help-28

[quaint prawn]

search on youtube, "how to integrate by parts"

[torn jolt]

ok

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[gusty carbon]

I got this problem correct but just by guessing and I have a quiz so I want to understand the question better

[quaint prawn]

for any one input, a function returns only one output

that should tell you all you need to know

[gusty carbon]

but for both the first and second table there are 2 2s on the y side

[quaint prawn]

but they are for different inputs

you can have the same ouput more than once

but for any given input, only one output

input 3, get 2

input 6, get 2

but you don't input 3 and get 2 and then input 3 again and get 7

[gusty carbon]

but what makes the first one y a function of x but the second one not

[quaint prawn]

in the 2nd one, you have

f(1)=2 and f(1)=14

one input, but 2 different ouputs

in the first one that doesn't happen

[gusty carbon]

ohhh i didnt see the 1

thank you for helping me

[quaint prawn]

np

[gusty carbon]

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[tawdry grove]

im not sure how to find F'(x) here

[gritty rose]

,tex .FTC2

[glossy valve]

dr. matlab plot

[tawdry grove]

what

[hot herald]

what's your issue with the fundamental theorem of calculus above?

[tawdry grove]

well

im not sure how to apply it

im struggling figuring out when to apply what

and how to apply that what

[devout valley]

If I said I wanted to call this $F(x)$ as $\int_{a(x)}^{b(x)} f(t) \dd t$, would you be able to identify what $a(x), b(x)$ and $f(t)$ are?

[glossy valve]

@devout valley

[tawdry grove]

yes

[hot herald]

what are:
f(t)
b(x)
a(x)

[tawdry grove]

f(t) = sin t

b(x) = cosx

a(x) =sqrtx

[devout valley]

Cool given that, can you then tell us what b'(x) and a'(x) are?

[tawdry grove]

b' = -sin

a' = x^(3/2)

[devout valley]

Careful, remember you're differentiating sqrt{x}...

Also better to write b'(x) = -sin(x), to keep the variables in, but you get how to do it at least

[tawdry grove]

hum

is x^(3/2) wrong?

i just added 1 to the exponet

[devout valley]

How do you differentiate something like x^n?

[tawdry grove]

drop n in front, and subtract 1 from exponet

[devout valley]

Yep, good enough, $nx^{n-1}$, and we're differentiating $\sqrt{x}$ of course - can you rewrite $\sqrt{x}$ as a power of $x$?

[glossy valve]

@devout valley

[tawdry grove]

.5x^(-.5)

[devout valley]

Nice, yep, that's the follow up question I would have asked you

But that's a'(x) for us

[devout valley]

Now, back to these, would you be able to tell me what f(b(x)) and f(a(x)) are?

[tawdry grove]

whya re we differntitating the integrals max and min numbers

[devout valley]

That's basically a result of the chain rule

Remember that one of the "original" FTCs say that if $G(x) = \int_a^x g(t) \dd t$, then $G'(x) = g(x)$

[glossy valve]

@devout valley

[tawdry grove]

but whu

those are not even part of the f(t)dt

[devout valley]

Why does the FTC say that? or?

Anyways, the explanation was that if you had something like $\int_a^{x^3} g(t) \dd t$ instead, you'd notice that you actually now have $G(x^3)$, with $G$ as per defined before

[glossy valve]

@devout valley

[devout valley]

So if you wanted to differentiate $\int_a^{x^3} g(t) \dd t$ with respect to $x$, you would need the chain rule because the "original" FTC didn't tell you anything about when you had anything other than just $x$ as the top power

[glossy valve]

@devout valley

[tawdry grove]

hum

[devout valley]

wait were you happy with the "original" FTC or?

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[sterile sphinx]

Hello, for this problem do you just find where y' and df/dy is continuous? or is there more to it

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[tepid glen]

Over the course of a few months I collected some data showing what time of day an event occurred (Each bar representing an event that occurred in that minute of the day). Sometimes an event happened at the exact same minute on two or three separate days. How can I smooth out this graph? and how would I interpolate areas where events rarely occur

[open zinc]

There’s many ways you can smooth this, the simplest is probably just a histogram with whatever bin width makes sense. A much prettier (and arguably better) way to do it would be kde and kernel smoothing methods such as Gaussian smoothing https://en.wikipedia.org/wiki/Kernel_density_estimation

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[copper sluice]

unable to do (v)

[narrow ermine]

If you can use the previous problems, I think you can get this result fairly easily using i).

Wait

(v) or iv?

[copper sluice]

red one

[narrow ermine]

Ok

Yes

[copper sluice]

ok i will try

ok thanks

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[waxen geyser]

What is the area of the shaded part

[sharp sinew]

How much information do you know

[waxen geyser]

Not that much in trig

[sharp sinew]

I meant do you know any other lengths?

Or angles?

