help-42

[jolly pilot]

so the sum would reduce to 2^(2k-1) * x^2k

[opaque palm]

alright i have no idea why but thank you guys

[sonic fern]

see simplify x power it doesnt depend on k so you can move it outside
then you can u see its the sum of all even coefficients of (1+1)^2k

[opaque palm]

oooh

thank you so much

.close

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[opaque palm]

.reopen

[calm coral]

✅

[opaque palm]

actually, the only thing i dont get yet is why its 2^(2k-1), not 2^(2k)

[sonic fern]

its only the even coefficients of the whole sum, and if u see binomial coefficient properties u can see that sum of even coefficients would be half of the sum of all so 2^(2k)/2

[opaque palm]

oh okay i didnt know that ill look it up

thanks

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[thorn parrot]

Find the smallest natural number n(n>1) so that: (n+1)(2n+1)/6 is a perfect square

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[blazing coyote]

Analyze the logical forms of the following statements. You may use the symbols ∈, ∉, =, ≠, ∧, ∨, →, ↔, ∀, and ∃ in your answers, but not ⊆, ⊈, P, ∩, ∪, , {,}, or ¬.

[blazing coyote]

$x \in \bigcap_{i \in I} ( A_i \cup B_i)$

[potent lotus]

What a wonderful world it is !

[blazing coyote]

So the logical form would be

$\forall i ( i \in I) (x \in A_i \lor x \in B_i)$

[potent lotus]

What a wonderful world it is !

[blazing coyote]

Is this fine

[topaz raft]

hi

[blazing coyote]

Hello!

[bronze adder]

Why did you simply not write $(\forall i \in I)$

[potent lotus]

neon

[blazing coyote]

I should have , yeah

but is this fine too

[bronze adder]

I don't know

[bronze adder]

I'd write this

idk about declaring i twice

[blazing coyote]

so $(\forall i \in I) ( x \in A_i \lor x \in B_i)$

[potent lotus]

What a wonderful world it is !

[bronze adder]

yeah

[blazing coyote]

Thanks

one more question

$x \in [( \bigcap_{i \in I} A_i) \cup (\bigcap_{i \in I} B_i)]$

[potent lotus]

What a wonderful world it is !

[blazing coyote]

<@&268886789983436800>

[clear delta]

yeah this seems like a scam

[blazing coyote]

so $(x \in \bigcap_{ i\in I} A_i) \lor ( x \in \bigcap_{i \in I} B_i)$

[potent lotus]

What a wonderful world it is !

[blazing coyote]

so $(\forall i \in I)(x \in A_i) \lor (\forall i \in I)( x \in B_i)$

[potent lotus]

What a wonderful world it is !

[blazing coyote]

Is thi fine

[teal quiver]

you shouldnt need to do define for all i twice

[blazing coyote]

yeah, but just want to be safe

[teal quiver]

is I a set or just some range?

[blazing coyote]

Some set

thanks

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[sweet agate]

b?

[sweet agate]

Is this enough:

f: R \ {3} -> [1,infty) \ {e²}

It is continuous on its entire domain because it consists of the following elementaire functions: |x| e^x 1/x.
Further more it is not continuous in x=3 since the limit does not exists there.
It is continuous in x=-1/2 since limit x-> -1/2 - =limit x-> -1/2+ = f(-1/2)

It is differentiable everywhere expect from x=3 and only from the right for x=-1/2 ?

[rigid cave]

yeah pretty good, but what did you mean by the elementary function part, just to clarify more
also it's not differentiable on -1/2 at all because the limit is not defined there

[sweet agate]

huh why is the limit not defined there?

we got told elementaire functions are continuous on their entire domain so if a function consists of elementaire functions than that function must also be continuous on its domain

[rigid cave]

oh not the continuity, that's fine, just the differentiability of the function at -1/2

[sweet agate]

aaah okay

hmmm

but why not at -1/2 tho?

limit x-> -1/2 - f(x) =limit x-> -1/2+ f(x) = f(-1/2) ?

[rigid cave]

youd need to evaluate (f(x) - f(-1/2)) / (x + 1/2)

youd find different values from the right and from the left

[sweet agate]

aaah I must use the definition of the limit at points like that?

i totally forgot that

[rigid cave]

yeah, just know when there is the absolute value, there's usually points where the function is non differentiable

[sweet agate]

aaah okay

imma do it rq wait please

wait so I just do this:

f'(-1/2) = f'+(x) =! f'-(x)

then i figure out if the left or right derivative equals f'(-1/2) and then it is differentiable from the left or the right?

[rigid cave]

f'(-1/2) doesn't exist in the first place

[sweet agate]

owh

[rigid cave]

if you graph the derivative out, you'd see a jump

[sweet agate]

ill just solve what u said

maybe i understand it better then

gonna take a bit of time tho

[rigid cave]

and if you graph f, you'd see a "sharp" turn

like a non smooth one

alright

[sweet agate]

ye I did know that

alright imma do what u said rq

you meant this?

since the right derivative does not equal the left one it does not exist?

[rigid cave]

yep

[sweet agate]

okay

but when does it happen that the derivative only exists from 1 side?

[rigid cave]

like the derivative is continuous from one side?

