help-28

[spice knot]

additionally, that does not at all address what I said

I provided an example f(x) and you made the same mistake within that example as well

[shy pilot]

there was only an A, maybe a B, in that sentence 😭

[spice knot]

the second derivative is positive, so its increasing when x =/= 3?
A B C

[shy pilot]

my guy

[spice knot]

it doesnt even matter because you then made the same mistake anyway

[shy pilot]

why are you convoluting a simple statement

????

[spice knot]

you dont seem to tell the difference

let me try again

[shy pilot]

?

[spice knot]

we expect that f'' > 0 means f' is increasing too

now to give an example of this, lets take a look at this separate function

[shy pilot]

ok

[spice knot]

now look at the left half of this parabola

is f'(x) increasing?

[shy pilot]

if f(x) = x^2, then f'(x) = 2x so yes

[spice knot]

well we're not going to be able to create an algebraic version of the function here

its a lot easier to just draw an example

also, f"(x) does have a visual example

there is a visual way, which we're going to prove is that concavity youre looking at

you looked at the left half of this parabola

[shy pilot]

it is positive everywhere where x =/= 3

[spice knot]

well I dont know how but youve managed to forget your own mistakes

Im going to pretend that by forgetting them, youre not going to make them anymore

[shy pilot]

i hope i dont 😭

[spice knot]

heres another example

now looking at this picture, can you tell when f''(x) is positive or is negative?

[shy pilot]

f(x) = x^3
f'(x) = 3x^2
f''(x) = 6x
so f''(x) is positive everywhere?

[spice knot]

❌

is that your process now?

just guess at f(x)?

[shy pilot]

no?

[spice knot]

look at what youre doing here

where did you get your f(x) from?

[shy pilot]

gimme a sec

[spice knot]

Im going to add you didnt give yourself that "sec" when you went out and said f''(x) is positive earlier

[spice knot]

and by earlier, hopefully you can intuit that I mean 1 minute ago

you almost immediately said that f(x) = x^3, which Im going to add isnt correct

[shy pilot]

sorry, f(x) = x^3 - 1?

[spice knot]

thats not correct either

[nimble ridge]

if you're looking for help put this in another help channel

[spice knot]

let me repeat

are you guessing?

[wary bridge]

where im new srry

[nimble ridge]

#❓how-to-get-help

[spice knot]

just trying out various cubics then seeing me say "no" to them is the proper definition of guessing

is it, or is it not, guessing?

[shy pilot]

i said x^3 cause it has the same "direction" at the ends

and since its not a negative cubic i assumed x^3 would be appropriate to find f''(x) to be a positive linear function

[spice knot]

f''(x) isnt even positive everywhere Id like to add

how about this

[shy pilot]

oh i see

its positive above the x axis and negative below the x axis

[spice knot]

please stop talking, I need to tell you something and I cant afford to have you try and interject

from my current point of view, you dont know what f'' > 0 exactly looks like

you first appealed to something you learned, which didnt work
because I showed you a picture of a concave up function

then you say this

there is simply no other way to twist those words other than:
where f(x) is incresaing, the derivative f'(x) is also incresaing
where f(x) is decreasing, the derivative f'(x) is also decreasing

the derivative f'(x) is also decreasing

then you insisted on this

"the slope of the derivative f'(x)" means f''(x)

[shy pilot]

.

[spice knot]

yea

youre supposed to notice the derivative at each point

is increasnig everywhere

it was a concave up function

[shy pilot]

and i told you, the derivative, at the points, where the function is increasing, and where the function is decreasing

[spice knot]

instead youre saying the derivative at each point is decreasing on the left half

[shy pilot]

i guess i was thinking about the tangent lines at those points

[spice knot]

were you?

[frosty gyro]

Is he confused about the derivative?

[spice knot]

I think he's confused that he even made a mistake in the first place

I dont think he remembers his own words

I cant exactly go forward if my man drops a freudian slip of a mistake in the middle then denies it

Im going to reiterate this one more time for the record

[shy pilot]

i was just answering YOUR quesion

[shy pilot]

that you asked at THAT point IN time

[spice knot]

you keep conflating f(x) decreasing with f'(x) decreasing

"they are the same"???

[spice knot]

and any time I get you to visually look for concavity, such as here, you dodge it and try to do it algebraically

[shy pilot]

you asked me what the derivative at each point of a parabola is
and i am basing my answers off of what you showed and told me

[spice knot]

so tell me

whats the derivative at each point you see

[frosty gyro]

Ooh does he think because f'(x) decreases, f(x) also has to decrease?

[spice knot]

I dont think so

but with the way he's saying it, I cant confirm it

[shy pilot]

i am basing my answer off of what they said here

[spice knot]

Im going to ask

the derivative increasing,

can you see that from the points?

[shy pilot]

that the derivative at one point IS NOT the same as it is at any other point

[spice knot]

or do you instead mean that the derivative is just negative?

[shy pilot]

the derivative of a postiva prabola is increasing

the derivative of the points on a parabola are all different

you even gave me this and said it

.

[spice knot]

❌

the derivative isnt decreasing in the red regions

its negative in the red regions

negative slope for decreasing

now compare that to the wording you used here:

[shy pilot]

so a derivative of a slope can be negative but still be increasing?

[frosty gyro]

Yes it can

[spice knot]

yes it can

[spice knot]

I showed you a picture of one

[frosty gyro]

It can go from -3 to -1 right?

[spice knot]

were you really looking at the derivatives here?

[frosty gyro]

Its still negative even though the graidnet is increasing

[spice knot]

at each point, the slope of the function is steadily rising

[shy pilot]

well then my bad for misinterpreting

[frosty gyro]

Oops

[spice knot]

blackslake

Im going to ask you to read what I said before

#help-28 message

then you tell me

but before that, you gotta read all of it

every single word, alr?

from there all the way to the present

Im gonna wait for around 10 minutes

[shy pilot]

im saying that because when i use the second derivative test to find concavity, if f''(x) > 0 its increasing or concave up, if f''(x) < 0 its decreasing or concave down. thats it

[spice knot]

"its"?

go read the words bro

actually go back and read them

[shy pilot]

its is referring to the second derivative

[spice knot]

the second derivative is increasing or concave up?

