help-33

[past estuary]

Sort of, it's saying it oscillating so much at x=0 that it it literally vertical there, meaning that it cannot be a function when x=0 meaning it cannot be defined at x=0

It's sort of bending the rules

I think this example works

Except that it doesn't

It says it is oscillating too much at 0 for it's to be continuous there, but for that to happen the function needs to overlap so much that it is not a function and not defined there, but earlier it said that the original function is defined at x=0

And it said it's differentiable

It seems like if the third statement is false it creates a contradiction for the first 2

So all three must be true

But common sense says pick choice B

Hehe

This is a weird question

[woeful root]

@past estuary

Could you ask it to graph the original fc?

The first one oscillates too

So its not continuous either

Idt continuity is dependent on oscillation

But ik finding the limits is

[woeful root]

I could be wrong tho

[marsh citrus]

@woeful root Has your question been resolved?

[marsh citrus]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[marsh citrus]

Please don't occupy multiple help channels.

[atomic quest]

is number 8 still a function if it doesn't cross the verical line, and one number 7 to find range does one only count the points or is it the lowest and higest point of the line and is it the same for domain

[still temple]

Yes it is

As long as there is only one value of the function for every x,it is a function

That came out wrong

[atomic quest]

so it is a fuction even if it doesn't cross the line??

[still temple]

Yes

[atomic quest]

but what if there is more that one line

[still temple]

Do it this way

Inorder form a curve to represent a function

There should be only a single value of y for an x

Like

This can't be a function

Because

[atomic quest]

yea but what if the lines aren't connected

[still temple]

There are two values of y for a single x

[still temple]

What lines ?
Do you mean the coordinate axis?

[atomic quest]

like this

[still temple]

Oh

[atomic quest]

would that still be function

[still temple]

Yep

[still temple]

This is the only condition for a curve to be a function

[atomic quest]

so if two of the line were to cross over the verical line would it still be fuction if they weren't connected

[still temple]

You mean this ?

[atomic quest]

yea

[still temple]

Well ,what do you think ?

[atomic quest]

would it still be considered function if they aren't connected??

becuase*

[still temple]

.

[atomic quest]

???

[still temple]

What do you think,
Does this represent a function or not?

[atomic quest]

i guess???

[still temple]

Guess what?

Yes or no?

[atomic quest]

yes

becuase they aren't connected

[still temple]

Now, what did I say the only condition was for a curve to represent a function?

[atomic quest]

for the same line not to cross over twice??

[still temple]

Yes

It's the ONLY condition

[still temple]

Now check it,
Is there a line which crosses the curve twice ?

[atomic quest]

there can't be more that one y value

[still temple]

Mhm

[still temple]

.

[atomic quest]

no it doesn't there is only one y value and two x vaule per line

so it would be function

[still temple]

🤨

Two x per line ?

Is your phone tilted by any chance?

[atomic quest]

one*???

[still temple]

Joke

[atomic quest]

lol

[still temple]

I can just keep on drawing the line

What do you conclude from this ?

[atomic quest]

i was thinking about the end of the line but it was still be considered one vaule

[still temple]

End of the line ?

[atomic quest]

where it started and ends

[still temple]

Ah about that

A line is infinite

A line segment is finite

[atomic quest]

wait what line segment

[still temple]

And I don't think you get the reason why are are drawing a line

We are*

Say we have a function f(x)

Okay?

[atomic quest]

ok

[still temple]

now it's graph looks like this

When we draw a vertical line

Say at x= 0

I.e the y axis

The point where the line intersects the curve,is the value of the function
In other words
f(0)= point of intersection of curve and vertical line passing through x=0

Understood?

[atomic quest]

as for what im understanding it would be considered function as long as the same like doesn't cross over the verical line twice

[still temple]

I'm explaining why we are doing this

[still temple]

.

[atomic quest]

im sorry i don't understand the f(0) what does that mean

does it mean it starts at 0

[still temple]

No!

f(x)=x^2

Do you understand this?

Like we are defining a function

[atomic quest]

kind of understand it

[still temple]

Try this
What is f(0) ?

[atomic quest]

f(x) is the same as x sqared

[still temple]

Yep,kinda kinda

[atomic quest]

so 0^

[still temple]

Yes

0^2

Which is just 0

[atomic quest]

so f(0)= 0

[still temple]

Yes

[atomic quest]

ok

[still temple]

The graph of f(x) looks like this

Now, here draw a vertical line passing through x=0

Where does it intersect the graph?

[still temple]

Imagine the curve touches the origin

[atomic quest]

the origin being the verical line

[still temple]

That statement is so wrong

[atomic quest]

or where the the y and x cross??

[still temple]

Origin is a point where the x and y axis meet

[still temple]

Yes

[atomic quest]

ok so it starts at 0

[still temple]

A line has no start nor an end

[atomic quest]

yea what i mean it that it cursve is touching 0

[still temple]

Yes

[still temple]

By 0, you mean origin

[atomic quest]

yea where the x and y meet

[still temple]

Yes

So

Man ,I forgot why I was explaining this

[atomic quest]

me too

[still temple]

Ah

[still temple]

Do you understand this now?

Note that this curve represents a different function

[atomic quest]

i think

[still temple]

Kinda kinda ?

Anyways

So let's see what happens if a vertical line crosses the curve twice ?

[atomic quest]

it wouldn't be a function

[still temple]

So here

[atomic quest]

it wouldn't pass the verical line test

[still temple]

Bro ,
You know the vertical line test ?!

[atomic quest]

yea i have been trying to explain

[still temple]

Bruh

What is it that you have a doubt with then?

[atomic quest]

it was that if there is two different lines and they both cross over the verical line how would i put it

if the two lines overlap would i put it in different () or would i just take the started of one and the end of the other line becuase they overlap

[still temple]

I lost you here

[still temple]

This is what you are talking about

[atomic quest]

wait let me draw it

[still temple]

Okay

[atomic quest]

let's say that im trying to find domain

how would i put it

if they overlap

[still temple]

Inorder to define the domain,
You have to first check whether it's a function or not

[still temple]

Is this a function?
Does it satisfy the vertical line test?

