help-49

[last slate]

integration byparts

[rose trout]

It's essentially about applying the chain rule

You have
$\int_{u(x)}^{v(x)} g(x) dx = G(v(x)) - G(u(x))$ from the Fundamental Theorem of Calculus.\
Then taking the derivative on both sides and applying the chain rule on the RHS will yield
that $$F'(x) = \frac{dv}{dx} \cdot G'(v(x)) - \frac{du}{dx} \cdot G'(u(x)) = \frac{dv}{dx} \cdot g(v(x)) - \frac{du}{dx} \cdot g(u(x))$$

Here we're assuming that g(x) is xe^x and G(x) has the property that G'(x) = g(x), i.e. it's the antiderivative.

[grand pond]

Azyrashacorki

[last slate]

FTC 2

[cosmic river]

is this problem using this

[rose trout]

No. You don't actually have to compute the integral here.

If you did, that method, integration by parts, would be needed to compute the antiderivative of xe^x. But you don't need to do that here

[cosmic river]

you mind checking this real quick

I think am getting a better understanding

but am still fucked

sec

[rose trout]

I think the reason this is confusing is that the choice of using the same variable inside the integral as in the bounds is making it look as the x in the bounds and the integration variable are the same. They're not.

[cosmic river]

find the area bounded between y = cosx and y = cos^2x over [0, pi]
firstly we evaluate points of intersection by setting equal to zero, I got 0 and pi/2

valid way to solve right

I have my exam tomorrow and I need to understand this really soon lol I fucked myself

[cosmic river]

could you show me how its done

for the f(x) given

x e^x

so I can see it

[rose trout]

I'll do the weird bounds problem then I'll come back to the area one

[cosmic river]

ok

[rose trout]

So we know that $F(x) = \int_{0}^{\sin(x^2)} t e^t dt$. Let's say that $g(t) = t e^t$. Then we could integrate that and get something like $G(t)$ such that $G'(t) = g(t)$, right?

[grand pond]

Azyrashacorki

[cosmic river]

yea

[rose trout]

We don't need to do it though. Since we have that, notice that with the fundamental theorem of calculus, one has that $$F(x) = \int_0^{\sin(x^2)} t e^t dt = G(\sin(x^2)) - G(0)$$.

[grand pond]

Azyrashacorki

[rose trout]

So then we can just differentiate to get F'(x) :

$$F'(x) = (G(sin(x^2)))' - 0 = G'(\sin(x^2)) * (sin(x^2))' - 0$$, but $G'(t) = g(t) = te^t$, so this yields $$F'(x) = \sin(x^2) e^{\sin(x^2)} \cdot 2x \cos(x^2)$$

[grand pond]

Azyrashacorki

[cosmic river]

Why is G(0) = 0

[rose trout]

G(0) is constant w.r.t. x, so since we're taking the derivative it vanishes.

[cosmic river]

where is the e^sin coming from

[rose trout]

Notice that g(t) = t * e^t, so then g(sin(x^2)) = sin(x^2) * e^(sin(x^2))

[cosmic river]

Ok

I'm still confusd why G(0) is constant

isn't G'(x) = g(x)

so G(x) still has an x in it

[rose trout]

We're taking the derivative with respect to x on both sides. G(0) doesn't have any x in it

It doens't have any t either

[cosmic river]

ok

if it was G(1)

you would have a t

but because its a zero its gone

?

[rose trout]

No it actually doesn't matter the variable in it.

We defined G(t) such that G'(t) = g(t). It's still just a function of t, so plugging in something constant will yield a constant through and through

[cosmic river]

Ok I see

can you have a look at this one

I did y^3 + 2y^2 + 1 = y^2 + 1

I got y^2(y^3+2) = 0

y = 0

y = ^3sqrt(-2)

[rose trout]

That's fine. Now that you have the intersection, you can integrate over their difference,

[cosmic river]

Ok ima go for now ty a lot for helping me both @rose trout @low moon

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[last slate]

hi again, can someone help me make a tree diagram or grid table for this question “if two dice are rolled, how many possible outcomes are there having a dot sum less than 10”

[last slate]

consider all the choices like last time

a dice can be 1 to 6

so what numbers between 1 and 6 sum to less than 10 together

oo okay

hold on so like compatible for 6 are only 1,2,3?

sure yes

okay and so 1 can be okay with 1,2,3,4,5,6?

yes

OKAY, THANK YOU 🙏🙏 you're really helpful

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[last slate]

idk what a "grid table" is supposed to be, but just write a table with the rows denoting the first dice's dots (from 1 to 6, so 6 rows) and the columns denoting the second dice's dots (similarily 6 columns) and just fill in all the possible combinations

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[gritty solstice]

i need help with geometry

[gritty solstice]

here a picture

given that AB = AC

i have to prove that the radius of the circle outscribed of ABD = to the radius of the circle outscribed of ACD

i did.

now i have to express the area of ACD using R and alpha.

i ll send the solution please tell me whats wrong

if the radii are equal of both circles why do i get different results for AC when using the sine theorem??