[shy pilot]

both

since its postivie

[spice knot]

thats not correct

[shy pilot]

f''(x) > 0 for x =/= 3

[spice knot]

youre saying f''(x) is increasing now

dont you mean f(x) or f'(x) instead is increasing?

you need to mean what you say

[shy pilot]

because, its positive

[spice knot]

what if f''(x) is decreasing?

it can still be positive

for example what if its 1, 1/2, 1/3, 1/4, 1/5, etc?

[shy pilot]

well then im wrong and please explain

[spice knot]

go click the link

[shy pilot]

i really dont have time to be going in circles

i have an exam tommorow that i have to get at least a 90% to pass the class

[spice knot]

you made me go in circles to explain a simple concept to you

[shy pilot]

and i have alot more things to study

[spice knot]

then when you finally figure out what Im asking you, you just go "oh okay"

[shy pilot]

so im sorry if im making a bunch of mistakes and not realzing them

[spice knot]

are you going to uh make up for that?

by reading what I was actually talking about?

because presumably now that you know what I mean, you can go back and see what I meant

[shy pilot]

do i? idk? what do you mean? i still dont understand how im supposed to use f''(x) > 0 for x =/= 3 to help me figure out the graph of f(x)

[spice knot]

DID YOU

oh my god bro

forget it

you dont even remember what we talked about

maybe youll figure it out if youre older lol

[spice knot]

not only did this misunderstand the question I asked

[shy pilot]

not really casue im stressed tf out trying to understand how im supposed to use f''(x) > 0 for x =/= 3 to help me figure out the graph of f(x)

[spice knot]

bro

can I vc you

this text is getting nowhere

[shy pilot]

sure ig

[full forum]

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[astral sinew]

Given f(x), f(x+2h), f(x+3h)

[delicate torrent]

aight?

[astral sinew]

Estimate f"(x)

[delicate torrent]

$f(x)$ can be any degree?

[glossy valve]

1 divided by 0 equals Infinity

[astral sinew]

f(x) is any general smooth function

[quartz flare]

Hint: consider the limit definition of derivative

[astral sinew]

Idt that's how you do it really

You need to use taylor expansions

But the problem is that the Taylor expansion of f(x) is just f(x)

So like why

And how

How do you use f(x)

[full forum]

@astral sinew Has your question been resolved?

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[median river]

hello. im having trouble with calculating area of shapes in the integral. It's a bit simple but I wanna clear up some problems i have

[thick hedge]

,rccw

[thick hedge]

part(b)?

[median river]

yes

[thick hedge]

okay

notice the curve b/w -5 and -3 is a semi-circle

[median river]

yes

[thick hedge]

so that's done?( it's area)

[median river]

also on -3 through 1 the area formula for triangle is 1/2bxh?

[thick hedge]

yes

[median river]

and semi circle is pi x r^2/2?

[thick hedge]

yes

[median river]

so would a1 and a2 be. pi/2 and 2?

[thick hedge]

the areas yes

but you need the signed area

[median river]

so -

[thick hedge]

yup

[thorny bluff]

also, its from -5 to 1, not -5 to -1, so your triangle area is incorrect

what are the base and height?

[thick hedge]

the triangle is from -3 to 1

[thorny bluff]

so 4 * 2 * 1/2 = 4, not 2?

[thick hedge]

oops

[median river]

oops

okay well I have what I wanted to clear up done

[quartz flare]

!done

[full forum]

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

[median river]

.close

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[cobalt rune]

Need help graphing trig functions

Got sick missed the class

Its x+sinx

I got the critical number pi

But how do i know if its increasing or decreasing

[hardy charm]

x+sin(x) right?

Draw a straight line y=x.

And draw your sine curve along that line.

Done.

[cobalt rune]

How do i actually do it tho

Less about the answer

I need to know the process

[hardy charm]

From 0 to 2π?

[cobalt rune]

Go and find the critical numbers

Find pts of inc dec

Then concavity

[cobalt rune]

Yes

[cobalt rune]

I got critical number for inc dec i think

Im just confused where to go next

[quartz flare]

To find critical numbers, find points where the derivative = 0
To find when increasing or decreasing, check when the derivative is positive/negative
To find concavity, check when the second derivative is positive/negative (positive = upwards concave, negative = downwards)

To graph a function in general I would choose some x values, plug them in, and then write the points on the graph

[cobalt rune]

Does my chart look right

I already know the basics of curve sketching

Im just congused how to know if 1+cosx is positive

Below pi

And above pi

[quartz flare]

Ah

You can plug in a value to check

[cobalt rune]

What if its a no calculator course

[quartz flare]

You're correct that 1 + cos(x) = 0 whenever cos(x) = -1, i.e. x = -pi, pi, 3pi, etc.

[cobalt rune]

1 +cos(1/2)

[quartz flare]

Choose an x-value you can evaluate by hand, like x = pi/2 or x = 0

[cobalt rune]

Is like

Thats the root3/2 one right

[quartz flare]

cos(pi/6) is root3/2

[cobalt rune]

Plus 1 is still positive

Right

[quartz flare]

Exactly

[cobalt rune]

Then for pi-2pi

Imma use 3pi/2

Isnt that 0

For cos

Plus 1

[quartz flare]

Yep

We can use a little reasoning to help us here
Since we know the least value of cosine is -1, and the maximum value is 1

The minimum value of 1 + cos(x) is 1 + (-1) = 0 and the maximum value is 1 + 1 = 2, so it will always be >= 0

[cobalt rune]

Ah ok

I was confused why it was always increasing

[cobalt rune]

Whys the minimum value -1 again?

Forgot all my trig

Esp the ratios

[quartz flare]

This one's by definition

Sin and cos are always between -1 and 1

[cobalt rune]

Oh

The second derivitivr i got is -sinx

Critical numbers pi

2pi

[quartz flare]

Good

0, 2pi, etc.