[atomic quest]

yes becuase the same line doesn't pass over twice

[still temple]

🤨

[atomic quest]

no

[still temple]

You sure ?

[still temple]

You circled it yourself
The region where the line passes the curve twice

[atomic quest]

so it doesn't matter if the lines are connected or not as long as there is two likes crossing over the verical line it wouldn't be function

[still temple]

Yes

[atomic quest]

lines* crossing over

[still temple]

Try watching a video on why we are doing this
I tried to explain but failed miserably

[atomic quest]

so that problem wouldn't be able to be solved

[still temple]

By problem you mean finding the domain,no

[atomic quest]

not at all i'm thankfull for your time and patience

[lofty gyro]

sorry to interrupt, it's been an interesting conversation and you both are great! 👍

One thing though, even if it's not a function (but if it's a relation) it has domain and range too

[still temple]

Fr

Didn't know that

Sorry for the wrong info

[lofty gyro]

no worries!

[atomic quest]

so i would be using Uinion

[still temple]

Can you continue @lofty gyro
I have a class to attend soon

[lofty gyro]

oh sure, i can, i have around 1.5 hours before i have to go to work

[still temple]

Thank you!

[atomic quest]

I am very thankfull for you trying to help me you are so kind no need to be sorry

have a good day and thank you sooo much

[still temple]

You too!

[lofty gyro]

hello there! @atomic quest
can you repost the question, it's been a very long talk and it's hard to scroll all the way back up

[atomic quest]

how would i put this

how would i find the domaind and range on a question like this

[lofty gyro]

to see, using vertical line test, we can see it is not a function

but if you got a question like this asking for domain and range, you'll have to check the valid x-values (domain) and outcomes y-values (range)

it doesn't matter if its overlapping or not, as along as the graph shows x-values that has their corresponding y-values, the x will be in the domain

[atomic quest]

for the range i think i just i would use U to show the difference

but what sould i do it it would laps for domain

would*

[lofty gyro]

let's say we have this

[atomic quest]

for range is would be (-1,0)

i think

[lofty gyro]

the domain will just be [-4,2]

and the range will be
[-2,-1]U[1,3]

[atomic quest]

ok

so we wouldn't add the overlapping like this (-2,3)

i would still use U

to show that there is more than one line

[lofty gyro]

well, if you wanna split into 2 curves, we still have
[-4,-1]U[-2,2] which is just equal to [-4,2]

[atomic quest]

so both are the same

[lofty gyro]

yep

[atomic quest]

ohhh ok i understand

[lofty gyro]

but that probably won't appear in the exams lol

[atomic quest]

so i just worried for nothing....

[lofty gyro]

however, for range, it may have something similar

[atomic quest]

for for range with there is a gap i would have to put it in different ( ) and use U in the middle of the parentheses

[lofty gyro]

yes

[atomic quest]

to show the difference

ok

[lofty gyro]

but sometimes it's again having overlapping y-values

and we don't need to double them in the intervals

[atomic quest]

so i would do the same and put it in one parentheses

so it would be (-infinity, infinity)

for the range and domain

ok i understand now thank you soo much

[lofty gyro]

cheers!

[atomic quest]

thank you soo much for you time your a live saver

you're*

[marsh citrus]

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

[lofty gyro]

[covert jungle]

Hi

Can I have some helps please

[marsh citrus]

@atomic quest Has your question been resolved?

Channel closed

Closed by @atomic quest

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[sinful ruin]

I have come across an interesting calculus puzzle: Can anyone give an example of a function whose derivative is defined at a point yet the derivative is noncontinuous.

[vernal forge]

take a noncontinous function and find its antiderivative

[sinful ruin]

Basically, an example of this would be: f'(x) = 5 (A derivative exists), however, the derivative of f is not continuous at x = 2.

[vernal forge]

that antiderivative will be what you are looking for

[devout mauve]

well you have to be careful. you have to pick a noncontinuous function without a jump discontinuity

[lucid zenith]

$$f(x)=\begin{cases}
x^2\sin\left(\frac1{x}\right)&\text{ when }x\ne 0\
0&\text{ when }x=0\end{cases}$$

ugh how do I align it properly

[devout mauve]

&

[elfin berry]

kheerii

[vernal forge]

true

[sinful ruin]

Is it usually a piecewise?

[lucid zenith]

there

[sinful ruin]

or can you get a plain function

[lucid zenith]

I mean..

[devout mauve]

well this one is barely piecewise

[lucid zenith]

yeah the extra branch here is just to make it continuous

at a single value

this is the classic example that i remembered but you can create arbitrarily many examples

[devout mauve]

you have to do some trickery with "dividing by 0" or absolute values or something somewhere

otherwise if you only use all the usual functions then the derivative will also be built from all those usual functions and therefore be continuous again

[sinful ruin]

Does this still involve the 'antiderivative' some guy mentioned earlier?

[devout mauve]

no

[lucid zenith]

this is an antiderivative of some function at all values of x, that's why it is differentiable

particularly it is the antiderivative of $$f'(x)=\begin{cases}
2x\sin\left(\frac1{x}\right)-\cos\left(\frac1{x}\right)&\text{ when }x\ne 0\
0&\text{ when }x=0\end{cases}$$

[elfin berry]

kheerii

[sinful ruin]

I lack too much knowledge for this

Thanks for the input

Ill try to figure out what all this sorcery means

[lucid zenith]

I just differentiated this piecewise

that's all

[sinful ruin]

oh

that makes sense

[devout mauve]

well for f'(0) you have to put in quite a bit more work

you actually have to compute the limit

and squeeze

[lucid zenith]

well, yeah

[red nimbus]

he did that in his head trivially

[devout mauve]

well I am just saying this sounds like if you took the pieces on their own and differentiated them as usual. which you cant do for the second piece

[sinful ruin]

Would something like this make sense?