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[ionic thicket]

idrk why is the shaded region that portion of the graph, can someone explain rq thx

[distant vigil]

The line is -1/2 - 2x which is the dotted line

[ionic thicket]

yup

[distant vigil]

And then the value of y is above (>) that line's value

[ionic thicket]

oh shit

😭

thx

[distant vigil]

XD

[ionic thicket]

did not see that

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[hard edge]

Find log21 9 if log3 7=a

I can't come up with a way to solve this

Looks so easy too

Please help

<@&286206848099549185>

[astral garnet]

try change of base

[hard edge]

Like?

[hard edge]

I should change base of second one right

[astral garnet]

i dont think it should matter

[hard edge]

@astral garnet thank you bro I just solved it

The answer is 2/1+a

[astral garnet]

no worries

yeah thats correct

[hard edge]

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[opaque timber]

Ok

[opaque timber]

<@&286206848099549185>

[lyric acorn]

explain your question first, then wait 15 minutes if nobody has showed up man.

[opaque timber]

Ohhhh sorry sorry I'm new here

[lyric acorn]

No worries man.

[opaque timber]

Very stuck on this question don't really understand pre algebra like that.

[lyric acorn]

Oh.

[opaque timber]

Like that?

[lyric acorn]

That's great.

So.

To help you, know that $8z$ is simply $8*z$

[grand pond]

the catmullah

[lyric acorn]

And when you know what z is, then you can simply plug it in the equation: 8z - 9

[opaque timber]

Ohhh.

[lyric acorn]

You there?

[lyric acorn]

?

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@opaque timber Has your question been resolved?

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[harsh cape]

how is this true?

[harsh cape]

how does lhs = rhs

[slender walrus]

what's (*)
is that referencing something that appeared earlier?

[harsh cape]

no

it isnt

• is not explained

[slender walrus]

are you sure?

[harsh cape]

yea

i checked

multiple times

[slender walrus]

is this the first part of the worked solution

what was the previous question

[harsh cape]

thats the first part

[slender walrus]

what was the previous question

wait

ok, its right there in the work

[harsh cape]

wait what it is?

[slender walrus]

second line

its just a manipulation of that equation

[harsh cape]

ohhhhhh

thank you so much

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[main grove]

i cant solve n 162

its supposed to result -5

[latent elk]

Its just long and painful

[alpine gyro]

Multiply common denominator on both side.

[main grove]

how

[alpine gyro]

By finding the smallest denominator, which is 24.

[latent elk]

Well, if u look at the denominator for x, its 2 3 4 so, u can try multiply each side to get the 24 term

[main grove]

oh yeah i tried that and it resulted x=1

[alpine gyro]

Left side, second expression, did you multiply 12 with all the terms in the bracket?

[main grove]

yes

[alpine gyro]

Remembered sign changes?

[latent elk]

Should be -5

[main grove]

YES IT SHHOULD

but when i do it its 1

[latent elk]

It would be nice if u can do step by step on paper then post it here

So we can see on which steps you made the mistake

[main grove]

ok

one sec

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[main grove]

i send it

bro

[latent elk]

Im pretty sure u made a mistake in this step

[torn compass]

that looks so much like a 7

[main grove]

nah the seven has a line

[main grove]

where

[torn compass]

if i were to show u this , is this 1 or 7?

[main grove]

a 1

because the 7 has a line

[torn compass]

oh u write 7 like that

[main grove]

yes

[torn compass]

ur mistake btw

[main grove]

wait how is it wrong?

i dont understand

[torn compass]

$$- \frac{x+3}{4} = +\frac{-1(x+3)}{4}$$

[main grove]

whats this

[grand pond]

JustToPro

[torn compass]

fractions with a negative can be a bit confusing , but that -1 is being multiplied to the entire numerator

$$- \frac{x+3}{4} \neq \frac{-x+3}{4}$$

[grand pond]

JustToPro

[main grove]

so if it aint -6x + 18

what is it?

[torn compass]

$$-1(6+18)$$

[grand pond]

JustToPro

[main grove]

-6 - 18?

where did the x go

is it -6x -18

😔

[uncut drift]

i think that's what he meant

[torn compass]

yea

[main grove]

but it still doesnt result

😔

[torn compass]

@main grove another mistake i found

x-3 just turns into x+3 randomly

[main grove]

how can i do so many mistakes

yes but it doesnt result

[torn compass]

isnt the answer -1?