[cobalt rune]

This will just always be CD right

Since -sinx will always end up negative

[quartz flare]

Not quite

Remember the maximum value for sin is 1

sin(pi/2) = 1, so -sin(pi/2) = -1

We can again plug in values again, or memorize that sin is positive between 0 and pi, and negative from pi to 2pi (you can use the unit circle to see this)

[cobalt rune]

I plugged in -sin(pi-6)

Cuz ik its negative -1/2 for that

[quartz flare]

Yep

[cobalt rune]

Oh yea

3pi/2

Sin is -1

[quartz flare]

Ah I mean to look at the minimum value XD

Yep

-1

[cobalt rune]

So positive

[quartz flare]

So -sin(3pi/2) = 1

Exactly

[cobalt rune]

How would i find the inflection pt

Does (pi + sinpi) work

[quartz flare]

The inflection point is exactly when the concavity changes direction

We know the second derivative goes from negative to positive at pi, like you said

And from positive to negative at 0 and 2pi

[cobalt rune]

Is there a way to find the y value

Like a exact value

Wait

Pi - 1

Is that a thing

[quartz flare]

The exact value is as you said, pi + sin(pi) = pi + 0 = pi

So (pi,pi) is an inflection point

[cobalt rune]

If the original function is y = x + sinx

Pi + (sin(pi)) is pi + (-1) right

Nvm

Its pi,pi mb

[tight cloak]

Not exactly relevant but there is a trick to drawing x+sinx
the range of this is (x-1 to x+1)
First draw the lines x-1 and x+1 and then draw the sinx curve between those lines.

[cobalt rune]

Got it

Looks close enough to the answer key

Imma try

Xcosx

Is the deriv -xsinx

[quartz flare]

Not quite

[cobalt rune]

Oh its product rule

[quartz flare]

Yep, 👍

[cobalt rune]

Missing the g

So at

Cosx - xsinx = 0

Am i missing a indentity

Or can i solve this

[quartz flare]

Yes

[quartz flare]

This

Hm

It's equivalent to tan(x) = 1/x but I'm not sure if there's a way to solve that by hand

[cobalt rune]

Theres no other way to solve for the 0 points is there

[quartz flare]

Well let me put it in the calculator and see if there even are any 0 points

I think there should be 1, but only 1

And then after that it goes positive to negative

Oh no there's a bunch

Ah right

Because sin(x) gets switches direction

So even as x increases you'll have 0's

[cobalt rune]

Yea i asked chatgpt, it said its more of a reasoning question

U cant do the math

[quartz flare]

But as for finding exact values yeah that may be impossible by hand 😆

[quartz flare]

Well there you go

[cobalt rune]

How would u think about it?

[quartz flare]

The graphing part or the calculus part?

[cobalt rune]

The graphing part

i just plugged in values for x=0 and x = pi

[quartz flare]

Good

0cos(0) = 0
picos(pi) = -pi

Sadly this function is crazy so there's a lot of different values you might plug in
I'd probably look at when xcos(x) = 0, which is either when x = 0 or cos(x) = 0
So 0, pi/2, 3pi/2
But yeah too much

[cobalt rune]

Ah

I doubt ill see a question like this on my test

My test is always doable without a calculator, but the homework is weird sometimes

[quartz flare]

Good, because if you did I would say to run from it 😂

[cobalt rune]

I think thats good for everything

Thanks!

.close

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[quartz flare]

No problem!

[full forum]

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[light olive]

!status

[full forum]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[light olive]

4

[hot kraken]

Final answer looks about right

[half drift]

Could i give me a min

I’ll check this

[light olive]

copy

[light olive]

alright thanks

[tacit siren]

Yeah it's right

[half drift]

Yep it’s correct

[light olive]

alright thanks guys

.close

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[tacit siren]

Your work seems good too. I'd just start off by writing "u =sqrt(e^(2x) -1)" at the very beginning if you need to show work, cause that's how you know that u is the positive square root rather than the negative one.

[half drift]

U could also take e^(2x) = cossec^2(y)

Which could make the transformation much easier

[light olive]

i got it
nice point

[full forum]

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[still current]

In a heavy machinery assembly plant, a machine is designed to lift enormous parts. To lift the part shown, the resultant force from the force applied by each of the lifters located at A and B must be vertical and have a magnitude of 5000 N.

What is the magnitude of the force resultant applied by each lifter in order to lift this part?

[still current]

how do i find the components without the norms of A and B?

oh the orientation is 90 and the norm is 5000

how do i find norm of b

[quartz flare]

We know that the resultant force is vertical and has a magnitude of 5000N, as you stated

Can you write the resultant force in component form?

We'll use the components of the resultant to find the components of A and B, then use that to find their norms

[still current]

yeah

[quartz flare]

What would that be

[still current]

(0,5000)

[quartz flare]

Good

Now how are resultant force and the individual forces related

In general

[still current]

the resultant force is a + b

a + b = r

[quartz flare]

Exactly

And how do we add forces?

[still current]

by their components

[quartz flare]

Good

So that means, if we write the horizontal component as x and the vertical component as y, then

A + B = R

means

Ax + Bx = 0
Ay + By = 5000

Since R = (0,5000)

Do you agree?

[still current]

ohh yeah

[quartz flare]

So now how do we find Ax, Bx, Ay, By?

[still current]

not sure because we got like 4 variables

[quartz flare]

Think back to vectors

Given direction, how do we find the x- and y-components of vectors

[still current]

with the orientation and norm

[quartz flare]

Good

So let's rewrite our equations with the orientation and norm

The x-component of A is what?

[still current]

Ax

[quartz flare]

Yes, but write it with orientation and norm

[quartz flare]

Since they give us the angle in the chart

[still current]

cos 113/1 = a/x

[quartz flare]

Almost

cos = adjacent/hypotenuse

So cos(113) = x/a

Therefore the x-component of A is |A|cos(113)

Can you write the x-component of B in a similar way?

[still current]

|B|cos(54)

[quartz flare]

Mmm is 113 the angle for B?

