you already have polynomials and can consider power series from there.

The way I would view an operator like 1/(D^2+3) is to manipulate it into a formal geometric series in D

[wispy stump]

Exactly

[paper stirrup]

Didn’t know you could do that

[wispy stump]

You can

[paper stirrup]

I always had it fixed that inverses mean solving differential equations and non inverses mean differentiating

But then i guess theres some way to prove that D behaves liek a real number or something then

[wispy stump]

No, there are always 2-3 methods but yeah upto you

Which one suits your interest

[marsh citrus]

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[void brook]

how do i start with this? i was thinking of using principle of mathematical induction but i am not sure if that would be a correct approach

[patent sandal]

hint: group the terms with lengths that are powers of two. like 1 + (1/2 + 1/3) + (1/4 + 1/5 + 1/6 + 1/7) + (1/8 ... 1/15) + (1/16... 1/2^n-1)

[indigo nest]

Induction would work in theory as well yes

Huh?

[devout mauve]

well technically you need induction but its such an obvious induction which can be done with ... so most people would skip that

[indigo nest]

H_j is the sum from 1/1 to 1/j

[noble geyser]

but i think we shouldn't do this

[indigo nest]

Why is the last term 2^n-1

[indigo nest]

True

[indigo nest]

The induction proof of this is actually easier I'd argue

But what Cherryman said works as well

[patent sandal]

@void brook so how would you go about it if you wanted to do an induction

[indigo nest]

I'll leave this to you then Cherry. Gotta get back to my work

[lucid zenith]

Because you're looking at H_(2^n)

[void brook]

hold on, i will write it down

is okay till here?

[patent sandal]

yes

continue

[void brook]

wait, the sum of the right has 2^k terms, each at least 1/2^(k+1)?

[patent sandal]

yes

[void brook]

okay i think i get it now

i will write it down

[void brook]

oh also, congratulations on getting the helpful role! (sorry for the ping)

[patent sandal]

wait

yeah

sorry

[void brook]

hm?

so, like this?

[patent sandal]

absolutely!

[void brook]

can we talk about this method, please? i did not get how this helps

[patent sandal]

right

basically its the intuition

if we do something similar to what you did

we get that H_2^n > 1 + (1/4 + 1/4) + (1/8 + 1/8 + 1/8 + 1/8) + (1/16..... + 1/16) ..... 1/2^n -1)

= 1 + 1/2 + 1/2 + .... + 1/2

with n many halfs

= 1 + n/2

so H_2^n > 1 + n/2

[void brook]

but it does not hold for every n, right? for instance, say, if we take n = 3?

H3 becomes 1 + 1/2 + 1/3 which is less than 2.5

[patent sandal]

oh sorry its supposed to be H_2^n

[void brook]

or, H_2^n >= 1 + n/2?

[patent sandal]

yeah what you just proved

[void brook]

all right, got it! thank you :)

[patent sandal]

also this has the interesting conclusion that the sum of reciprocals diverges to infinity

even if it does so very slowly

[void brook]

right, i see that

anyway, thanks again!

[patent sandal]

welcome

[void brook]

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[indigo nest]

. Thank you! And don't worry about pinging me lol. It's good to see you around again 🌺

[indigo nest]

. H_j goes up to 1/j though?

[silent ledge]

Find the area of the sector formed by the circumference of a circle with a radius of 12cm, and two lengths which intersect at the centre of the circle, whose lengths are 12cm each, and the angle between which is 120 deg.

[silent ledge]

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[finite forum]

i know how to work out FP:PE

[finite forum]

but idk how to prove that P lies on EF

maybe show that the vector FP is a scalar of CG?

[harsh scroll]

prove EP||AO and also EF||AO

[finite forum]

|| means parallel?

[harsh scroll]

That would make E,P,F collinear

[harsh scroll]

yeah

[finite forum]

yeah thats sort of what i thought abt

[harsh scroll]

This's euclid axiom

[finite forum]

okay thanks

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[fickle shell]

Discrete maths question

[fickle shell]

Is this a valid solution?

[stoic saddle]

do you think it's valid, or is there a point of doubt for you

[fickle shell]

The solution they gave was weird

They took x=1 and y =1/pi

[fickle shell]

So I'm wondering if there was a reason I couldn't take these more obvious numbers for x and y

I can't see how it would be invalid but I'm new to this topic

[stoic saddle]

ok so they took a different counterexample than yours

how bad could that be

sometimes theres multiple different counterexamples to disprove the same claim

[fickle shell]

So its correct, right?

My solution

[limber hearth]

Its been a long time odie

[fickle shell]

Yesssss haha I'm back!!

In uni this time with discrete maths

(Its hard)

[limber hearth]

I want you to trust what you did

Say to me its correct and i know it and go to the next problem

[fickle shell]

Hi chartbit!!! ❤️

[limber hearth]

Fancy but why not

[fickle shell]

Yesss its correct and I know it 🤩

Imma move on to the next

[limber hearth]

Wunderbar

[fickle shell]

Thanks everyone!! You're all still just as sweet as I remember you being

❤️

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[marsh citrus]

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[limber hearth]

Hf in university !

[marsh citrus]

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[summer escarp]

Hello! I'm struggling with this related rates question, but mostly the first part. I've found the Volume of the pool as a whole to be 144 m^3 but I'm having trouble finding the Volume of the water. My approach is to use Volume formulas of both and then put the volume of the water over the volume of the pool for the % but I'm struggling with finding the distance from the pool's wall to the point at which water stops

[glass silo]

You can make a similar triangle/solid, comparing the water to the whole pool, if that helps?

[summer escarp]

do you mind elaborating? sorry

[glass silo]

So, bearing in mind that the shallow end of the pool is 1m deep, we can like focus on everything "under" that, if you get what I mean?

[summer escarp]

right

[glass silo]

At which point, from the bottom of the deep end of the pool, you know that to the very bottom of the deep end is 2m, and across is 12m

But also with how much water you have, the water is 1m tall at the deep end, so you can use that to figure how far you go to the side closest to the shallow end?

[summer escarp]

so you're saying (please correct me if i'm wrong):

• the deep end is 2m under the shallow part
• we know that the water is 1m deep at the deeper end

[glass silo]

Yea give me like a few moments

[summer escarp]

allg

OH i think i see what you mean

find this triangle?

[glass silo]

Yep that's it saved me, I was struggling to get this drawn out

[summer escarp]

great, thank you so much! just found it to be 6

[glass silo]

Cool cool, from there getting the volume of water should be good, I hope?

[summer escarp]

yes, if you don't mind I'm just going to input what i think is the answer before i close the thing

[glass silo]

Yea sure thing

[summer escarp]

alr it was correct