[devout mauve]

well the derivative is continuous here

[sinful ruin]

True.

I forgot nondifferentiable != noncontinuous

[sinful ruin]

Would you agree with this claim?

[devout mauve]

complete bullshit

dont use ai

[sinful ruin]

makes sense.

[marsh citrus]

@sinful ruin Has your question been resolved?

[sinful ruin]

I think I got it finally

$$f'(x)=\begin{cases}
x^{2}+3x\
5x-5\
x^{2}+3x+\left(x-2\right)^{3}$$

[elfin berry]

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

[sinful ruin]

cooked

f(x) =\begin{cases}
x^{2} + 3x & \text{for } x < 2, \
5x - 5 & \text{for } x = 2, \
x^{2} + 3x + (x - 2)^{3} & \text{for } x > 2.
\end{cases}

[devout mauve]

thats not how derivatives of piecewise functions work

you cant just differentiate each piece individually

thats what I was talking about earlier

derivatives cant have jumps. if they do then at the point of the jump the original function is not differentiable

[sinful ruin]

Isnt the discontinuity in my function a removable discontinuity?

Since the left and right approaches the same value, but the value is something else.

[elfin berry]

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

[sinful ruin]

YES

[devout mauve]

I am just saying that if you get a jump then you need to have made a mistake cause derivatives cant have jumps

note that if I just replaced the 5x-5 by for example 7x-7, then suddenly you would claim that the derivative at 2 was 7

even tho the function didnt actually change

[sinful ruin]

Oh I see

[devout mauve]

that should tell you that you cant just write down some formula for a single point and then differentiate that formula at that point

[sinful ruin]

thank you

Ill try again

[devout mauve]

and the fact that the derivative need to be discontinuous while having no jumps should also tell you that you are looking for some weird thing

cause jumps are what we usually think of for discontinuities

[sinful ruin]

My math teacher is not a good person

I just started learning derivatives last week

[devout mauve]

well thats rip but nothing I can do about that. I would still suggest to try to understand the example mentioned earlier

there is the reason its the go-to example

[sinful ruin]

Apparently, oscillations dont count

[devout mauve]

wdym dont count

[sinful ruin]

I proposed that exact function

and he mentioned something about it being continuous still

[devout mauve]

ok so your teacher apparently doesnt know what they are doing

great

[sinful ruin]

it could be that I misunderstood

but he is a man who got accepted to Harvard and denied it

so I wouldnt jump into such conclusion

All I am sure about is that periodical functions do not count.

[devout mauve]

the function from earlier is not periodical

[sinful ruin]

oh

[sinful ruin]

Using the example, something like this then:

$$f(x)=\begin{cases}
x^2\sin\left(\frac1{x-2}\right)&\text{ for }x\ne 2, \
5x & \text{for }x=2.
\end{cases}$$

[elfin berry]

Artificial Stupidity

[devout mauve]

I mean you can just write 10 for x=2

amounts to the same thing

[sinful ruin]

ah ok

[devout mauve]

but that example isnt differentiable

[sinful ruin]

even if I change it to 10?

[devout mauve]

have you checked for the original example why it is differentiable?

with the limit def?

[sinful ruin]

The LHD and the RHD are equal.

and the value exists

at x = 0.

[devout mauve]

uhh

I am not sure you are talking about the correct limits

this sounds more like you checked for continuity

[sinful ruin]

this

given that I just pulled it from the internet

[devout mauve]

ok good

there is no actual reason to separate left and right limit here

you never used that h is positive

but ok whatever

now try that same approach with the other function you had

[sinful ruin]

alright

I just realized

I mistyped the function

$$f(x)=\begin{cases}
\left(x-2\right)^2\sin\left(\frac1{x-2}\right)+5&\text{ for }x\ne 2, \
5 & \text{for }x=2.
\end{cases}$$

[devout mauve]

make it 5 instead of 5x. otherwise its not continuous

[elfin berry]

Artificial Stupidity

[sinful ruin]

like that?

oh Ive been looking at it wrong

I am completely wrong

[sinful ruin]

if I understood correctly now, the antiderivative is simply the derivative of a function?

which means, there is a way to reverse engineer the derivative back into the original function

is that correct?

[devout mauve]

antiderivative means "undoing the derivative"

[sinful ruin]

oh

So if I undo the follow, I should finally get the correct function?

[devout mauve]

whats the +5 doing there

youll get the function from above

[sinful ruin]

Im trying to get f'(2) = 5

[devout mauve]

why

in your function up there you can have 5x instead of 5 if you want

[sinful ruin]

wouldn't that turn f'(2) into 0?

oh wait

right, limits

hold uon

isnt sin(1/x) undefined at x = 0?

oh but xsin(1/x) is 0

makes sense

[devout mauve]

thats why we do the whole piecewise thing

we never get sin(1/0)

0*undefined isnt 0

[sinful ruin]

I still dont really understand why

x^2 sin(1/x) behaves pretty much like xsin(1/x)

Cant you just use the Squeeze Theorem for that

which means the lim = 0, as x -> 0 either way

aaaaa, but right, plain x plugin is undefined

I keep forgetting

[sinful ruin]

.close

[marsh citrus]

Channel closed

Closed by @sinful ruin

Use .reopen if this was a mistake.

[sinful ruin]

.reopen

[marsh citrus]

✅

[sinful ruin]

Brother this is a headache

I learned the basics of this function

But this is not what I needed

the derivative is still continuous

The goal is to find a function whose derivative is defined at a point, to which said derivative is noncontinuous.

[devout mauve]

the derivative is not continuous

I dont know why you think that

[sinful ruin]

oh

where is it noncontinuous?

[devout mauve]

at 0

[sinful ruin]

apparently, it is though https://math.stackexchange.com/questions/232672/show-that-the-function-gx-x2-sin-frac1x-g0-0-is-everywhere

[devout mauve]

and?