[still current]

54

[quartz flare]

Good, yeah the x-component of B is |B|cos(54)

So now we have our first equation

|A|cos(113) + |B|cos(54) = 0

Now we have to get the y-components

This part should be easier

What should those be

[still current]

its |A|sin(113)

|B|sin(54)

[quartz flare]

☑️

So now we have our second equation

|A|sin(113) + |B|sin(54) = 5000

[still current]

ohhhhhhhhh

and the cos with the x

0

[quartz flare]

Since it's two equations and two variables, we can solve by substitution

[still current]

yeah

[quartz flare]

And this will give us the norms directly 👏

[still current]

how do we substitute

[quartz flare]

Let's take the first equation and solve for one the variables, say |A|

It would look like this

[still current]

wait i know

i have to divide 5000 by sin 113

then divide 5000 by sin 54

[quartz flare]

Mmm not quite

[still current]

hmm

[quartz flare]

We have to substitute before we do any division, because we need only one variable in the equation to solve

Here's what solving the first equation looks like

\begin{align*}
|A|\cos(113^\circ) + |B|\cos(54^\circ) &= 0 \
|A|\cos(113^\circ) &= -|B|\cos(54^\circ) \
|A| &= -\frac{|B|\cos(54^\circ)}{\cos(113^\circ)}
\end{align*}

[glossy valve]

Coolempire93

[quartz flare]

I moved the B term to the right side, and then got |A| by itself

That allows me to plug it into the second equation

\begin{align*}
|A|\sin(113^\circ) + |B|sin(54^\circ) &= 5000 \
-\frac{|B|\cos(54^\circ)}{\cos(113^\circ)}\sin(113^\circ) + |B|sin(54^\circ) &= 5000
\end{align*}

[glossy valve]

Coolempire93

[still current]

damn

[quartz flare]

Now that I'm here, do you think you can solve for |B|?

[still current]

was there an easier way or was this the easiest way

[quartz flare]

The only easier way would be to calculate the sin and cos values as numbers rather than working with them symbolically

But the process is the same

[still current]

alright

[quartz flare]

I prefer to use the symbols because the numbers sometimes introduce error

[still current]

first thing i'd do is multiply everything by -cos 113

[quartz flare]

Sure, that gives us $$|B|\cos(54^\circ)\sin(113^\circ) - |B|sin(54^\circ)\cos(113^\circ) = -5000\cos(113^\circ)$$

What do you plan to do next?

[glossy valve]

Coolempire93

[still current]

wait whys there sin 113 there

oh nvm

hm

now i would divide by cos 54

wait no

wait how did 5000 become -5000

[quartz flare]

Because you divided by -cos(113)

[still current]

5000/-cos113 is -5000?

[quartz flare]

I'll give you a hint: ||since we have two |B|'s, we want to make it so that there's only one||

[quartz flare]

Sorry multiplied by

[still current]

oh

[quartz flare]

Haha as it says in your original message

[still current]

it says it gives 1953,65

in the calculator

5000 x -cos113

[quartz flare]

Yeah, once we put in the numbers that's what we get

[still current]

oh

[quartz flare]

$$|B|\cos(54^\circ)\sin(113^\circ) - |B|sin(54^\circ)\cos(113^\circ) = -5000\cos(113^\circ) \approx 1953,656$$

[glossy valve]

Coolempire93

[still current]

oh i see you put the - on the 5000

instead of doing 5000(-cos113)

?

[quartz flare]

Yeah 👍

Same thing though and same answer 👌

[still current]

alright

ok now we divide 1953,65 by sin54?

[quartz flare]

We still have to deal with this before we can divide again

Otherwise we'll just be creating more fractions

[still current]

oh i do cos 54 - sin 54

[quartz flare]

We can't subtract directly, but we can pull out the |B| so that we can

$$|B|\cos(54^\circ)\sin(113^\circ) - |B|sin(54^\circ)\cos(113^\circ) = |B|(\cos(54^\circ)\sin(113^\circ) - sin(54^\circ)\cos(113^\circ))$$

[glossy valve]

Coolempire93

[quartz flare]

Oh, it's too long to read

$|B|\cos(54^\circ)\sin(113^\circ) - |B|\sin(54^\circ)\cos(113^\circ) = |B|(\cos(54^\circ)\sin(113^\circ) - \sin(54^\circ)\cos(113^\circ))$

[glossy valve]

Coolempire93

[quartz flare]

Do you see how I turned the two |B|'s into a single |B|?

[still current]

ok listen its alr ty for helping me i gtg to sleep but yeah i gotta revise what u taught me

like that part where u got the 2 equations

by using the sin cos equations

for components

[quartz flare]

Okay, I'll just demonstrate the last step then for you to see before you go

[still current]

alright

[quartz flare]

,wolf calculate cos(54)sin(113) - sin(54)cos(113)

[glossy valve]

Results provided by WolframAlpha

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

Requested by coolempire93

[quartz flare]

So I'll take that number

And have

[still current]

yeah

[quartz flare]

$|B|(\cos(54^\circ)\sin(113^\circ) - \sin(54^\circ)\cos(113^\circ)) \approx 0.85717|B| = 1953,656$

[glossy valve]

Coolempire93

[quartz flare]

Dividing, I get the norm of B

,wolf calc 1953.656/.85717

[glossy valve]

Results provided by WolframAlpha

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

Requested by coolempire93

[quartz flare]

So |B| = 2279.193

[quartz flare]

Finally, to get |A| I plug back into my expression for A that I solved for from the first equation

,wolf calc -2279.193*cos(54)/cos(113)

[glossy valve]

Results provided by WolframAlpha

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

Requested by coolempire93

[quartz flare]

So |A| = 3428.639

And I'm done

[still current]

so first

[still current]

we did this

then i dont get it

after u got a number

oh ait

alright so that number u got

x(b)

then u divide 1953,65

and got b then u replace b in the equation

gotcha alright thanks

[quartz flare]

Exactly

1. Factor out |B|
2. Plug the sin and cosines into the calculator
3. Divide to find |B|
4. Plug back into the equation from part 1

Like you said you can revise this chat and of course you can always come back and ask again for help for more practice!