[main grove]

no

its supposed to be -5

[torn compass]

wait ima solve this and check

[main grove]

but it only results -1 to me

[torn compass]

ok ok i found it

ok i got -5 yea

[main grove]

HOW

[torn compass]

x+2 changes into x-2
-6 randomly appears

[main grove]

ok so lemme re try

[torn compass]

for sure u getting -5 on this one

[main grove]

still not resulting 😢

[torn compass]

tf thats further from -5

k

[main grove]

what am i doing wrong

[torn compass]

thats a +18 not -18

[main grove]

ofc

[torn compass]

$$-1(x-3) = -x + 3$$

[main grove]

yes because it changes sign

[grand pond]

JustToPro

[main grove]

it changes sign if you move it in the other thing

[torn compass]

before moving

its +18

[main grove]

didnt you tell me it was -18

[torn compass]

its -18 when u wrote it wrong

when u wrote that x+3 then it was -18

but it was actually x-3 , that means its +18

[main grove]

yes

but it still doesnt result

[torn compass]

howww

[main grove]

you said you got -5

[latent elk]

u 2 still stuck on this?

[main grove]

yes

[torn compass]

ok yeah i found it

another mistake

[latent elk]

lol

He does make quite a lot of mistakes

[main grove]

💀

[torn compass]

there are supposed to be 3 -6x

2 from the fractions which cancel out and another from the other fraction

[main grove]

thing is i looked at this for half an hour and said "shit theres no mistake "

[torn compass]

tbh im quite dumb :p

so its a mix of both thats taking this long

[main grove]

so what do i have to change

[torn compass]

$$6x-6x-6x+18+4x+8+3x = 2-6x-11$$

[grand pond]

JustToPro

[torn compass]

this is what u have now

simplifying gives

$$ x + 26 = -6x-9$$

[grand pond]

JustToPro

[torn compass]

further simplifying gives
$$7x = -35$$
$$x = -5$$

[grand pond]

JustToPro

[latent elk]

start

[torn compass]

ez

[latent elk]

Stop

3 mins to finish it! Yay

[main grove]

finally

tho

i have to do ANOTHER one

[latent elk]

Just be careful bruh

A miss is as good as a mile 💀

[torn compass]

the next one looks a bit harder tho , but u can get it

it has x^2 so more terms , but it would be fine

[torn compass]

what is blockstate btw? is it like black plasma studios that make mc animations?

they changed thier name to square media btw

[latent elk]

No, we make Minecraft Addon for Bedrock Edition, like mods on Java

[torn compass]

oh ok

[latent elk]

Bruh

[torn compass]

yes ik

[latent elk]

It got cut out

[torn compass]

ima guess its a 3

[latent elk]

Assume its 0 then write "NaN" for solution 💀

[torn compass]

xd

btw each question is solvable in under a sec

if u just go "multiply both sides by 0"

ez 0=0

[latent elk]

"Clearly the answer is ...." 😫

[torn compass]

well i gtg , gotta study some conic sections for a test 🤞

[main grove]

wait

imma do the 163, if it doesnt result imma send what i did

[latent elk]

@main grove For the love of god please do it step by step

Not mess by mess

[torn compass]

its more fun that way tho

it proofs how smart u are

[main grove]

GUESS WHAT

didnt result

but its actually pretty close

help

[midnight plank]

@main grove Has your question been resolved?

[midnight plank]

@main grove Has your question been resolved?

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[zealous mauve]

For this do i need to use lagrangian to see what k actually is

[zealous mauve]

Like before i do the diagram

I got k = 1/5a but not sure if that’s right

Markscheme says that k is large so that means my answer is wrong?

[dull yoke]

where is k coming from

[zealous mauve]

lol my bad k = x^2 + y^2

Not sure why my prof did that

I think to just show its a circle idk

[sage helm]

can't you just sub in the condition and use regular derivatives

[dull yoke]

i mean, u can do it with lagrange, but u can do single var calc if u want too

makes it super easy tho

[zealous mauve]

Do u mean treat as simultaneous equations?

[dull yoke]

solve ur constraint for x or y, then plug into ur objective function, then differentiate ur new objective function, and find critical points

[zealous mauve]

Yeah have a feeling my prof wont want me to do that which sounds so dumb

I don’t really get the luxury of choosing methods

Now im getting a = 5/2lambda which also seems wrong

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[zealous mauve]

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[loud dawn]

I have a 2d pattern (for a spatial hashing algorithm), and I need to make 2 functions that would reverse an index (the numbers in the pattern) into the x (top bolded row) and the y (left bolded column). I do have a function that can generate the index from the x and the y (https://www.desmos.com/3d/9qdbtpknyt), but I can't figure out how to reverse it at all. I do have the liberty of using a programming language, so piecewise functions and such are available, but the function basically has to have O(1) time complexity (eg. it can't use iteration or anything)

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@loud dawn Has your question been resolved?

[little jungle]

if x>y: f(x,y) = x^2 + y

so i think that you can just take floor(sqrt(n))

that should be x, and y would be n-x

if y>x: then it becomes y^2+2y-x

[loud dawn]

I don't know what x and y are beforehand, so I can't see which one is greater

[loud dawn]

and would there be a more efficient way for this computation wise?