[mellow crag]

that shows that g'(0) = 0?

[devout mauve]

g'(0) exists and is zero, sure

[sinful ruin]

Thats what Im wondering

[devout mauve]

but it doesnt show that lim x->0 g'(x)=g'(0)

[sinful ruin]

yes

[mellow crag]

this is what the derivative looks like for other values of x

[sinful ruin]

But how does that make the graph discontinuous?

[sinful ruin]

there are no jumps or breaks

so it is continuous

[mellow crag]

although it doesn't fit our usual idea of 'a function is discontinuous if it jumps from a to b'

it's still discontinuous

[sinful ruin]

oh

how so?

[mellow crag]

thats...not the definition of continuous

what is the definition of continuous?

[sinful ruin]

I am truly a confused man now

A continuous function, as its name suggests, is a function whose graph is continuous without any breaks or jumps. i.e., if we are able to draw the curve (graph) of a function without even lifting the pencil, then we say that the function is continuous.

[devout mauve]

the limit $\lim_{x\to 0} g'(x)$ does not exist

[elfin berry]

Denascite

[sinful ruin]

yes, the limit does not exist

[devout mauve]

no thats the intuitive def. that wont bring you far

[mellow crag]

ok so that's like an intuitive definition

[mellow crag]

but the intuitive definition fails for cases like these

[mellow crag]

^

[sinful ruin]

I think thats what my evil teacher was using when he gave us this challenge

[devout mauve]

a function $f$ is continuous at a point $c$ iff $\lim_{x\to c} f(x)=f(c)$

[elfin berry]

Denascite

[sinful ruin]

oh interesting

[devout mauve]

or we can also do the eps delta def but you will like that even less. and its less suitable here

[sinful ruin]

so you also have to be able to determine the limit.

[mellow crag]

yh

[mellow crag]

was this seriously the definition ur teacher gave lol?

[sinful ruin]

yea

[mellow crag]

bruh

[sinful ruin]

but then I think he was trolling me

because I recall him telling us the 'calculus' definition

and you in fact have to check the limits aswell

thank you guys

I think that finally concludes my journey

.close

[marsh citrus]

Channel closed

Closed by @sinful ruin

Use .reopen if this was a mistake.

[mellow crag]

nw!

[marsh citrus]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[hazy dragon]

can i get an example?

[hazy dragon]

.close

[marsh citrus]

Channel closed

Closed by @hazy dragon

Use .reopen if this was a mistake.

[devout mauve]

(x-2)^3

[marsh citrus]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[vocal cloud]

how do I even approach this

this is what I found:

P_300 = k * e^(-0.5/K_b*300)

P_310 = k * e^(-0.5/K_b*310)

then taking their difference

so basically taking their difference i get

(1-e^-10)*P_300

[tawdry rampart]

classic heisenberg asking a chem ques

is this thermo chemistry

[vocal cloud]

physics

thermal physics

[tawdry rampart]

ohh

lol

i dont know this sorry

i have only done first law of thermodynamics rn

[tawdry rampart]

maybe i can help

eV means electron volt?

[vocal cloud]

yea

[tawdry rampart]

P proportional to e^ -1/t

does E act change with T?

[vocal cloud]

nope, we can assume its constnat

[tawdry rampart]

yeah

maybe they want rate of change at T = 310

[vocal cloud]

.close

[marsh citrus]

Channel closed

Closed by @vocal cloud

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[frosty thicket]

i do not understand these truth values. i need someone to help explaining it to me

[quartz oracle]

Let $ P $ and $ Q $ be continuous and have continuous first partial derivatives in a region $ \mathcal{R} $. Let $ C $ be any simple closed curve in $ \mathcal{R} $ and suppose that for any such curve $\int_C Pdx + Qdy = 0,$

$(a)$ Prove that there exists an analytic function $ f(z) $ such that $$ \text{Re}{ f(z)dz } = Pdx + Qdy $$ is an exact differential.

$(b)$ Determine $ p $ and $ q $ in terms of $ P $ and $ Q $ such that $\text{Im}{ f(z)dz } = pdx + qdy $ and verify that

$$
\int_C pdx + qdy = 0.
$$

$(c)$ Discuss the connection between (a) and (b) and Cauchy's theorem.

[elfin berry]

Franklin244

Let $ P $ and $ Q $ be continuous and have continuous first partial derivatives in a region $ \mathcal{R} $. Let $ C $ be any simple closed curve in $ \mathcal{R} $ and suppose that for any such curve $\int_C Pdx + Qdy = 0,$

$(a)$ Prove that there exists an analytic function $ f(z) $ such that $$ \text{Re}\{ f(z)dz \} = Pdx + Qdy $$ is an exact differential.

$(b)$ Determine $ p $ and $ q $ in terms of $ P $ and $ Q $ such that $\text{Im}\{ f(z)dz \} = pdx + qdy $ and verify that 

$$
\int_C pdx + qdy = 0.
$$

$(c)$ Discuss the connection between (a) and (b) and Cauchy's theorem.

[marsh citrus]

@frosty thicket Has your question been resolved?

[marsh citrus]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[marsh citrus]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[neon tinsel]

what'd i do wrong

i found the integral of 8/t+1 then subbed in x^2

[mellow crag]

derivative :)

not integral

[neon tinsel]

find the derivative of the integral?

[mellow crag]

yeah

[neon tinsel]

how do i do that

like

integrate then the derivative of that?

[mellow crag]

well you could but the integral and derivatives are opposites

so 'a derivative cancels out an integral'

you have to do a bit more work here because we hvae x^2 rather than x

[neon tinsel]

so can i say d/dx integral 8/t+1 and thats just
d/dx 8/t+1

so -8/(t+1)^2

[mellow crag]

uhh no

i think i saw u posted a question here about integral and derivative cancelling each other out, did that ever get resolved?