[still current]

alright tysm man

[quartz flare]

No problem 👍 have a good night

!done

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[still current]

you too

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[still current]

how can i find the norm and orientation of the resultant vector

cos 191 = a/7
sin 191 = b/7

cos 127 = a/3
sin 127 = b/3?

then add both vectors

[spiral vigil]

probably want to figure out the angle that second vector takes relative to the xy plane

[still current]

why

[still current]

i cant do that? since we got orientation of bot vectors and norm

[spiral vigil]

it's a little hard for me to tell but it looks like v is 191º relative to u, right?

[still current]

yeah

oh wait

its like this

[spiral vigil]

oh ok

then yeah your initial approach should work

if you want to be precise about it you could write something like

$\vec v_x = 7\cos191\deg$

[glossy valve]

two wuggen in a trenchcoat

[still current]

yup

alright gotcha ty

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[native matrix]

I think im missing something like an formula but not sure spent 30 mins ish trying to solve couldnt find it

[spiral vigil]

are you meant to assume BD is parallel to the ground?

[slate violet]

that can't be possible though by alternate angles

[spiral vigil]

oh true

[native matrix]

A E and D are on a line

A is bottom left

no more info

[slate violet]

okay I constructed it on GeoGebra

[full forum]

@native matrix Has your question been resolved?

[native matrix]

ight

geometry gaming

[slate violet]

alright, I think I have a solution by angle chasing

reflect point B across DA and mark point F
reflect point B across DC and mark point G

then you have DB = DF = DG

the key idea is that DBG is an isosceles triangle

but then by the reflection property, angle DBC = angle DGC
so that allows you to conclude angle CBG = angle CGB

wait that still doesn't get it

[native matrix]

how do you see these stuff

[slate violet]

I'm praying that somehow it'll pop up from a construction

[native matrix]

hopefully

[slate violet]

ah got it

it works

[slate violet]

first step is to draw my construction out, of course

[native matrix]

sec

[crisp thicket]

Is answer 20 ?

[slate violet]

no, x = 100

[native matrix]

forgot why were the ones on the right 80

[slate violet]

we'll get to that later

if you've drawn it out, do you agree that angle FDG = 180 - 40 - 40 = 100?

[native matrix]

whys db=df

[slate violet]

by definition of reflection

[native matrix]

can you define reflection rq

[slate violet]

we defined DB = DF and DB = DG

[native matrix]

ah wait no nvm

see it

lul

[slate violet]

okay, tell me when you understand this step

[native matrix]

i dont

[slate violet]

okay, do you agree that angle ADF = 10?

cause we reflected the angle

[native matrix]

yup

[slate violet]

right, so what must angle DFC be?

[native matrix]

feel like im lacking in geometry

[native matrix]

not sure

[slate violet]

you have 2 angles in a triangle and you need to find the 3rd one

so what's angle AFD?

[native matrix]

found bfd as 80 cant find afd tho sec

[slate violet]

80?

that's correct but how did you even find that

[native matrix]

10 90 80 since reflections need to be 90 degrees? Did i do that wrong

[slate violet]

no, reflections don't need to be 90 degrees

[native matrix]

actually

[slate violet]

oh I think I know how you did it

[native matrix]

can you define reflection rq

[slate violet]

did you figure out that BF was perpendicular to AD

[native matrix]

using the reflection must be 90 but if thats wrong then i got it from the wrong way

[slate violet]

the following must all be true if B is reflected to C:

1. AB = AC
2. angle BAD = angle DAC
3. BE = EC

[native matrix]

gotcha ty

[native matrix]

got something similar to this and found 80

[slate violet]

yes there is another way of proceeding from this

[slate violet]

note that BDC is an isosceles triangle here (DB = DC)

[native matrix]

how do you find afd

[native matrix]

İght

wait sec

140?

[slate violet]

180 - 30 - 10 = 140

okay yeah let's stick with my method honestly

yes, so now you can figure out angle DFC

[native matrix]

does the dot f come exactly on the line ac

[slate violet]

yes

[native matrix]

40 dfc?

[slate violet]

yep!

okay, by reflection we know that DB = DF and DB = DG right?

[native matrix]

yup

[slate violet]

so that means DF = DG

[native matrix]

sec

nvm

[native matrix]

yes

[slate violet]

triangle DFG is also isosceles

[native matrix]

yup

[slate violet]

so angle DFG = angle DGF right?

[native matrix]

is g on the ac line aswell

[native matrix]

yes

[slate violet]

yes

[native matrix]

Sec

How do i get the reflection drawing right on the first try

is that possible

[slate violet]

it doesn't need to look perfect

I literally drew this and I'm still fine

[native matrix]

on a line tho wouldnt have found fdg with the way i have drawn

fdg is 100 btw see that now

120 30 30 10

found 80 80

[slate violet]

the better thing to do was to let point F be the point on AC such that AB = AF, so then it follows that DB = DF

[slate violet]

yeah so what I did was to split BDG = 10 + 10 + 100 = 120 in half

120 becomes 60 and 60

[queen flame]

w drawin

[native matrix]

found x

[slate violet]

that means angle BDC = 60, so angle FDC = 40

cool, once you find x = 10

just look at triangle ABG

[native matrix]

what way is that

[slate violet]

the missing angle is the one you want to find

[slate violet]

I didn't work it out fully haha

[native matrix]

how do i improve at geometry

[native matrix]

yeah but then what if its not on the line ac is what i thought

[slate violet]

you need to recognise this isosceles triangle setup

if DB = DC then AB = AC
if AB = AC then DB= DC (assuming that AD is perpendicular to BC)

[slate violet]

yeah okay, I see that was the big issue

[native matrix]

Was there an specific reason you drew it like this and not the other way for the first drawing btw

[slate violet]

the main construction is DB = DF = DG, so AB and AF are irrelevant for this problem