I did some research on integer square rooting, and they all had their downsides (floats lose accuracy at 2**53, other methods were all not O(1))

[little jungle]

Precomputed table i guess

[loud dawn]

actually

I'll just round down a float I guess

x and y can only reach 2**16

[loud dawn]

and i can figure out this myself then

[loud dawn]

thanks for the help

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[forest ginkgo]

ive been trying for a very long time trying to figure this one out but i just dont know how to approach it

[forest ginkgo]

would the power reducing identities help?

i would reckon, but those werent taught when this homework was assigned

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[tribal tartan]

just expand the left hand side

it’ll become quite obvious when you do that

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[last slate]

obviously the distance is positive but the equivalence of a & b with the remark isnt clear how we get positivity

we just show 0 iff d(x,x)

could be negative if not d(x,x)

[visual tiger]

no, d:X x X -> [0,+inf)

[last slate]

ik that

[visual tiger]

so it can't be negative

[last slate]

but then why call b positivity

[visual tiger]

because it's positivity

[last slate]

just say neq 0

[visual tiger]

we only know d is non negative

[visual tiger]

it's the same

"just say" it's the same thing

[last slate]

they make these definitions less clean than they could be

d(x,x)=0; d(x,y)!=0

[visual tiger]

d(x,x)=0; d(x,y)>0

what's the difference in "cleanness"?

[last slate]

you are talking about a single case vs a set

[visual tiger]

and?

[visual tiger]

how is this less complex

[visual tiger]

than this

[last slate]

they are opposites without talking about the codomain of d

[visual tiger]

?

i don't see how they would be "opposites"

[last slate]

=0, !=0

[visual tiger]

ok?

Maybe it makes more sense to you but really I think we're diving into subjective territory

[last slate]

x=x => d=0
x!=x => d!=0

we are

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[crisp obsidian]

How did they get from $[T(x)]{\gamma} = [T]{\gamma}^{\beta}[x]_{\beta}$ to $T(x) = Cx = Lc(x)$ for all x? Like how do they know they are equal for all x? I can’t seem to make the connection, as one of these are matrices, while another are transformations

[grand pond]

LXDL

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[gloomy helm]

how did the ^-2 become a -1?

[abstract bridge]

power rule of integration

[gloomy helm]

could you explain it pls?

[subtle blaze]

It’s the opposite of the power rule in differentiation

Verify this by differentiating what’s inside the brackets

[gloomy helm]

ohhhh

[subtle blaze]

You should get the integrand back

[gloomy helm]

omg i really forgot that

XD

thanks!

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[tiny gate]

How can you tell which of these the q function is

[tiny gate]

In the textbook the method is subbing in at least one value as zero?

I’m assuming that if it’s neither then when you sub in different values eg x1 =0, x2 = 1, x3 = 1 and then x1=1, x2=0, x3 =1, you get positive and negative answers?

Just trying to get my head around the more nuanced examples :/

Is the q function positive definite?

<@&286206848099549185>

[still nacelle]

I'm not sure what the definitions are for you but the way I learnt this was to write it as a 3x3 matrix $A$, and check that the determinants of three principal minors $A^{(1)}, A^{(2)}$ and $A^{(3)} = A$ are all positive, then the matrix is positive definite. Principal minors are the upper left 1x1, 2x2 and 3x3 matrices of $A$

[grand pond]

lgkoo

[tiny gate]

Yeah that’s a good method, what’s really strange is though is that our prof told us to skip that section

But is still asking these questions on past papers bruh

Is there any other way?

[still nacelle]

not that I can recall soz

[tiny gate]

I just found this from my profs notes

In your opinion would u even say this is valid

Icl this is a joke that he made us skip the solid method

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[unique stream]

Hm I have a matrix
0 -1 -1
0 2 0
1 4 -1

[unique stream]

and determined its characteristic polynomial x³-x²-x-2

[unique stream]

Its elements are from a set F3 btw, but I consider it isomorphic to Z/3Z

The eigenvalues I found are 1 & 2 but I'm not certain whether they are correct

Since for 1 I get an empty eigenspace

[midnight plank]

@unique stream Has your question been resolved?

[odd solar]

If its over F_3 the you can’t have negatives

The only numbers available are {0,1,2}

[unique stream]

Hm why would that be

[-1] would be in the equivalence class of [2] for instance

[odd solar]

Thats exactly right

So you have to begin by transforming the matrix into its equivalent form over F_3

[unique stream]

yh but I still receive 1 & 2

[odd solar]

Can you show your working

[unique stream]

shouldn't change under equivalence properties

sure a sec, different setup

For the matrix we'd have
0 2 2
0 2 0
1 1 2

and the characteristic polynomial is
x * (x-2) * (x-2) - 2 * (x-2)

inserting 1:
1 * (-1) * (-1) - 2 * (-1)

= 1 + 2 = 0

making 1 a valid eigenvalue @odd solar :/

[robust isle]

how did you get the eigenspace ?

[midnight plank]

@unique stream Has your question been resolved?