[neon tinsel]

nope id be happy to go back to it

[mellow crag]

yeah i think it's best to start with understanding this first before u try to tackle problems

[mellow crag]

right so the statement of the theorem has a lot of jargon

you have all ur conditions etc. etc. etc.

don't worry about those for now

let's just say our function f is a 'nice' function

as in it's one of those functions u'll encounter in the wild, not some monstrosity that technically exists

well we can define a function g(x) by

$g(x) = \int_0^x f(t) dt$

i.e. the signed area below the curve

[elfin berry]

LY

[mellow crag]

so g(2) would be the area between x=0 and x=2

g(5) would be the area between x=0 and x=5

now let's say we wanted to differentiate this function g

now i have no idea what g(x) is, it's some weird function that tells me the area under a curve between 0 and x

but this theorem states that if i differentiate this function g

i end up with my original function f

this formalises what it means for 'the integral and derivative to be opposites'

[neon tinsel]

and your original function f is x^2 here right

[mellow crag]

well in the example image i gave, yeah

but the point is it doesn't matter what function f is

it could be f(x) = e^cosx ln(x) + arctan(x) or whatever

[mellow crag]

i have no idea what g(x) equals!

but i know that it's derivative is f(x)

there's an intuitive reason why it works out like this

[neon tinsel]

like we're given the answer now we have to find the question

[mellow crag]

so let's say we're at a and we wanna find the rate of change of g(x) at a

then the area effectively changes by f(a) as i've shown in this diagram

(the green bit)

hope that cleared some things up

[neon tinsel]

a is the red line on the right?

[mellow crag]

yeah forgot to label it

[neon tinsel]

so the area we're looking for is between a, a+delta

and we want a function to represent that area between a,a+delta?

[mellow crag]

as in loosely speaking, that's 'the rate of change' of g

(well we really should divide by delta x etc.)

[mellow crag]

so loosely speaking, the function changes at a rate of f(a) there

[neon tinsel]

so the rate that f(t) under a curve grows is described by that curve

i think i found a good definition to help

i think i understand whats going on here

first you evaluate the integral at [a,x]

then take the derivative of that

[marsh citrus]

@neon tinsel Has your question been resolved?

Channel closed

Closed by @neon tinsel

Use .reopen if this was a mistake.

[marsh citrus]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[humble river]

first off

its best that i should do

2.54 lb to g?

then get the volumee

by v=m/d

[grand radish]

yes convert

[humble river]

ok thnxs

[cold pecan]

If you have that many questions to ask you can ask in one channel, you have closed and opened a lot of channels for each question

[humble river]

Oh

ok

thanks

[marsh citrus]

@humble river Has your question been resolved?

[marsh citrus]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[marsh citrus]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[gusty kindle]

Hi guys I'm new

[wheat rover]

yeah have any questions?

[marsh citrus]

@gusty kindle Has your question been resolved?

[marsh citrus]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[marsh citrus]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[eternal fog]

I worked it out to be (4,6)

but what is the length of the line segment?

[main idol]

do you know distance formula

[eternal fog]

just minus smaller from larger?

[main idol]

and some worked examples here:
https://www.mathsisfun.com/algebra/distance-2-points.html

[eternal fog]

btw

in the answers

it just says the mid point

so are they the same or did they just not say

[main idol]

yea i don't see the length in the answer

so they just didn't say

[eternal fog]

thx for forumula

[eternal fog]

ah okay

.close

[marsh citrus]

Channel closed

Closed by @eternal fog

Use .reopen if this was a mistake.

[marsh citrus]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[terse skiff]

GIYS

[terse skiff]

Help

[limber hearth]

Post your problem

[terse skiff]

How do to question 6

How to do

Espically part i

[limber hearth]

Well, when you throw it, time = 0

[terse skiff]

Oh ok

Ohhh

I get it

So time is 0 for both equations

[limber hearth]

So what is the height ?

For each

[terse skiff]

First is 20m and second 50m

Ok now ima try draw the grpah

[limber hearth]

Yes

[terse skiff]

I did the graph look

In part iii wat da hell does it mean by numeric

[marsh citrus]

@terse skiff Has your question been resolved?

[marsh citrus]

@terse skiff Has your question been resolved?

[terse skiff]

<@&286206848099549185>

[marsh citrus]

@terse skiff Has your question been resolved?

[past estuary]

@woeful root Hello, I was reading through the messages and I think @sinful ruin may have found an answer, I think we were using the wrong definition of continuous also

Original Question:

#help-33 message

Final Point
#help-33 message

[marsh citrus]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[humble river]

idk why symbol looks confusing

[humble river]

But

Ik that it will be element F right

like im looking at the other symbols

what is that 2- and that 3+

like ik 109 is mass number

52 is proton

[thorny goblet]

Charge

[humble river]

oh

is that found in periodic itabl

[thorny goblet]

Typically not

[humble river]

ah

[still temple]

this aint maths?

[clever spade]

charge is protons-electrons

[humble river]

how to find charge

oh

perfect ok

so it wil always be 0 ??

da heck

oh in neutrals

ok

got it

thxn

[clever spade]

that's for neutral yeah

[mellow crag]

charge isn't something that's 'inherent' to an atom

an atom is neutrally charged

and then you can have different ions of that atom when you add/remove electrons

[humble river]

ok

thnx

.close

[marsh citrus]

Channel closed

Closed by @humble river

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[stark glacier]

This might be physics heavy, maybe someone is familiar? lol. Sooo..

Bungee Jumper has a mass of 50kg. And will jump from a platform 30m above the ground. The bungee cord is 6.0m long when no forces is applied to it, and has a spring constant of 150N/m.

a) How far below the platform is the speed greatest?
(Actual answer: 9.3m)

b) How far below the platform does the jumper get before she is pulled up again?
(Actual answer: 16m)

c) What is the acceleration of the jumper? (Use the max displacement from question b)
(Actual answer: 21m/s^2)

[stark glacier]

I finished a and b. I'm stuck at c. I need to find the acceleration of the object at its maximum displacement from the platform.

Equilibrium position: 9.27m below platform.