I just saw it worked (GeoGebra really helps)

in hindsight I should have explained why we were allowed to assume F and G were on AC

[native matrix]

you still should

[slate violet]

I'm sorry as I told you

[native matrix]

unfortunate cant use geogebra when needed for now

[slate violet]

well GeoGebra isn't enough on its own

it's just a way to test out various constructions

for the ideas you need to practice

[native matrix]

makes sense

[slate violet]

like, if you see the marked angles being equal, immediately try to find a way to construct an isosceles triangle

it helps they have extended line AC to the right

so the first thing I tried was to make a copy of triangle DCB reflected across DC

and then when that wasn't enough, I reflected triangle DBA across line DA

those two ideas together were enough

[native matrix]

good tip

ty

[slate violet]

that's how I did it

I should have explained it like that, I know

it's harder to explain than it is to convince yourself

I thought it'd be easier to imagine reflecting a single point rather than an entire triangle

and then that way of thinking of it really didn't help explain

[native matrix]

this solves the problem i had

makes more sense

Where can i find similar questions which can be solved this way btw

[slate violet]

Art of Problem Solving, make an account

and then go to Alcumus

https://artofproblemsolving.com/alcumus/problem

wait it doesn't have these problems

probably find a book then, ask in #book-recommendations

[native matrix]

made

how do i word the question what kind of a book does this count as

[slate violet]

there are practice problems for AMC8/10/12

[slate violet]

Euclidean geometry

make sure to specify you're a beginner, so not Olympiad geometry

[native matrix]

will euclidian geometry be enough to specify

[slate violet]

it's hard to specify

say not Olympiad geometry as well

and you should be good

[native matrix]

now i wait

[native matrix]

said beginner should be good enough

Ty x2

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[native matrix]

.open

.reopen

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✅ Original question: #help-28 message

[native matrix]

what does amc 8 10 12 mean

is it the focus one

[slate violet]

no, those are competitions

AoPS has questions for at least 20 years

search up 'amc 8 AoPS'

[native matrix]

ty

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[fathom drift]

im trying to follow a proof of stokes theorem, stated as follows:

heres what ive written so far

can i get help finishing off the proof?

[leaden ermine]

pull

[full forum]

@fathom drift Has your question been resolved?

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@fathom drift Has your question been resolved?

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[manic bramble]

A book has a derivative calculation like the following.

[manic bramble]

Notice between the 3rd and 4th line, the h at the bottom is extracted and multiplied as 1/h (in the 4th line) to the stuff on the right. But, that's not the same as the 3rd line.

[quartz flare]

It's 1/h * what was in the numerator

[manic bramble]

I would expect it to be like this:

[quartz flare]

Correct 👍 but anything over 1 is ||itself||

[manic bramble]

But, in this case, it would be significant, because that multiplication would be 1 x (...) and h x 1.

Whereas, what's in the book, I read it as 1 x (x+h)(...) and h x (h+h-1)(...)

[quartz flare]

Yes; we can simplify fractions like this because multiplication and division have the same priority in order of operations (I can swap the order of multiplication and division and still be fine)

If I have $\displaystyle \frac{1}{2}\cdot\frac{2}{4}$ that is the same as $\displaystyle \frac{1 \cdot 2}{2\cdot 4} = \frac{2}{8} = \frac{1}{4}$, or I can write $\displaystyle \frac{\frac{1}{2}\cdot 2}{4} = \frac{1}{4}$

[glossy valve]

Coolempire93

[manic bramble]

interesting. Did not know that was ok.

[quartz flare]

Yep! Fractions are special

[manic bramble]

I feel both good and bad. Good because I learned this new thing, and bad because this seems like something "basic" that I should have known a long time ago.

[quartz flare]

There are always things we'll miss in every class, basic and advanced (which will at some point also become basic compared to something else)

Can't expect to know everything, but as long as you're always learning, you'll fill in the necessary gaps with experience 🙂

[manic bramble]

Thanks person. You answered my question, and thought me something new.

I appreciate it.

[quartz flare]

No problem! Good luck with your calculus 🙂

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[manic bramble]

.done

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[drowsy niche]

This is my attempt to show that given a field K of zero characteristic and a field homomorphism f:Q->K where Q denotes the rational numbers, f(x)=x for all x in Q(that is, 1_K • x)

[drowsy niche]

First, note ker(f) is an ideal of Q as it is the preimage of the zero ideal, since Q is a field, it has no proper ideals, so either ker(f)={0} or ker(f)=K, however, in the second case, f(1)=0≠1_K, since K has characteristic 0, as such, we have that ker(f)={0}, so f is injective, which implies it is an isomorphism onto its image f:Q->Im(f), so K contains a copy of Q, which is precisely the image of the given homomorphism, as such, we may assume, without loss of generality, Q⊆K, and f reduces to an automorphism of Q, as such, all we need to show is that Q has no nontrivial automorphisms, let m/n∈Q for m,n integers, then f(m/n)=f(m•1/n)=f(m•n^-1)=f(m)•f(n^-1)=m•f(1)•(f(n))^-1=m•(n•f(1))^-1=m•(f(1))^-1 •n^-1=m•n^-1=m/n, as such, f fixes all points in Q

My main gripe with this is that I haven't used the characteristic 0 condition all that much, but don't see where this could go wrong

[fast peak]

f:Q->K, so ker f would be an ideal of Q

[drowsy niche]

🤦🏻‍♂️thank you

I think that doesn't really change much though, despite making that misstep, the rest of the argument doesn't rely on that, I was actually using the consequences of ker(f)={0} in Q

[fast peak]

yes rest is fine

paragraphs would be good

or at least actually finishing a sentence

instead of continuing with a comma

[drowsy niche]

Yeah, I apologize if it was a little hard to read

Thank you for your help

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[half verge]

Hello, I’m looking for help for a research paper on power and water usage efficiency. I’ve run into a problem with WUE (water usage efficiency) where the projections use the formula ‘consumed L/ IT energy kWh’ but my sample data uses ‘withdrawn L/ IT energy kWh’. I’m working with very limited samples. Because of this, I’m calculating the values based on published data as well but finding limited information here as well. For example, I have withdrawn water and IT energy from individual locations, as well as total amounts for all. Is it reasonable to ratio the totals to find my missing consumption amounts? Based on the discrepancy in the screenshot after doing that, does it even matter or should I just use the actual values and make note of the small uncertainty the formula differences cause?