[unique stream]

Well Av = 1v

Which gives me only a set containing the zero vector

[autumn canopy]

Can confirm, using Sage

[unique stream]

Hm yeah but isn't it odd that 1 would be a valid eigenvalue then

[robust isle]

ofc lol

I was wondering about how you solved the system

maybe there's some reduction mod 3 you didn't notice somewhere

actually

A - I is
[-1 -1 -1]
[ 0 1 0]
[ 1 4 -2]

if you reduce the last row mod 3, you get [1 1 1]

so row 3 = - row 1

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[wanton shore]

Range of sin^-1(x)+cos^-1(x)+tan^-1(x)

[wanton shore]

Isn't this (0,π)

??

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[dim shard]

anyone can help with chemistry balancing redox equations

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[last slate]

]

[last slate]

the area needs to be in teh form of aln(2) + b

however after integrating this between the points ln 3 and ln 6 (which are the points at which the graphs intercept), the area isn't coming in that form

$$[A = \int_{\ln(3)}^{\ln(6)} |(18e^{-x} - 6) - (3 - e^x)| dx]$$

[grand pond]

Emploice Muswashans

[last slate]

this woudl evaluate to

$$[A = \int_{\ln(3)}^{\ln(6)} |18e^{-x} - 9 + e^x| dx]$$

[grand pond]

Emploice Muswashans

[vague remnant]

Take an integral?

[last slate]

hmm yeah I did that

But I didn't quite get the answer

in that format

oh wait nvm I think I got it

[vague remnant]

Lemme see

Ok

[last slate]

sorry I think my brain was just being dumb

Thank you so much

sorry to trouble ya

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[north escarp]

.

[north escarp]

This is coulombs law i got this problem how should i solve it without prefixes

Should i input it without prefix im not really sure can somebody help me out this problem

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[vocal steeple]

how do i do this?

[crude pier]

try multiplying out the right hand side and then comparing coefficients to the left hand side

[vocal steeple]

i dont get what to do

i dont get what you mean

[crude pier]

do you know how to distribute?

for example (x+1)(x+2)

[vocal steeple]

yes

[crude pier]

alright

now do the same thing for the right hand side of the equation, (x-1)(x+a)(x-b)

[vocal steeple]

how do i do it in this case

i havent done it with 3

[crude pier]

start with the first two

(x-1)(x+a)

then once you get that result, multiply the result you just got like -> (...)(x-b)

it's the same procedure but you do it an extra time

[vocal steeple]

ohhh

what do i do no

w

[midnight plank]

@vocal steeple Has your question been resolved?

[iron raven]

What did u get

[vocal steeple]

x^3+ax^2-x^2-ax-bx^2-abx+bx

[iron raven]

Seems correct now solving for coeffs should give the value of a and b

[vocal steeple]

and how do i do thyat

[iron raven]

For x^2 -1+a-b should equal -5

Do the rest like that

[midnight plank]

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[warm widget]

I need help with question 1(ii) i am not too sure how to get the Phase current Ica

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<@&286206848099549185>

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.reopen

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✅

[warm widget]

Hi i need help for this <@&286206848099549185>

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[warm widget]

Not yet

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[burnt prairie]

Hey how do i paratetmerize #3?

I got something like

r(x,y) = <x, y, 3x-2>

then i got something like this for my integral

[midnight plank]

@burnt prairie Has your question been resolved?

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@burnt prairie Has your question been resolved?

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@burnt prairie Has your question been resolved?

[odd solar]

I believe you want to parameterize wrt r and theta

[burnt prairie]

hmm

<2rcostheta 2rsintheta r>?

[odd solar]

so the param should look like
[ \sigma (r, \theta) = (r\cos\theta, r\sin\theta, 3r\cos\theta -2 ) ]

[grand pond]

shsgd

[burnt prairie]

in this case what is r

[odd solar]

radius

[burnt prairie]

ah

is this like from the origin to the plane?

[odd solar]

no, it would be the radius of the cylinder

[burnt prairie]

is the radius of the cylinder like predetermined

so r = 2

[odd solar]

yes, but pre-emptively subbing in r=2 will lead to different result

[burnt prairie]

ohh

how come?