[waxen geyser]

Kind of

That is a right triangle

That's a tangent

It forms 90°

It's on the diameter

[knotty grail]

Hmm

Oh there are two people

I didn't see that lol

[waxen geyser]

Does this require trig?

I feel like lol

[knotty grail]

Is the hypotenuse of the right angled triangle passing through the center of the circle?

[waxen geyser]

Nope

[knotty grail]

Huh

That's weird

[sharp sinew]

It doesn't even feel solvable lol

[knotty grail]

Since the radius can be almost anything

And there's no way to solve it

[waxen geyser]

Oh yeah it does pass through the center

[knotty grail]

Ah

[waxen geyser]

I miss draw it

[sharp sinew]

That's better

[knotty grail]

So the radius is 6

And one of the legs of the triangle is 6

hmm

that's still not enough info

do you have the original question?

[waxen geyser]

w8

Oh I missed 3 aswell

I think that's the original question

[waxen geyser]

There that's better

[knotty grail]

Ok

That's better

You can find the triangles area now right?

[waxen geyser]

Welp, yeah i think

[knotty grail]

Ok what is it

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[sharp flame]

is there a neat and tidy formula for the number of transitive relations on a set?

[sharp flame]

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Please don't occupy multiple help channels.

[next basalt]

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[manic mulch]

A pool is 3/5 full. When another 48 liters of water is added to the pool, it carries 1/3 of the water that was initially in the pool.

Accordingly, how many liters of water are in the pool initially?

C) 48

B) 72

A) 80

D)

[pseudo cape]

Do you have an idea on how we could describe 3/5 full mathematically

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[soft folio]

Proving that W is a subset of Finfinity

Subspace

<@&286206848099549185>

🙏

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[ashen tulip]

2^-
What u mean?

Okay

but

sqrt(-0.04) is not undefined

just use imaginary(complex) numbers

like, √-1=i and i²=-1(btw (-i)²=-1 too)

but it`s right

X=1.9999 u get number very close to 0

1.999999999... u get very very close

And

1.(9)=2

Because 0.(9)=1

So, √2²-4 without limit`s is zero

2²=4 and 4-4=0

√0=0

Doesn`t matter, x→2- or x→2+

It is 0

but u can - in sqrt

But it doesn`t matter(answer is very close to 0)

Bruh

interesting

Then, what answer for this limit?

Okay

Here

Okay

1 sec

damn

My discord can`t open this

okay

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[fresh terrace]

how do i solve this?

[hard veldt]

Do you know vieta's formulas?

[deft tulip]

if the sum of the roots is 0 and product is -9, what are the roots?

[fresh terrace]

3,-3?

[hard veldt]

Or that

Yes

[fresh terrace]

nope

[fresh terrace]

ic... just fancy way of writing that?

[deft tulip]

yes, so a polynomial with roots 3 and -3 would be what in factored form?

[hard veldt]

Alr u got this

[fresh terrace]

(x-3)(x+3) = 0?

[deft tulip]

yep, now how can you find a and b from that?

[fresh terrace]

expanding?

[deft tulip]

yeah FOIL it

[fresh terrace]

x^2 - 9 = 0?

[deft tulip]

yep

[fresh terrace]

a = 1

wait so whats b?

[deft tulip]

well whats the coefficient of the x term?

[fresh terrace]

oh 0

[deft tulip]

yep

that's it

[fresh terrace]

gatcha thanks!

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[hexed sphinx]

I need help

[hexed sphinx]

So I find the answer EC is 15

But the answer key is 25

Idk how they got 25

Either I found the missing length wrong

All I did is 8/10 = 12/x then cross multiply then divide what ever by 8

[twilit leaf]

both of those statements are correct

read the problem again

[hexed sphinx]

Find the length B C

So AC and DE are parallel to each other

B and E is 10

EC is 15

Omg

I’m stupid

10 + 15

You get 25

[twilit leaf]

yeah

[hexed sphinx]

Which is 25

Omg

How did I miss that

Alr thx

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[civic tartan]

I need help with part ii)

[civic tartan]

this is what i got for i)

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[frigid gulch]

square root of (4x^2+5y) dx dy

[frigid gulch]

can someone help

@terse terrace

<@&286206848099549185>

[dusty wigeon]

Double integral?

[frigid gulch]

yes

[dusty wigeon]

What are the bounds?

[frigid gulch]

square root(y) to 2

for dz

dx

[dusty wigeon]

You mean dy

[frigid gulch]

no dx

[dusty wigeon]

Can you please screenshot. If it’s bound problem there’s usually two bounds.

I’m mostly help at advance topology. Calculus 3 and DE I need to refresh.

[frigid gulch]

the

theres more bounds

soryr

square root of (4x^2+5y) dx dy
square root (y) to 2 dx
0 to 4 dy

@dusty wigeon

[lofty merlin]

hi

do u mean under root (4x^2 + 5y) ?