[quartz flare]

This might be a better question for a specific subject chat to get experts on it?

Not sure

[half verge]

oh ok

thank you

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[fathom saddle]

Sad

[twilit heart]

do you have a legitimate question

[hollow wharf]

Nope they were a spambot lul

[delicate torrent]

<@&268886789983436800> troll

[quartz flare]

Was about to do the same thing xd

[umbral dome]

,purge 30 --from 1017215960857456681

[glossy valve]

Couldn't find a member matching 1017215960857456681!

Dec 18, 2025, 11:43 PM

Couldn't find the requested user, cancelling purge.

Dec 18, 2025, 11:43 PM

[spiral vigil]

I hate texit

[quartz flare]

!done

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[umbral dome]

.close

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[quartz flare]

(There's another one they opened as well)

[supple bloom]

Rahhhhh @umbral dome not thumbs upping the ping

[devout valley]

(to be fair, at least for me, I react to the pings after I've taken all/most of the action I've intended, which, well, )

[delicate torrent]

Rahhh @supple bloom not lance lancing the ping

[supple bloom]

I like to do it after the fire is put out but before doing paperwork/purges

[devout valley]

Yea that's pretty much my view, I always have the (probably illogical) fear that I might get hit by lightning or a bus or something before completing what I wanted to

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[brisk minnow]

Hi is this enough to say that a sequence can’t both diverge to negative and positive inf?

[rare dock]

yea

[brisk minnow]

Ty

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[brittle sun]

Actually slight subtlety

.reopen

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✅ Original question: #help-28 message

[brittle sun]

x ≥ L and x ≤ L is a contradiction, but you do have to explain why

Nvm he's gone

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[rare dock]

hm true i guess that’s not 100% done

[brittle sun]

Yes

[rare dock]

should have picked two different L’s

[brittle sun]

Technically could just make the inequalities strict and be done

Depends on your definition tho ig

[rare dock]

yea

they are all equivalent ofc but

i think i have a story related to this

[brisk minnow]

.reopen sorry

[full forum]

✅ Original question: #help-28 message

[brisk minnow]

Im here

Was doing a different exercise sorry

[rare dock]

#「helpers-lounge」 message

[brisk minnow]

It says no access

Could I choose L to be -1 and 1?

To be more precise ig

[rare dock]

yea (name them differently though)

[brisk minnow]

Ah alright fairs

[rare dock]

secret for green names only

[brisk minnow]

Can I ask a different question on here now?

[brittle sun]

Sure

[quartz flare]

(how does one get a green name I always wondered)

[rare dock]

you can also do what xavier said and write strict inequalities if you expect your grader will be ok with that

[brisk minnow]

For this question, I understand the intuition but idk how to rigorously phrase it, could I have a slight hint

[rare dock]

if you need to like, explain why the < and <= definition is equivalent i wouldn’t bother with that though

[brisk minnow]

Hm I think that should be assumed clear

[rare dock]

type in help channels and forum posts (that aren’t yours)

[rare dock]

what does it mean for the sequence to not eventually be constant?

[brisk minnow]

Well it diverges?

Or not within epsilon of the limit

[rare dock]

||no matter how far out you go in the sequence, you can find n such that x_n and x_{n+1} differ by at least 1||

[brisk minnow]

Ohhh that’s tempting 😭

Does to give it away?

[rare dock]

not completely but it is a big hint

[brisk minnow]

Ahhh it’s only been 5 mins since I started the q, maybe I’ll put it off until a another 5 before I look at a bigger hint

Yh checked

oh xn and xn+1 differ by at most 1

Since thr sequence comprises of integers

[rare dock]

yea, so they can’t both be close (within e.g. 1/4, or 1/2 with strict inequality) to L

it might take a little thinking to write it up properly but that’s the idea i would use

[brisk minnow]

Yh I think writing it formally will be the issue, I’ll think abt it for a bit

Idk if this sounds like yap 😭

[rare dock]

yea this is yap

[brisk minnow]

😂 alright I’ll try again

[rare dock]

oh i missed that we also need to prove L is in Z

[surreal hull]

couldn't you just use the cauchy criterion

[rare dock]

that would make my proof idea very easy but i’m guessing we can’t use it

[rare dock]

@brisk minnow is this what you’re working on, or the prove it is constant part?

everything i’ve said has been about the constant part

[brisk minnow]

Yh have to prove it’s eventually constant and l is in Z

[brisk minnow]

L is in Z first

[rare dock]

oh ok

ok my differ by 1 stuff wasn’t for that

just fyi

[brisk minnow]

This might help actually 👍 lemme see what I can do

[brisk minnow]

Got it

I think Cauchy w triangle inq

[rare dock]

if you can use cauchy criterion, the eventually constant part is very easy with this

[brisk minnow]

I think this is valid?

[brisk minnow]

Yh that’s it 👍

[rare dock]

i think we want a < 1 somewhere

not a <= 1

[brisk minnow]

Oh hm

[tiny raft]

[rare dock]

or just like

use 1/4 instead of 1/2

[brisk minnow]

Ohh lol i think I missed it there

[rare dock]

but yea this is good

[brisk minnow]

Should be here due to my assumption

Got careless there

Tysm

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[twilit osprey]

I came across a puzzle which goes as follows.