[odd solar]

not sure to be honest, but I computed it both ways and got a slightly different result

[burnt prairie]

interesitng

what did u get for both

[odd solar]

12sqrt10 pi and 20sqrt10 pi

[burnt prairie]

ah

im guessing the 2nd way is pre-emptively subbing in

cuz thats what i kinda did

[odd solar]

yeah

[burnt prairie]

12pirt(10) is correct

[odd solar]

I guess its because in the parameterization, the z component is a function of r

[burnt prairie]

hmm

[odd solar]

but i'm not certain

[burnt prairie]

or maybe if we lock r we are only saying its a circular projection?

actually that doesnt make too much sense

[odd solar]

anyway

[burnt prairie]

alr for the future ill try to rememeber that

im guessing its also because if we lock r we only have 1 variable

and that shouldnt give a surface

[odd solar]

well the only difference in the calculation I got was from the

$\abs{\sigma_r \cross \sigma_{\theta}}$

[grand pond]

shsgd

[burnt prairie]

ah

[odd solar]

root 10 vs r*root10

[burnt prairie]

how do you even caluclate a cross product if you only have theta

[odd solar]

[ \sigma (r, \theta) = (r\cos\theta, r\sin\theta, 3r\cos\theta -2 ) ]

[grand pond]

shsgd

[odd solar]

or do you mean when subbing in r=2

[burnt prairie]

when r = 2

ye

[odd solar]

I used the same param as you

but I guess thats where the problem lies

[burnt prairie]

ohh

[odd solar]

since its a change of coordinates you would need to account for the Jacobian

which is rdrdtheta

[burnt prairie]

ic

changing rectangular -> polar right

[odd solar]

yes

[burnt prairie]

alr ty

[odd solar]

I'd just stick to not subbing in just in case

[burnt prairie]

ye

[odd solar]

easier to spot mistakes

👍

[burnt prairie]

kk

!solved

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[heavy zealot]

how do i approach this problem, i tried seperating them into 3 surface and used stokes but got 0 as the answer for all of them. the answer should be 104pi/3
Image
am i suppose to use stokes in this case or just evaluate the flux

[midnight plank]

Please don't occupy multiple help channels.

[odd solar]

Evaluating the integral over 3 separate surfaces and summing should get you your answer

[heavy zealot]

like

using stokes theorem or no?

or just literally do the siurface integral

@odd solar

[odd solar]

Either way should result in the same andwer

[heavy zealot]

whattt

im so confused

i got 0

so for the cylinder

i did parameterization of (theta, z) = 2cos(theta), 2sin(theta), z

i then find the nromal vector by

and then change the vector field with the parameterization

then dot product and evaluate the double integra;

is this how it goes?

[odd solar]

I believe your parametrisation is wrong

[heavy zealot]

why

[odd solar]

Using cylindrical coordinates

[heavy zealot]

so (r, theta)?

how does the z come in then

[odd solar]

Ah no sorry, i’m tripping

The parametrization is correct

Then you need to evaluate
[ \int_{\sigma} \mathbb{F}(\sigma (r,\theta) \cdot \abs{\sigma_r \cross \sigma_{\theta}} rdrd\theta ]

[grand pond]

shsgd

[heavy zealot]

this is what I did

[odd solar]

Hang on let me get some paper

[heavy zealot]

ok

i tried to compute the line integral of hemisphere with the boundary using stokes theoreom, i got 0

for that

for the disk i tried to get the curl and dot it with the nomnrla vector with parameteization (r, theta) = (rcos(theta), rsin(theta), 0)

also gave 0

oh btw the answer should be 104pi/3

[heavy zealot]

once i try to solve by getting a curl and dot with nromal vector, all teh answer are 0, which is so confusing

oh btw i think its helpful to mention this problem is under the gauss theorem section

[odd solar]

I also ended up with 132pi for the cylinder…

Hmm.

[heavy zealot]

hmmmm

[odd solar]

And you say the surface integral over F of the cylinder should be 104pi/3?

[heavy zealot]

the whole thibg

my bad on miscommunication

[odd solar]

Then 132pi may be fine

Since flux on some patch can be negative too

[heavy zealot]

oh btw

for the hemisphere

is the parameterization of (theta, phi) = (2cos(theta)sin(phi), 2sin(theta)sin(phi), 2cos(phi) + 3)

correct?

[odd solar]

For what bounds of theta and phi

[heavy zealot]

theta should be 0 to 2pi

phi should be 0 to pi/2 since its a upper hemisphere

[odd solar]

Yeah I believe that should be correct

[heavy zealot]

okay

[odd solar]

Have you calculated the integral over the hemisphere yet?

[heavy zealot]

not yet im doing rn

the normal vector is tough to calculate

its so long

is it supposed to be this long

[odd solar]

Yeah

Stokes theorem would be better for hemisphere

In retrospect

Sorry its very late so my brain isnt 100%

[heavy zealot]

hmm i see

ill try stokes

it still gets kinda shitty

[odd solar]

Stokes should give a line integral

[heavy zealot]

oh what

oh the boundary just the circle

[odd solar]

Yeah

[heavy zealot]

I got 0

What

this is so weird ive tried using strokes

and it just gives me 0

[odd solar]

I’m falling asleep and my brain is not working, i’ll write up a solution in the morning if no else gets to it

[heavy zealot]

okay

thanks for the help tho

[midnight plank]

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[serene skiff]

do i solve the second problem the same w ay as i solve the first problem even though the first is a power and the second is a maclaurin?

[sharp coral]

a maclaurin series is a particular type of power series

[glossy vortex]

https://youtu.be/BDFbOwkd0KA?si=B3zxnP8A9flFRdgC

[midnight plank]

@serene skiff Has your question been resolved?