[frigid gulch]

yes

@wicked seal

<@&286206848099549185>

[lofty merlin]

what do u want

[frigid gulch]

double integral

[lofty merlin]

its derivative ?

[frigid gulch]

square root of (4x^2+5y) dx dy

[lofty merlin]

can u

send pic

[frigid gulch]

square root (y) to 2 dx
0 to 4 dy

those are the two bounds

no

[lofty merlin]

why not

im not getting any of it

[frigid gulch]

wdym

square root (y) to 2 dx
0 to 4 dy

these are bounds

square root of (4x^2+5y) dx dy

this is function

[lofty merlin]

do u mean d^2 y / dx^2

[frigid gulch]

no

its oduble integral

[lofty merlin]

thats literally double integral

[frigid gulch]

oh

no

first integrate dx

then integrate dy

[lofty merlin]

integrate ?

or

derivate ?

[frigid gulch]

double integral

<@&286206848099549185>

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@torn jolt Has your question been resolved?

[stiff musk]

your majesty, did you ever get anywhere with this 😁

[twilit leaf]

respectfully, go check a book or video about the basics of complex numbers first

[twilit leaf]

you need to learn from the beginning

most of us learned from a book or video to start

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[snow nymph]

Hello, I noticed that when solving trig equations, such as this one, squaring is an easy way to get the same trig function to solve. I got pi/2+2*pi*k, pi+2pi*k, and 0+2pik. However, I got an extraneous solution, which is 0+2pik, which I figured was from squaring. I was wondering if anyone had a deeper explanation into this, and if anyone had any methods on how to avoid extraneous solutions or any tips.

[rapid rain]

pi + 2pik is not extraneous? 0 - 1 = -1

[snow nymph]

oh sorry I meant 0+2pik

[rapid rain]

you will almost be guaranteed to have extraneous solutions when squaring an equation

here's a trick to get rid of them for sure on this one

sin(...)- 1 <= 0

so cos(theta) has to be <= 0

and you can then square, knowing anything in the quadrants where cos(theta) > 0 is invalid

[snow nymph]

but I would still have to substitute theta into cosine right?

[rapid rain]

?

not here

[snow nymph]

like how do I apply that piece of information

[rapid rain]

once you guarantee that sin(theta) - 1 and cos(theta) are of the same sign, sin(theta)-1 = cos(theta) and (sin(theta)-1)^2 = cos^2(theta) are equivalent

so write sin(theta) - 1 = cos(theta) <=> (sin(theta)-1)^2 = cos^2(theta) AND cos(theta) <= 0

[snow nymph]

< 0 right?

[rapid rain]

oops

[snow nymph]

all good

[rapid rain]

there you go

cos(theta) <= 0 will lead to theta in [pi/2,3pi/2] + 2piZ

[rapid rain]

and keeping equivalences, extraneous solutions are immediately removed if they contradict this statement

[snow nymph]

ah I see what you mean now

from the problem only angles in q2 or q3 as well as pi/2 3pi/2 and pi will qualify

any tips or other strategies to solve without extraneous for other trig equations?

[rapid rain]

in general, a = b <=> a^2 = b^2 AND sign(a) = sign(b)

if it ever comes to squaring, you can use this

here it was easy to use the trick because sin(...) - 1 is always non positive

[snow nymph]

thanks for that tip

but for like

this problem

sinx-1=cosx

do you know of any other way to do it

[rapid rain]

ah, there is another way without squaring

[snow nymph]

besides squaring

[rapid rain]

wait might have been mistaken

[snow nymph]

that's alright

if no other method exists that's pretty much all I needed

[rapid rain]

oh I do remember

has to do with writing sin(theta) - cos(theta) = 1

and then surprise surprise

multiply by sqrt(2)/2

[snow nymph]

oh is this the converting to one expression w/ sine

ah yeah

[rapid rain]

so cos(pi/4)sin(theta) - sin(pi/4)cos(theta) = sqrt(2)/2

meaning sin(theta-pi/4) = sin(pi/4)

there we go

[snow nymph]

damn

alright thank you so much for the help

🙏

appreciate it

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[soft agate]

how do i do this?

[quaint prawn]

https://youtu.be/McDdEw_Fb5E?si=i1hVVzvrh4pFYjRs

Watch this video

Also, if you still don't get it, he has more videos on completing the square with more examples.

[soft agate]

ok ty

what did I do wrong

i followed the video steps

[grim skiff]

Where did 10x come from?

[soft agate]

good question

oops i think i wrote that down from the video

6x*

[full forum]

@soft agate Has your question been resolved?