Let x1, x2, x3 be three independent random variables drawn from a continuous distribution on [0, 1]. What is the probability that x3 lies between x1 and x2 so that x1 < x3 < x2 or x2 < x3 < x1

I came to the conclusion that it should be a fifty percent chance using the following logic:

i first noticed that the order that the numbers are chosen doesnt matter because they are independent.
first, you choose x3, the middle one. next i will choose x1. It doesnt matter where x1 goes, it can be on either side of x3. Now, when i choose x2, theres a 50 percent chance on average that its on the right or left of x3, and since x1 is already on one of those sides, theres a 50 percent chance x2 is on the other side.

i then checked the provided answer which said 1/3 because of the following:

There are 6 possible orderings,
x1 < x2 < x3,
x1 < x3 < x2
x2 < x1 < x3,
x2 < x3 < x1
x3 < x1 < x2
x3 < x2 < x1
each one should have the same probability, and since exactly 2 of them satisfy the conditions, theres a 2/6, or 1/3 chance of x3 being in the middle.

Could someone point out the flaw in my reasoning?

[warped frost]

you are fixing x1

[twilit osprey]

hm?

[warped frost]

x2's 50/50 chance of being in the right position is relative to x1 already being fixed

so you turned it into a conditional probability

[twilit osprey]

hm

ok

[warped frost]

at least that's how I see this

[twilit osprey]

i still dont exactly see how that changes anything

[solemn ermine]

i think you're fixing x3

(to beg the question, why is there a 50% chance that x2 is on the other side of x1?)

[twilit leaf]

if you pick in reverse order so x3 goes first, thats a good way of thinking about it

however consider the example of a uniform distribution on [0,1]

say your x3 is 0.2

think about what happens when x2 is on each side of x3

[twilit osprey]

yeah, then thered be a bias towards one side, but i think that on average x3 would be close to 0.5

[twilit leaf]

x3 is no more likely to be closer to 0.5 than one of the ends

[twilit osprey]

sure i get that

but if you take the mean over some large number of random numbers from [0, 1] it would be close to 0.5

am i wrong?

[twilit leaf]

no

[twilit osprey]

and is that even relevant

[solemn ermine]

to put it in a pithy little saying, "the expected value isn't always the value you should expect"

[twilit leaf]

if you are not at 0.5, your probabolity of a successful trial is always less than 0.5

so you should expect a n average of less than 0.5

[twilit osprey]

ok yeah that makes more since

is there a way i coudlve come to the 1/3 conclusion without looking at the combinations?

[twilit leaf]

if youre good at integrals, you can use one of those

[twilit osprey]

i see

well i think i have a better understanding of the problem now, and why its not 50%, so i think im done here

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[quartz flare]

.reopen

[quartz flare]

A question about the previous problem asked in here

[twilit osprey]

what is it ?

[quartz flare]

Oh sorry I didn't mean to ping you 😅

[twilit osprey]

all good

[quartz flare]

I saw it when you posted it and went to try it myself, wanted to make sure my logic was sound

I did get the expected answer

Essentially I said first place x1 and x2

The probability that x3 lies in between should be |x1-x2|

Therefore we should be able to sum over all the |x1-x2| for all possible x1 and x2 and that's the probability across all scenarios

(Excuse my language I haven't taken classes in probability so I don't know the right terms)

Which gives me

I assume this is what they were alluding to when they said if you know about integration that's another way to solve?

Or did I do it the wrong way and end up with the right answer

[twilit osprey]

thats an interesting way to solve it i wouldve never come up with lol

[quartz flare]

It's funny because combinatorics is what I do but this was the only idea that came to mind 😅

[solemn ermine]

does the math work out the same if x1, x2, x3 instead are uniformly distributed across some other interval? (i.e., what happens if they instead have the uniform distribution on [3, 5]?)

[quartz flare]

Mmm I would think so? Let me think 🤔

Just needing to divide by 2

Let's see

After I place x1 and x2

The probability x3 lands in between is |x1 - x2|/(5-3)

part/total

And then I would integrate both from 3 to 5

Oh that probability seems a bit high

Just a tad

But why 🤔

I have to think about why I chose the first way in the first place

Which was setting x1 and then choosing all possible values of x2, adding the probabilities

By examination I have to divide by the size of the interval to the third power why

(still here thinking and coming up with no fruit)

[solemn ermine]

what is the probability that some x1 is in the interval? what about the probability it is between two points contained in the interval?

(i was thinking for a while too--i'm really good at confusing myself)

[solemn ermine]

perhaps more explicitly, ||how might we relate this probability to an integral?||

[quartz flare]

Just writing it here because it was my train of thought that I'm leaving off to consider what you're saying

My idea was we can reduce the 3to5 interval to an interval of size 1 like so

x1 -> x1/2
x2 -> x2/2
x3 -> x3/2

Because this transformation is one-to-one and onto the probability should be the same for all sized intervals

Okay now back to what you were saying

[quartz flare]

Some x1? like any point x1 or an x1 that is u.d. on [3,5]

Because I would say 1

But that seems silly

[quartz flare]

Oh and the probability that it's in between two points I would think this

[solemn ermine]

it's not silly, it just feels 'obvious' to say--our random variable x1 (which is uniformly distributed on [3, 5]) is always going to be between 3 and 5

[solemn ermine]

[quartz flare]

Right 🧐

[solemn ermine]

and if we want to compute a single integral to compute this probability, what would feel appropriate? what should bounds look like? what should we integrate?

(to circle back to this question)

[quartz flare]

Oh I never opened that spoiler up

Should I?

[solemn ermine]

it's just asking "how would this probability become an integral", don't want to bombard like 8 unanswered questions at once, lol

[quartz flare]

Ah XD

How would it become an integral

A single integral

As in not a double integral

Mmm

If we're talking about the probability that a single variable lies between two set points then I would integrate $$\frac{\int_{x_1}^{x_2}1dx_3}{\int_{3}^{5}1dx_3}$$

[glossy valve]

Coolempire93

[quartz flare]

Hm 🤔 that's interesting actually

Even for the 0 to 1 interval that's interesting haha

[solemn ermine]

this is basically the idea--what do we get if we compute the bottom integral?

[quartz flare]

There was a hidden third integral inside

[quartz flare]

We get the 5-3

[quartz flare]

Like here

[solemn ermine]

(it's 'traditional' to bring the third probability up into the integral; this relates to the idea of a "probability density function")

[quartz flare]

To bring it up into the integral?