[serene skiff]

@sharp coral iz this right so far? Is y prime of 0 negative 2

[sharp coral]

seems right so far

[serene skiff]

i got that the first four terms are 1 -2x + x^2 -(x^3)/3 does that seem right

[serene skiff]

actually wait aren't the equilibrium points just (0,0) and (-1, 0) then?

<@&286206848099549185>

[olive granite]

Can u help me my homework? @sharp coral #help-42

[midnight plank]

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[serene skiff]

<@&286206848099549185> can anyone help pls?

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[pastel rivet]

im doing part c

but i dont know how to make the subject

and then put it into y

[wary epoch]

Hint, what is $x^2$

[grand pond]

Xwtek

[pastel rivet]

root? of x

so like p part

all square root

or am i bugging

[wary epoch]

No, I mean express x^2 in terms of p

[pastel rivet]

ngl

how do i do that

[wary epoch]

Just plug in x = ... into x^2

[pastel rivet]

wait

so it becomes p squared + 1/p squared + 2?

then i just put it into y?

[wary epoch]

Yup

[pastel rivet]

oh ok

thx

.close

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[old citrus]

how to differentiate it with respect to (B-V}

[old citrus]

PLEASE HELP

anyone

[atomic jolt]

differentiation happens with respect to a variable

you should give more context for what you're doing

[old citrus]

b-v is the variable

so this is for error propagation of astrophysics

i need to find the standard deviation

of T

where

[atomic jolt]

do B and V have some kind of error bounds?

[old citrus]

yeah

0.9+_0.01

i just didnt know how to differentiate T

the fractional parts

[atomic jolt]

Usually what people do in physics is they approximate

$$|\Delta T|=\left|\frac{\partial T}{\partial B}\Delta B+\frac{\partial T}{\partial V}\Delta V\right|\le \left|\frac{\partial T}{\partial B}\Delta B\right | + \left | \frac{\partial T}{\partial V}\Delta V\right|$$

where the inequality is the triangle inequality

which should not be confused with "the standard deviation"

because the standard deviation relates to random variables

[old citrus]

oh sorry i didnt explicitly mean standard deviation

by standard deviation i meant the error range of T

umm theres been another mistake too

sorry

B-V

[grand pond]

Stipendi

[old citrus]

arent different variables

(B-V) the entirety itself is the variable

[atomic jolt]

So B-V is thought of as a single variable that has some error bound?

[old citrus]

yeah

[atomic jolt]

well you should probably call it something else

[old citrus]

its called colour excess

[atomic jolt]

In that case you don't even need the triangle inequality

if you wish to do a linear approximation

[old citrus]

yeah i was wondering that this shouldnt be that hard

[atomic jolt]

the triangle inequality isn't hard

So I suppose your question is how you differentiate the function 1/(x + 1)

[old citrus]

i dont know what that is

kinda yeah

i suppose

[atomic jolt]

1/(ax+b) = (ax+b)^(-1)

and thus the derivative is -a(ax+b)^(-2)

[midnight plank]

@old citrus Has your question been resolved?

[old citrus]

so for the derivative of the whole thing

i just keep 4600 on the outside of brackets

and inside

i keep sum of two fractions

of -a(ax+b)^(-2)

?

sorry for replying late

my electricity went out

[midnight plank]

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[plucky thorn]

e

[plucky thorn]

i cant help but notice

I mean i cant help but question

Where did the formula
if f(x) = ax + b / cx + d
Then f^-1(x) = -dx +b / cx -a
Came from?

[fresh sparrow]

you can derive it yourself

let x = (ay + b)/(cy+d)
solve for y

[plucky thorn]

so x(cy+d) = (ay + b)?

[fresh sparrow]

yes

[plucky thorn]

do i multiply the x with the cy+d?

wait so uh

cxy + dx -b = ay

[fresh sparrow]

yes

[plucky thorn]

(cxy + dx - b)/a = y

[fresh sparrow]

uh

i mean sure but that wont help you

[plucky thorn]

how do i remove the why on the left

[fresh sparrow]

do cxy - ay + dx - b = 0

you get (cx-a)y + dx-b = 0

(cx-a)y = -dx+b

[plucky thorn]

oooohhhh

That make sense

also one more thing

What is it called again when the power is on the left of the number

so its like

[fresh sparrow]

$^3 x$ ?

[grand pond]

artemetra

[plucky thorn]

yeah that one

[fresh sparrow]

tetration

[plucky thorn]

whats that called again?

ah alright thank you mate

[fresh sparrow]

you are talking about $x^{x^x}$ right

[grand pond]

artemetra

[plucky thorn]

yeah like that one yeah

I forgot the name lol

[fresh sparrow]

i see, glad to help

if you are done type .close

[plucky thorn]

Thanks once again mate

imma close it now

.close

[midnight plank]

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[midnight plank]

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[raven frost]

solved everything until question D

I just don't get the coordinates of the stationary point when I revert back

[midnight plank]

@raven frost Has your question been resolved?