[soft agate]

<@&286206848099549185>

[lime ether]

ok so

we see

b=6

so we immediately take b/2

and add it to x

then square the whole term

so we get (x+3)^2

that gives the x^2+6x

but now we have an extra +9

so

in order to keep the +13

we need it to be (x+3)^2+4

so that

if we expanded it

it’d be the same

x^2+6x+9+4=x^2+6x+13

[soft agate]

wait

[soft agate]

i got that but it equaled to 38

[lime ether]

u added 25 for some reason

[soft agate]

oh

i think i mixed the video numbers

[lime ether]

yea

u did

[soft agate]

oops

[lime ether]

oops

[soft agate]

it was supposed to squared

[lime ether]

try this one

x^2+10x+1

complete the square

well what’s going to be the

(x+a)^2

@soft agate

hello

do u need help

[soft agate]

SORRY

IM BACK

[lime ether]

hello

@soft agate

ping me

if u reply

[soft agate]

@lime ether hi

[lime ether]

hi

ok

[lime ether]

^

try this

[soft agate]

so i use the video method

?

[lime ether]

nevermind my langauge

ignore the video

take the b term

take half of it

and keep the sign

so

(x+5)^2

if it was x^2-6x

we do

(x-3)^2

if it was x^2-7x

we do (x-7/2)^2

x^2+14x

we do (x+7)^2

do u see?

[soft agate]

ok

[lime ether]

does this make sense th

tho

so

[soft agate]

yes it gets halfed

[lime ether]

for x^2+10x+1

what’s the (x+a)^2

we will figure out the constant after

[soft agate]

10x = 5x

[lime ether]

like what’s the a

[soft agate]

1?

[lime ether]

no

(x+5)^2

notice for each of the examples

i take x

and add or subtract half of the b term

then square the whole thing

if it was x^2-10x+1

id do

(x-5)^2

but

we still have to work out the constant term

[soft agate]

o

[lime ether]

do u see now?

[soft agate]

where did the 1 go

[lime ether]

we r gonna worry about that now

so

now that we have the

(x+5)^2

in order to keep the expression the same

we need to end up with x^2+10x+1

but

(x+5)^2 is x^2+10x+25

notice

the constant term is 25 here

not 1

so how do we make the constant term equal 1

[lime ether]

from here^

just subtract 24 right

because 25-24=1

does this make sense

[soft agate]

ohh

[lime ether]

we already had

(x+5)^2=x^2+10x+25

but in order to make the right side look like the original

we have to subtract 24

but whatever we do to one side

WE HAVE TO DO TO THE OTHER!!!

to make sure we don’t change the equation

so

subtracting 24 from both sides

we get

(x+5)^2-24=x^2+10x+25-24

and the right side simplifies

to x^2+10x+1

which is the original!!

wait

now it’s good

see how there’s -24 on both sides

so it doesn’t change the equation

they cancel

it doesn’t change anything

and notice the x^2+10x+25 is the same thing as (x+5)^2

tell me if this makes any sense

or if ur confused

@soft agate

r u here

[soft agate]

im so lost

😭

let me reread

[lime ether]

all good take ur time

another way to see this is

we want to create the perfect square

so when we started with x^2+10x+1

we know (x+5)^2 is x^2+10x+25

so we can just rewrite 1

x^2+10x+25-24

and basically separate the 24 from the rest

(x^2+10x+25)-24

=(x+5)^2-24

maybe just watch the videos then

if this isn’t helping

do whatever makes sense to u

[soft agate]

thank you so much for your help

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[charred ermine]

Hello

[charred ermine]

Could I get help with this

Im unsure how to start

[gritty rose]

Use fundamental theorem of calculus

[charred ermine]

?

With what values???

F(5) - f(2)?

[gritty rose]

You're given one other piece of information

[gritty rose]

f(5)-f(2)=?

[charred ermine]

hmmm 26-14

?

12

[gritty rose]

No

[gritty rose]

Answer this

[gritty rose]

By using this

[charred ermine]

i'm kind of confused ngl i'm rusty on calc😓

[gritty rose]

Look up fundamental theorem of calculus in your book then

[charred ermine]

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[swift fulcrum]

Hello

[swift fulcrum]

Do I just substitute 1, 0 respectively?

[stiff musk]

no, that would give you 0/0 which is indeterminate

a typical way to show that a limit doesn't exist is to show that you can get two different limits by approaching the target point (1,0) by different paths

[swift fulcrum]

Something like this?

[stiff musk]

yes

[swift fulcrum]

Ohhh okay

Tysm

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[charred ermine]

Hello could I get help with this

[rough plaza]

you didn't change the bounds while doing u-sub

[charred ermine]

OHHHH

Thank you very much

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[visual furnace]

Need help with a proof. I'm trying to find the possible ending positions of a path of a n x n grid. A path can start from any square, and it can go up, down, left or right (not diagonally). For an example, see the image.