Total Energy in the system at 9.27m = 3210.32J

Maximum displacement from equilbrium position: 6.5425m.

So max displacement down is 15.8125m below platform.

What to do.

At 15.8125m, I tried to to plug in "ma=kx" into "Ep=0.5*kx^2" to get 0.5max, but a=19.6275ms^-2, not 21.. badbad.
I tried a=kx/m, got 19.63ms^-2 too.
I tried (kx+mg)/m, this one I got 29.44ms^-2 on,

I'm stuck

and for "x", i used "6.5425" bcuz thats max displacement

im actually stuck, need to sleep. but plz try if u are familiar with physics.

[marsh citrus]

@stark glacier Has your question been resolved?

[marsh citrus]

@stark glacier Has your question been resolved?

[near pecan]

I'm looking at this.

(Not a physics expert; and I need to re-read the question!)

[marsh citrus]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[dreamy lintel]

Need help working through this problem

[near pecan]

OK. I think if the max displacement is 16m and the equilibrium is 9m, the jumper is 7m below the equilibrium. Hope that helps. if you need something more, feel free to post again. Also a = (kx)/m looks right. Just make sure you're using the right values.

Hi Egg

[dreamy lintel]

Hi

[near pecan]

Gimme a sec to look this over

OK. Which part are you stuck on?

[dreamy lintel]

all of it lol

[near pecan]

OK. Looks like you have a line in $\mathbb{R}^2$ parameterized as a vector equation.

[elfin berry]

Stefano

[near pecan]

How familiar are you with vectors in the real plane?

[dreamy lintel]

very little

[near pecan]

Do you know how to graph $\vec{c}$?

[elfin berry]

Stefano

[dreamy lintel]

its just a vector going through 1,3 right

[near pecan]

It might be better to think of it as going to (1,3)

[dreamy lintel]

gotcha

[near pecan]

What do you think of when you think of a vector in the real plane?

(I just want to clear some things up so we can make sure you're able to answer questions like this in the future.)

[dreamy lintel]

a line with direction and magnitude

[near pecan]

Sure. So, it doesn't have to end at (1,3), but by convention we start at the origin.

Does that give you enough information to try to graph c?

[dreamy lintel]

right

yes

[near pecan]

Nice!

Do you find analogies helpful?

[dreamy lintel]

yes

I do

Im having trouble graphing K and the projkc

[near pecan]

Cool, a vector in space can be thought of as a relative position. Let's say you're playing a video game, like a first person shooter and your team mate is in front of you and a little to your right. If the two of you keep that formation as you look around, you can describe their position with the same vector

They're always 1m to the right and 3m forward from you

Even if you move around

[dreamy lintel]

makes sense

a parallel vector

translated 1,3

?

[near pecan]

Even better. The same vector.

[dreamy lintel]

the same vector in two places at once?

[near pecan]

No.

Your teammate is 1m to the right and 3m forward from you and you're on the second floor of a building. (Maybe looking for zombies, gold, enemies, whatever!)

Then you two go outside, but keep the same formation. Where are they relative to you?

[dreamy lintel]

1,3?

[near pecan]

Of course. You're well trained and keep a tight formation. That relative position from you and to your team mate is a vector

[dreamy lintel]

I see

I understand

[near pecan]

It's the same vector as long as you don't break formation

OK. Now, where's the one place neither you nor your teammate should aim your guns? (Sorry if you're not into shooting games)

[dreamy lintel]

towards eachother

[near pecan]

Exactly

And if you have a third teammate who's another metre to the right of the second one and 3 metres farther to the front of the second one, are they still in that direction you don't want to aim?

[dreamy lintel]

yes

[near pecan]

You could have a fourth, fifth, sixth... etc you could have teammates all along that same line of fire

That's a vector equation for a line

We start with your coordinates and then we say whoever is in this direction ahead or behind is on the same line of fire

[dreamy lintel]

thats vector form?

[near pecan]

Exactly! In our example, we might say your position is (5,2) and your teammate one's relative position is (1,3) ahead of you. Teammate two is (1,3) ahead of teammate one.

[dreamy lintel]

I see

[near pecan]

Do you know what the vector from you to teammate two is?

[dreamy lintel]

its the one shown in the first sentence right?

the x=[1,2]+[1,0]?

[near pecan]

No

First teammate is 1 metre to the right and 3 metres ahead, second teammate is another one metre to the right and another 3 metres ahead for a total of?

[dreamy lintel]

2,6 meters?

[near pecan]

Perfect! 2 metres to the right and 6 metres ahead

[dreamy lintel]

yes

[near pecan]

Notice that 2*(1,3) = (2,6)

[dreamy lintel]

yes

[near pecan]

If there was a teammate a metre to your left and 3 metres behind you, they'd be at (-1,-3) relative to you

[dreamy lintel]

right

[near pecan]

If there was another teammate ahead of the _third) and they were also spaced (1,3) from the second, they'd be at 3*(1,3) relative to you (3,9)

[dreamy lintel]

yes

[near pecan]

So, starting at your position we could say anyone from your position plus or minus any fraction of (1,3) is on the firing line

[dreamy lintel]

yes

[near pecan]

$L = t \begin{bmatrix} 1 \ 3 \ \end{bmatrix} + \begin{bmatrix} 5 \ 2 \ \end{bmatrix}$

[elfin berry]

Stefano

[near pecan]

OK. So that's the firing line. Starting at (5,2) and going any multiple of (1,3) away from there.

[dreamy lintel]

how do we know 5 2 is the starting line?

is that just given?

[near pecan]

Exactly! In our example, we might say your position is (5,2) and your teammate one's relative position is (1,3) ahead of you. Teammate two is (1,3) ahead of teammate one.

I just made up that example

And technically, it's a starting point

But let's look at the example $\vec{x}$

[dreamy lintel]

gotcha

[elfin berry]

Stefano

[dreamy lintel]

that makes sense

[near pecan]

Can you identify the starting point? Can you identify the vector that defines the line?