[midnight plank]

@raven frost Has your question been resolved?

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@raven frost Has your question been resolved?

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@raven frost Has your question been resolved?

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[golden vigil]

I worked out a and b, and I just realized I was wrong with part b but idk why

Sending the work rn

Ok I think it’s because I solved for c after I already had an equation which is wrong

[midnight plank]

@golden vigil Has your question been resolved?

[golden vigil]

Yea ok I figured it out

[midnight plank]

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[golden vigil]

That was the issue

[midnight plank]

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[wheat notch]

Hey guys, I need help with a calculation:
A vector, 𝑏⃗
, is given below
𝑏⃗ =(−15 over −8)
What is the length of 𝑏⃗?
In this picture, I followed the hint for calculating the length of this vector, but as I calculated it I got so confused because, I started with calculating -15^2, but every calculator I used said it was -225, I don't understand why?

[lament fox]

you should do (-15)^2 and not -15^2

this is because $-15^2$ implies $-1\cdot 15^2$

[grand pond]

The د

[wheat notch]

ok. thanks

is it the same with the -8?

[lament fox]

yes

[wheat notch]

ok thanks

.close

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[midnight plank]

@last slate Has your question been resolved?

[exotic musk]

Do you know what an area scale factor is? If you don't try to find out and then try tyhe problem again

[midnight plank]

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[naive sonnet]

Help

[naive sonnet]

This is a lot to ask

But can someone check my linear algebra homework

[naive sonnet]

I don’t have any questions in particular

[hexed birch]

just post those and you gonna find out

it will take faster for sure instead of saying

[naive sonnet]

I’m not necessarily looking for a right answer but I’m curious if you spot anything that may look wrong

It’s mainly the first page I’m worried about

[midnight plank]

@naive sonnet Has your question been resolved?

[naive sonnet]

.closes

.close

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[coral bobcat]

Hii

[coral bobcat]

How do i find sin and cos

[tribal temple]

3 = 3cos(theta)

[coral bobcat]

Is that part right? 😭

[dusty turret]

Well you need to solve

[coral bobcat]

I pur in 3sin(theta) in for x

[grand pond]

Niklaus

[dusty turret]

Sorry

That was supposed to be sin

[tribal temple]

More that you can find cos(theta) very easy from here-

[coral bobcat]

Oh wait 3 = 3 - sin 💀

[grand pond]

Niklaus

[coral bobcat]

Wait so

3 - sin

3 = 3 - sin

[tribal temple]

she's already done the simplification so just needs to consider this

[coral bobcat]

Idk why i put cos 😭

[tribal temple]

Oh?

[coral bobcat]

🥲

Wait so after that

Can u minus the 3

To just get -sin

[tribal temple]

after getting this, or?

[grand pond]

Niklaus

[coral bobcat]

OH I KNOW WHY I PUT COS

[dusty turret]

You cant separate the 9 from sin theta like that

[coral bobcat]

3 - COS

[dusty turret]

You end up eith

[tribal temple]

So, if you know that 3cos(theta) = 3, then cos(theta) = ?

[coral bobcat]

What if u factor it

[coral bobcat]

1?

[grand pond]

Niklaus

[coral bobcat]

No but sin is squared

So u use the trig property to turn it to cos^2

[dusty turret]

Yes squared

Yup then you used the identity

[tribal temple]

(and we've said @coral bobcat has done the first part already...)

[dusty turret]

1-sin^2

[dusty turret]

I thought they wanted to verify. That part

Sorry

Long story short

Yes

Ur that part is good

[coral bobcat]

No worries! Thanks

Wait how would i find sin(theta)

[dusty turret]

1=

[tribal temple]

You know cos(theta) = 1, and you can relate cos(theta) to sin(theta) easily

[coral bobcat]

You can? 😭

[tribal temple]

Yes, you can, and you already have in multiple places!

[coral bobcat]

._.

Huhh

[tribal temple]

[coral bobcat]

OH

So if i put it back in the sqrt

Then sqrt(9-(3sin)^2)

And then

Sqrt(9-(1-sin^2))

Oh

😭

[dusty turret]

Solve for theta my friend

[tribal temple]

(easier to e.g. notice that you know the sin^2(theta) + cos^2(theta) = 1, and that you deduced cos(theta) = 1, so...)

[dusty turret]

That too

[coral bobcat]

So if i substitute 1 for cos^2 and the subtract it

That gives us sin^2 = 0?

[grand pond]

Niklaus

[coral bobcat]

Wait

Whys it wrong

[tribal temple]

Why did you change it

[coral bobcat]

Huh?

[tribal temple]

Said it was correct originally

[dusty turret]

No

[coral bobcat]

When it was sin?

[dusty turret]

Sin

Sin doesn't turn into cos randomly