I have experimented with 3 x 3 and 4 x 4 versions, and came to this conclusion: We can colour the squares like a chessboard. If the path starts on a light square and n^2 is odd, it will end up on any of the dark squares and vice versa. If the path starts on a light square and n^2 is even, it will end up on any of the light square(except for the starting square ofc).

I understand how to prove that starting from a light square will end up on a dark square when n^2 is odd and a light square when n^2 is even, but I'm struggling to explain why it could end up on any of the squares.

[deft tulip]

well i guess each time you advance a square, you cross from dark to light and vice versa

if there is an even number of squares, just mathematically you have to end on the opposite color since it just goes DLDLDL....DLDL till the end etc..

since you visit each square once

[visual furnace]

yep but why can it end up on any of the dark squares for example?

im just struggling with proving that part

why any, why not just a select few dark square

[deft tulip]

oh i see

im thinking you could prove this by induction maybe? 🤔

like the cases for a 2x2 grid is trivial

[visual furnace]

mmhmm

[deft tulip]

wait one sec

and maybe whichever square you pick as your "end one" you might be able to come up with some predictable path that includes that destination square as a smaller "Sub-board" and visit other squares in a predictable way before you enter that sub-board maybe?

but no that doesnt really work hm

intuitively though i feel like it must be possible to deconstruct the guaranteed path into something that could use induction

[visual furnace]

hmmm

alright then

thank you

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[deft tulip]

.reopen

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✅

[spiral vigil]

once your board is bigger than like 4x4 it's pretty easy to construct a path that will always work

you just have so many options

[visual furnace]

yep but could i prove that?

i dont need to

but just for fun

[spiral vigil]

if you have a set of instructions for making a path and you can argue convincingly that they'll always work

seems good to me

[visual furnace]

coolio

ill close this now

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[clear panther]

i have a calculus/differential equations/complex exponential question:

[clear panther]

my end goal here is to have a non-linearly decaying differential equation to start from, for which no solution exists

[full forum]

@clear panther Has your question been resolved?

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@clear panther Has your question been resolved?

[hoary wyvern]

Okay so thats not the answer that fits here
Think about the i" you put
You use "i" when you have linear second ordinary diff equation

Some thing else should be your typical answer for this type of diff equation

[full forum]

@clear panther Has your question been resolved?

[hoary wyvern]

You want the typical answer for this question?

When you see a linear diff equation
Just put e^pt
You will get a equation (its not diff anymore)
That would give you the p

So
For x= e^pt
Dx/dt=p * e^pt
This is equal to ax
Thats a * e^pt
You can cancel e^pt
And a = p
So x is e^at

Check it yourself

[clear panther]

I'm not sure you understood the question?

[clear panther]

I'm not trying to solve the differential equation. I'm trying to transform it from an exponential decay to an exponentially decaying complex sinusoid without solving the differential equation

[spice knot]

how did you "deduce" this new differential equation?
are you aware thats d/dt of x times e^iwt, which must be distinctly different than multiplying the real-valued x by a complex factor (a + iw)?

[clear panther]

i deduced it trivially, since the new solution is just the old solution with a different exponent factor

second approach is correct in theory

[spice knot]

thats not how derivatives work

[clear panther]

but in practice i'm seeing the reverse

yeah i'm baffled too

[spice knot]

you know (fg)' isnt f'g'

[clear panther]

i know

[spice knot]

so you cant multiply by a certain factor and expect it to work

[clear panther]

see the second approach :)

the baffling thing is that the first approach yields the correct output

[spice knot]

let me repeat, how did you justify this deduction

[clear panther]

i know that it should be wrong

but it isn't

[spice knot]

in other words, you have no words to justify this deduction

get some words out

[spice knot]

dx/dt = ax fundamentally has no complex sinusoid solution because a sinusoid rotates and "dx/dt = ax" doesnt ever rotate because a is real
it only rotates if a is complex: dx/dt = Ax where A is complex will decay (re(A) < 0) and will rotate (im(A) ≠ 0)
this matches your requirements for a differential equation of a decaying complex sinusoid without resorting to inserting factors that may work

[clear panther]

how do you explain that numerically integrating the result of the first approach gives me a perfect match with xe^{iwt} though?