(brb)

[dreamy lintel]

for the problem I posted?

[near pecan]

Yes

[dreamy lintel]

It would be 1,0

[near pecan]

There are two questions there. Can you try to answer both? And be specific about which answers go to which.

[dreamy lintel]

the 1,2 defines the direction and the 1,0 defines the starting point

[near pecan]

Nice!

You think you can graph that?

[dreamy lintel]

yes

[near pecan]

Good

OK. Let's talk projections

[dreamy lintel]

sure

[near pecan]

Do you have any thoughts about what $\text{proj}_{\mathbf{k}}\vec{c}$ might mean?

[elfin berry]

Stefano

[dreamy lintel]

no idea

something to do with the relation of c and K?

[near pecan]

OK. It's read as the vector projection of 'c' onto 'k'

Remember how we were talking about the vector from you to your teammate being the same (if you maintain formation) regardless of where you are?

[dreamy lintel]

yes

[near pecan]

And you were able to draw the vector c, right?

[dreamy lintel]

yes

[near pecan]

Nice! So "move" c (you can draw it again) so it's on the line K

[dreamy lintel]

but arent they different directions?

1,3 and 1,2

[near pecan]

Good point.

We want c to start at any point on k

Recall, we used the convention of starting c at the origin, but that's not how we define the vector. Just like you and your teammate moving around. (In fact any teammates where one is 1m to the R and 3m in front have the same vector as you!)

[dreamy lintel]

ok so how do we start c on any point of k

[near pecan]

1. Pick any point on K

2. Pretend that point is the origin and count out x and y coordinates of c

3. Draw an arrow from the point on K you picked to that endpoint you found

[dreamy lintel]

so when sketching projkc I just draw any line with c direction starting anywhere on k?

[near pecan]

That's the first part. We're finding the projection; we haven't drawn it yet

But yes. The vector C is always the vector C

Just like 1m to the right and 3m ahead doesn't change if you start in Sydney, Australia or Lagos, Nigeria

[dreamy lintel]

gotcha

[near pecan]

How many units to the right or left of that point on K and how many units up or down from that point on K will you go to find the endpoint of C?

[dreamy lintel]

-2,-2

[near pecan]

Why?

[dreamy lintel]

I chose the point 2,2 on K

and had to go down and left 2 to reach the origin for c

[near pecan]

No

[dreamy lintel]

oh

[near pecan]

We move C so that it starts at that point

What is C?

[dreamy lintel]

1,3?

[near pecan]

Right

[dreamy lintel]

so down 1 and right 1?

1,-1

[near pecan]

Hold on

If you're teammate is 1m to the right and 3m ahead. And that vector always stays the same regardless of where in the universe it is, how far to the right is your teammate and how far ahead?

[dreamy lintel]

1,3

[near pecan]

OK, but what if you stay in formation and go to Tokyo?

[dreamy lintel]

still 1,3?

[near pecan]

OK, but what about in London?

[dreamy lintel]

same

[near pecan]

What about New Mexico?

[dreamy lintel]

same

[near pecan]

So you're telling me that we define the vector by its relative coordinates regardless of where it is?

[dreamy lintel]

yes

[near pecan]

OK, just lemme make sure. What about if you stay in formation but go to Sao Paulo?

[dreamy lintel]

same

1,3

[near pecan]

OK, so an ant walking from your feet could walk 1m to the right and 3m ahead to get to your teammate, any where in any possible universe or multiverse as long as you stayed in formation?

[dreamy lintel]

yes

[near pecan]

Even if you were standing on line K?

[dreamy lintel]

yes

[near pecan]

That vector from you to your teammate is (1,3) right?

[dreamy lintel]

yes

[near pecan]

And C is (1,3) right?

[dreamy lintel]

yes

[near pecan]

So if you're on line K, say at (2,2), where does the ant go to get from you to your teammate?

[dreamy lintel]

3,5?

[near pecan]

Yes!

Because it's 1 to the right and 3 up!

[dreamy lintel]

yes

[near pecan]

in any possible universe or multiverse!

Awesome

[dreamy lintel]

yes

[near pecan]

Imagine the sun is shining right above K. What shadow would C make on K?

(To be precise, when I say above, I mean as if the sun's rays make an upside down T when the hit K)

[dreamy lintel]

a 1,3 shadow?

[near pecan]

Not quite

Notice that the shadow might not be the same length as the original vector

[dreamy lintel]

im a little lost

im not seeing the shadow

[near pecan]

OK

Do these images make sense?

[dreamy lintel]

i see

[near pecan]

In the last image in particular, can you see that the shadow (in grey) is shorter than the vector (in yellow)

(the line in blue would correspond to K)

[dreamy lintel]

it looks the same length as the yellow vector

[near pecan]

It's a little shorter. If you use a compass, it might be easier to see

Hold on

Can you see the other yellow vectors now? They're all the same length

[dreamy lintel]

yes i see them

[near pecan]

Do you see that they are longer than the grey one?

[dreamy lintel]

yes

[near pecan]

Here's another with the vectors and lines labelled

[dreamy lintel]

i see

so the shadow is it sketched?

[near pecan]

I'm don't understand.

Are you asking if it was sketched in the image above?

[dreamy lintel]

yes

[near pecan]

I didn't sketch it, but I suppose someone did

[dreamy lintel]

i mean thats it conceptually though right?

[near pecan]

That's literally the projection

[dreamy lintel]

gotcha

[near pecan]

The vector projection is that vector, in this case that grey arrow, but the grey arrow is really defined by a start and end that stay in formation.

You could move it wherever

The scalar projection is the length of that grey arrow

[dreamy lintel]

i see

so i just put c on k and shade the projection?

[near pecan]

Exactly!

But remember the direction

[dreamy lintel]

right

[near pecan]

If the yellow vector, in the above example pointed towards the centre of the circle, the grey vector would reverse as well

[dreamy lintel]

Like this?