[spice knot]

explain your deduction, Im certain the only issue here is a typo that you are inadvertently defending

find this typo and fix it

"deduction" is not enough

[clear panther]

let me check my matlab code to be sure

[spice knot]

the typo cna be seen in the image

its there from step 2 to step 3

[clear panther]

clear; clf;
hold on

Ts = 1/100; % sample time (Hz)
T = 10; % time length (s)

t = 0:Ts:T; % time

a = -0.1; % feedback gain
f = 4; % resonance frequency (Hz)
x0 = 1;

w = 2*pi*f; % radial frequency (rad/s)

% linear, real

dxdt = @(t, x) a*x;

xn = [x0, zeros(1,length(t)-1)];
for i = 1:length(t)-1
    [~, s] = ode45(dxdt, [t(i) t(i+1)], xn(i));
    xn(i+1) = s(end);
end

plot(t, xn, "DisplayName", "linear, real");

x = @(t) x0*exp(a*t);

plot(t, x(t), "DisplayName", "linear, real");

% linear, complex

dxdt = @(t, x) (a + 1i*w)*x;

xn = [x0, zeros(1,length(t)-1)];
for i = 1:length(t)-1
    [~, s] = ode45(dxdt, [t(i) t(i+1)], xn(i));
    xn(i+1) = s(end);
end

plot(t, real(xn), "DisplayName", "linear, complex");

x = @(t) x0*exp((a + 1i*w)*t);

plot(t, real(x(t)), "DisplayName", "linear, complex");

axis([0 T -1 1]);
xlabel("time (s)");
ylabel("amplitude")
legend();
hold off

this is the matlab code

[spice knot]

yea youre not finding the error in here my guy

its up there in the doc

and so if its in the code, its buried

[clear panther]

no, this code works

and uses the result of the first approach

[spice knot]

thats even worse

that means the code skips the typo

so the code wont help you here

the typo is here:

[clear panther]

i understand your point lol

[spice knot]

so you found the typo?

[clear panther]

nope

[spice knot]

think of it this way

[spice knot]

the x in the lower part of this image, what does it represent

$x_0e^{at}$ or $x_0e^{(a+i\omega)t}$

[glossy valve]

matt07734

[spice knot]

the typo is when you essentially typed one of these for the other

[clear panther]

yes, i understand your point

and i know this shouldn't just work

second approach should be correct

[spice knot]

its just a typo, you need to find the typo

you see me talking about approaches here? the method of inserting factors and vaguely saying "deduction" is risky because it allows typo that dont care about what you know is correct

what you can also do is consider $X(t)=x_0e^{(a+i\omega)t}$

[glossy valve]

matt07734

[spice knot]

youve correctly figured out that the differential equation for this is dX/dt = (a + iw)X

however in your 'deduction' haste you didnt write this too correct

but youre still very close, do you see the typo

[clear panther]

$${dX\over dt}=x_0(a+i\omega)e^{(a+i\omega)t}$$

[glossy valve]

antiprosynthesis

[clear panther]

$$=e^{i\omega t}(a+i\omega)x$$

[glossy valve]

antiprosynthesis

[clear panther]

aka the result of the second approach

but hm

[spice knot]

break down this "deduction" of yours

[clear panther]

ok X vs x mismatch there

i see

[spice knot]

yea thats the typo

[clear panther]

so that's why first approach is actually correct

[spice knot]

yes I was about to say

naturally the code wouldnt do this since youd be in the weeds attempting to numerically use X

neither would the second approach

[clear panther]

the code gives me the correct result, but the "deduction" was indeed lackluster :)

[spice knot]

should be a better word than deduction

cant think of one

[clear panther]

yeah, just had to use some word to jot this down into a question

guess i'll mark this one as solved. or at least as "back to the drawing table"

thanks for the help

[spice knot]

you can have it be a "turns out this approach isnt different after all, so I do have one approach and its solid on all fronts"

np

[clear panther]

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[pearl vault]

v42 is that

hello my friends, my senseis, I’m looking to make a statistical matrix for my paper role-playing game. I tried to simplify the data as much as possible to obtain a balanced matrix, but after 4 months I am at version 43 and it is still not functional or balanced. I seek to make statistics without bonus features, race, or equipment just pure stats of class lvl 1. I understood over time that it should make a system of increase of stats per fixed level based on precise figures but I would like to obtain methods to quantify and visualize my balance. I did a lot of testing on rpg proggrams to get a vvisuel of the results but I’m never getting a perfect result or approaching it. the stats to take into account are the basic physical and magical bonus draws, life and mind, armor and magic armor, endurance and mana (in 50 class)

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[pearl vault]

shit

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[lapis tangle]

part iii) :(

[lapis tangle]

<@&286206848099549185>

nvm i am so dumb

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[lapis tangle]

part c

[spice knot]

two vectors are perpendicular <==> dot product is 0
OP and AB are perpendicular <==> OP * AB = 0
then solve for t

[lapis tangle]

i get the concept but

how would i solve for t here

[spice knot]

OP * AB = 0
(OA + tAB) * AB = 0
OA * AB + tAB * AB = 0

[lapis tangle]

thank you i hate math

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[spice knot]

np

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[limpid ocean]

What am i doing wrong here?

[limpid ocean]

To find the derivative of the function

[spice knot]

maybe write ln(√x) as ln(x)/2 if the answer looks off from what it should be

[nova sigil]

Also always factorise exponentials

You can pull out $e^{\sqrt{x}}$

[glossy valve]

Max

[limpid ocean]

oh yeah i just didnt continue factoring cause i assumed it was wrong mb

Thanks

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[digital dock]

I dont know how to start on finding where the particle is parallel to i, this is what ive done so far but I have no clue where to go from here, i know there should be no movement in the i direction but no clue how to actually make it there

[digital dock]

wait i did it wrong let me fix that and repost it

updated working

<@&286206848099549185>

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[digital dock]

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[torn jolt]

hello. do i have to open the brackets or is there a direct way to do this

[torn jolt]

yes you have to open the brackets i think

oh

😭

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[shrewd seal]

proof that sec^4-tan^4=1+2tan

[shrewd seal]

i havw no clue about this one

[grave elm]

That doesnt feel right

are u sure it's correct

[glossy valve]

MethIsAlwaysRight

[shrewd seal]

yeah

i only know that 1+tan^2=sec^2

[grave elm]

Okay

[grave elm]

This could be useful

[grave elm]

try simplyfying this RHS expression using that

Try rewriting $1+2\tan^2(x)$ using that identity you just wrote

[glossy valve]

MethIsAlwaysRight

[shrewd seal]

no,u have to prove that lhs=rhs

i cant take rhs

[grave elm]

Yes, that's right

[grave elm]

You can replace the RHS with something that is equivalent to the RHS, cant you?

and what is equivalent to 1+2tan^2(x)?

if 1+tan^2=sec^2

[grave elm]

sorry, rhs i had a typo there

[shrewd seal]

ok so, what i did is that squared the lhs like (sec2)^2 - (tan2)2

and then i put the formula a+b a-b

[grave elm]

That's correct

[shrewd seal]

and i got (1) (sec2+tan2)

[grave elm]

perfect

[shrewd seal]

then i put tan2=sin/cos

idk what to do after

and 1/cos

[grave elm]

I have a different idea

1+tan^2=sec^2

[shrewd seal]

cus reciprocal of sec

[grave elm]

try using this ^

[shrewd seal]

how i make it tan 2+sec2

[grave elm]

With this, the identity reduces to $\sec^2(x) + \tan^2(x) = 1 + 2\tan(x)$

[glossy valve]

MethIsAlwaysRight

[shrewd seal]

yeah what after

[grave elm]

Let me flip it around, sec^2(x) = 1+tan^2(x)

now it might be slightly more obvious

there is a nice substitution you can do

[shrewd seal]

no it isn’t

ok

wait

OHHH

[grave elm]

You got it?

[shrewd seal]

1+tan2+tan2

YEAHH

[grave elm]

Perfect

[shrewd seal]

THANKSSSSS ALOTTTT

[grave elm]

You are welcome

[shrewd seal]

i should have attempted it 😭💔

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[plush egret]

Could someone give me a hand with what I'm pretty sure is a simple problem?

Sketch the characteristics for $$2 \pdv{u}{x} - 3 \pdv{u}{y} = 0$$ subject to $u(x,1)=6x$ for $x>0$

[glossy valve]

jan Niku

jan Niku

jan Niku

[plush egret]

we should be able to track out an ODE right? that has the relationship between y and x

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@plush egret Has your question been resolved?

[plush egret]

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[cunning ledge]

How will I do a and b

Like there is no path which takes me there

Is AB = 5/2a - a

[restive ore]

oh do you mean AB

[cunning ledge]

Ye

AB

[restive ore]

https://www.mathematicsdictionary.com/english/vmd/full/v/vectordifference.htm#:~:text=vector difference&text=The result of subtracting two,the rule on vector subtraction.

Be careful that you don't find BA instead of AB

[cunning ledge]

So b - a

[restive ore]

I think so yeah :)

[cunning ledge]

Thanks

[restive ore]

nw

u might wanna .close now

[cunning ledge]

Same for question a right

[restive ore]

wut

[cunning ledge]

Question a here

[restive ore]

do u mean part A?

No, its not the same

[cunning ledge]

Ye

[cunning ledge]

So I simply write this

[restive ore]

it's just ratios for that part

[cunning ledge]

OY = 2b

That’s all?

[restive ore]

mhm

and OX = (5/2) * a

[cunning ledge]

Kkkk

Thanks mate

https://tenor.com/view/whos-awesome-reaction-you-are-you-are-awesome-pointing-gif-15769676

[restive ore]

then you can do part C with vector difference of OX and OY

aww thanks

[cunning ledge]

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[wanton basin]

I need help with this

[wanton basin]

I need to find F'(x)

Isn't it just the same function but replaced with x?

So F'(x) =

Because the d/dx of the integral cancels out? Or am I tripping

[spiral vigil]

,tex .FTC2

[glossy valve]

hayley!

[wanton basin]

So this?

Just plug and play?