[near pecan]

That's pretty good. Remember that K is that line of fire we were talking about. It goes on forever in both directions

[dreamy lintel]

right

[near pecan]

And you should be able to draw a perpendicular line from your vector c to the tip of the projection on k

[near pecan]

(a line perpendicular to K)

[dreamy lintel]

gotchaok nice

i understand 13.1

what does 13.2 mean?

[near pecan]

Hold on, just want to check with you

[dreamy lintel]

ok

[near pecan]

Just a quick check. Which projection is correct?

[dreamy lintel]

doesnt that depend on the direction

[near pecan]

The direction of what in this picture?

[dreamy lintel]

both k and v

[near pecan]

Good question.

K is a line. It has no start and no end. Does it have a direction?

[dreamy lintel]

yes

[near pecan]

How do you define its direction?

[dreamy lintel]

x and y

[near pecan]

More precisely, how would you use x and y?

[dreamy lintel]

for example 1,2 is the direction of k in the form x,y

[near pecan]

Hmm. I see.

Not quite

Let's say that in this photo, K is flat along the x axis, ok?

[dreamy lintel]

yep

[near pecan]

(Just for now)

So, what's its direction?

[dreamy lintel]

1,0

or x=1

[near pecan]

So, K goes one step to the right?

It looks a bit longer than that

[dreamy lintel]

infinitely in both directions

[near pecan]

both directions

Does it have a direction

[dreamy lintel]

yes

[near pecan]

More precisely, does it have only one direction?

[dreamy lintel]

no

[near pecan]

That's it

[dreamy lintel]

but we describe it as if it does right?

like with k its 1,2

[near pecan]

No

[dreamy lintel]

despite being infinite?

[near pecan]

Because it's infinite

A direction means to and from

[dreamy lintel]

but then why was k in the problem make k 1,2

[near pecan]

There is no "from" with (this type) of infinity

but then why was k in the problem make k 1,2
To position the line

[dreamy lintel]

but isnt that the direction in the vector form we were given?

[near pecan]

No the line "goes in both directions". Unlike a vector, it does not have (only) a single direction.

[dreamy lintel]

but its described as xvector=[1,2]+[1,0}

[near pecan]

In fact, from your teammate's perspective (-1,-2) is the vector used to describe the line

t[1,2]

that t could be negative

To your teammate you are behind (negative ahead) and to the left (negative right)

Literally the opposite direction, but the same line

[dreamy lintel]

right but

[dreamy lintel]

why is it described like this then?

[near pecan]

So if we go back to the first person shooter example:
Your teammate is 1m to your right and 3m ahead (then you are -1m to the right and -3m ahead of your teammate)
But

Your opponents are also in formation their formation is one is -2m to the right and 5m ahead of the other. They will be on a different line.

Both lines go on infinitely. They are not the same lines

Each line is parallel to a vector (1,3) in your case and (-2,5) in your opponent's case

So the vectors have specific directions. They are oriented with from and to. We can define them with a start and a stop (regardless of where we put them)

The line has no start or stop

But those lines are still defined right?

[dreamy lintel]

I am lost now

[near pecan]

Forget all the math and just picture 4 soldiers (we do have to imagine that bullets go on forever), two are in a certain formation along a line, two are in another formation along another line

[dreamy lintel]

ok

I mean I understand its infinite in both directions

[near pecan]

Hold on

"in both directions"

[dreamy lintel]

I just dont understand why we are stating this now after we just described and sketched the line k with a direction

[near pecan]

I realize this may seem overly precise, but we're differentiating between one direction:

And two directions

If I tell you to go in one direction, you know where to go, right?

[dreamy lintel]

yes

[near pecan]

If I tell you to go in two directions, where do you go?

[dreamy lintel]

two directions

[near pecan]

You can do that?

[dreamy lintel]

isnt that what you've been telling me?

[near pecan]

Not at all.

You can, right now, walk in two different directions?

[dreamy lintel]

I feel like at this point I'm becoming more confused

yes

[near pecan]

I am confused.

[dreamy lintel]

how so

[near pecan]

You can get up, maybe divide your body in two, and walk to the left and to the right at the same time?

[dreamy lintel]

no

but you can travel in both directions

[near pecan]

Ah. I think I see the confusion

You mean, I can travel, for example, both north and west, right?

(northwest)

[dreamy lintel]

whats confusing me is that we used 1,2 to draw the line k but you said its not the direction of the line

[dreamy lintel]

yes but in the context of the line north and south makes more sense

[near pecan]

Right, you cannot go north and south at the same time

Does that sound reasonable?

[dreamy lintel]

yes

[near pecan]

Good. That's what a line is like, it's like a highway that runs along the north south directions (and again, note in the interest of precision, it's "directions" plural, not "direction" singular). You can go one direction or another on the highway, but not both at the same time. You can do that because you start somewhere

The highway has no start.

[dreamy lintel]

right

but

[near pecan]

So, why do we use north?

Well, we could have used south

[dreamy lintel]

that doesnt make the vector form irrelevant tho

[near pecan]

But we cannot use east

that doesnt make the vector form irrelevant tho

Not at all

[dreamy lintel]

right so isnt this all semantics though?

like

[near pecan]

Yes

Absolutely

[dreamy lintel]

ok

so

whats been confusing me

[near pecan]

Please, go ahead....

[dreamy lintel]

is that we are talking semantics like what we did was wrong

because you can solve it by describing the line by the vector form in the direction 1,2 and just extend in both directions right?

[near pecan]

Yes

[dreamy lintel]

ok so thats all we really need to know right?

sorry if that comes off as rude I enjoy the theory but

[near pecan]

Yes. In a sense you're completely right. Your instincts are spot on

[dreamy lintel]

gotcha

[near pecan]

It's OK. Semantics can really help hone those instincts

[dreamy lintel]

again not trying to be rude just Ive been working on this first part of the first problem for the past 4 hours lol

and this is due tomorrow

[near pecan]

Your drawing had an inccorect angle