A little odd I didn't see that stipulation for the ftc2 in my textbook

I just have this

and I used integral properties to flip the 0 and x^3 to negative integral of x^3 to 0 of the same thing

[spiral vigil]

good idea

but yeah ftc2 is useful

[wanton basin]

so which version do I use?

[spiral vigil]

they should be the same

[wanton basin]

both would be correct right?

I get a different result for both though

[spiral vigil]

oh i see well um

[spiral vigil]

this isn't going to help you that much because you don't have x on your integral

you have x^3

[wanton basin]

oh right

I would need to use u

[spiral vigil]

yeah

[wanton basin]

hmmm

[spiral vigil]

and figure out the d/dx situation

[wanton basin]

it's a bit weird with t though

since usually I replace dx with du/something

but now its dt

but maybe its irrelevant?

[spiral vigil]

you'd replace the dx on the outside

the one in the d/dx

but also like. this is going to be very tedious lol

[wanton basin]

I hate calc II

I'll be honest

[spiral vigil]

just use ftc2 that i posted

[wanton basin]

alright

I'll just go with that

and come back later to it

[spiral vigil]

ye

[wanton basin]

1 last question

something is bugging me about a trigonometric properties problem

This thing

[spiral vigil]

yeah

[wanton basin]

It looks so close to one of these

#1 is almost dead on

Except for the fraction issue

[spiral vigil]

but very different because denominator haha

[wanton basin]

yes

[spiral vigil]

the way i usually do these is like

start with $\sin^2 + \cos^2 = 1$

[glossy valve]

hayley!

[spiral vigil]

and then divide by sin^2 or cos^2 if i feel like it

and then try to manipulate it so that i have something matching the pattern i want

[wanton basin]

How do come up with a sin cos manipulation in this case though? I don't see the connection immediately

do I*

[spiral vigil]

try subtracting sin^2 from both sides ^^

[wanton basin]

cos^2 = 1 - sin^2

[spiral vigil]

yeah

multiply by 4?

[wanton basin]

I had thought an idea would be 4-x^2 as u

and try to get it into another form

but that leads nowhere lmao

[spiral vigil]

nah that's not going to help you here

[spiral vigil]

this is almost there

just multiply by 4 to get 4cos^2 = 4 - 4sin^2

and then notice that 4sin^2 = (2sin)^2

[wanton basin]

mmmmm

[spiral vigil]

so $(2\cos\theta)^2 = 4 - (2\sin\theta)^2$

[glossy valve]

hayley!

[wanton basin]

I see, oh that's a suspicious substitution

I would never see cos^2x + sin^2x = 1 for this on my own

[spiral vigil]

this is where you should pretty much always start

[wanton basin]

alright cool I'll try to work it out

thanks for the help

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[tall crypt]

When doing this question would I get the derivative of f(x) first?

[harsh pilot]

I don't know, what is f(x)?

[tall crypt]

(x^2 - x + sinx)/2x

[harsh pilot]

In that case, no. Perhaps you are thinking about L'Hôpital's rule.

But in L'Hôpital, you find the derivative of the numerator and denominator separately.

[hot herald]

personally I'd split it into 3 fractions and avoid derivatives

[quick cairn]

yeah no need for l'hopital here

lim sinx/x is a known limit

[hot herald]

limit identity can be used for the sin(x)/x

[tall crypt]

So, (x^2)/2x-(x)/2x+(sinx)/2x?

[harsh pilot]

Yep

[tall crypt]

I got 0 as the answer

Thank you!

[harsh pilot]

I also got 0.

Your thinking is right, but your steps look fishy.

[tall crypt]

How?

[harsh pilot]

Like you start out with a limit in the first line, but in the next line, you have no limit, but you still have x's.

Also you have no equality signs.

And this bit looks funny:
It says that -1/2 = 0

[tall crypt]

Is that better?

[harsh pilot]

Yeah, much better. 👍 I would also pop in some equality signs to be clear what is going on.

Like

[tall crypt]

Okay, that makes since. Thanks

[harsh pilot]

np

You could also expand on the step from the limit of all three terms, if you want to be super clear.

Like $$\lim_{x\to 0} \left(\frac{x^2}{2x} - \frac{x}{2x} + \frac{\sin{x}}{2x}\right) = \lim_{x\to 0}\frac{x^2}{2x} - \lim_{x\to 0}\frac{x}{2x} + \lim_{x\to 0}\frac{\sin{x}}{2x}$$

[glossy valve]

OneTrackPony

[tall crypt]

Yeah, I don’t know why I didn’t do that😂 Thankyou

[harsh pilot]

Might just be your personal notes, who knows. 😄 np

There's usually a difference between people's personal scribbles and something they need to be readable for other people

[tall crypt]

Yeah, that’s why I usually make a first draft of my homework then clean it up a bit