[sweet agate]

yes

[rigid cave]

i don't know what that means actually, cause it causes problems definition wise

derivative exists means the limit should exist and be well defined

[sweet agate]

aaah okay

thank you though!

btw uhm

1 last question

if i have a function with uhm

lim x->1+ f(x) = 2

lim x->1- f(x)= 9

f(1)=2

then the function is continuous in x=1?

[rigid cave]

no not really, you'd need the limit on the left to also be equal to f(1), to extend the function's continuity at 1

[sweet agate]

aaah okay

thank you 🙂

[rigid cave]

no problem!

[sweet agate]

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[gusty cliff]

I don’t understand guys

[wintry hinge]

Help me with this

[velvet osprey]

!occupied

[calm coral]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[velvet osprey]

(sorry, the other person got here just a second earlier)

[gusty cliff]

The mark scheme says the enclosed area is the green part I just colored

[wintry hinge]

Okay

[gusty cliff]

Sorry

[velvet osprey]

it doesn't mention the triangle right above it?

[gusty cliff]

Nope

[velvet osprey]

can you show the ms for this question in full

[gusty cliff]

That’s what I thought at first

Ok

[velvet osprey]

it's weird that it wouldn't

[velvet osprey]

Adds area of triangle

[gusty cliff]

Omggggg

I am fkn stupid😭😭

Ight sorry for this💀🙏

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[gusty cliff]

What is wrong with my triangle area💀

Mark scheme says it’s 4sqrt3

But isn’t that just half only?

(Ignore that 11/6pi is wrong, I forgot to add the rest)

[unkempt drift]

so r = 3 + 2 sin(pi/6) = 4, yeah that's correct

then you can just do area of triangle = 1/2 * 4 * 4 * sin(120)

[gusty cliff]

Huh

I don’t get it🧍‍♂️

[unkempt drift]

1/2 ab sin C

[gusty cliff]

Why’s my method wrong tho?

[unkempt drift]

ah one sec then

[gusty cliff]

Okayy

[unkempt drift]

yeah you're forgetting the 1/2

it's also 1/2 * 2 * 4sqrt(3) = 4sqrt(3)

[gusty cliff]

Omgggg

Bro I keep doing shit like that 24/7💀💀

[unkempt drift]

yeah 4sqrt(3) from 2 * 2sqrt(3)

[gusty cliff]

Bro the amount of marks I lost cuz of stupid shit like that😭

anyways thanks for the help🙏

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[sudden oyster]

hi, problem:

[sudden oyster]

what i tried:

I think I am missing something simple since my professor only showed the formula in class and did not do a trapezoid, only a square

[calm coral]

@sudden oyster Has your question been resolved?

[calm coral]

@sudden oyster Has your question been resolved?

[calm coral]

@sudden oyster Has your question been resolved?

[sudden oyster]

I ended up sending my professor an email because myself and a sophomore industrial engineer spent 2 hours on the problem, got 6 answers, none were right

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[manic vault]

💀

Damn

[wooden socket]

wth

why did I get simpler answer

351032.1N

try it@sudden oyster

I am sure I made a mistake it cant be just
F=roh g A hcg

[sudden oyster]

Not correct

[wooden socket]

yea it shouldn't be that simple..

do you know what's wrong?
I used the formula
F=pgAh
p and g are gived
area of a trapezoid is (h(b1+b2))/2
and the depth of centroid is
Hbase + h/3 × (2b2+b1)/(b1+b2)

[sudden oyster]

I don't know what is wrong. I tried many things

this is the formula we are given to use

[wooden socket]

try 349860

@sudden oyster

[sudden oyster]

Not correct

[wooden socket]

if your prof answered DM me the solution please

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[cedar ether]

In Rudin's PMA, he has this theorem that a linear map in L(Rn,Rm) is continuous (Theorem 9.7). Later in the text, there's a linear map S which maps from subspace of Rm into Rn. I have figured out that any finite-dimensional subspace is isometric to Rk for some k, however, I don't see how to apply what Rudin has already proved to the map S to show it is continuous. How can I proceed?

[glass heart]

composition of continuous maps

[cedar ether]

Are you saying S is a composition of continuous maps? Call the subspace Y. Then we have an isometry Y -> Rk. Next we have S itself which maps Y -> Rn, but we don't know that S is continuous. Which more maps do you have in mind?

[glass heart]

find a map Rk->Rn

linear map

[cedar ether]

ok 👍 sorry to bother again, but I'm still a little confused. Can it just be any linear map from Rk -> Rn? It has to be a map such that the isometry, call it T, composed with the map Rk -> Rn gives S, though I'm not so sure how to make that happen.

[glass heart]

T is linear

obtained by sending relevant bases to each other

let L be the linear map you want, you only need to define what it equals on the basis vectors of Rk

[cedar ether]

ok, makes sense

thanks!

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[cedar ether]

.reopen

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✅

[tall moon]

?

[cedar ether]

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[orchid palm]

for this question in particular is the image I attached acceptable for part a?

[orchid palm]

or would I need to do something like {x | x is a non-negative multiple of 3 less than or equal to 12}

[pale blade]

lgtm, there are multiple ways to write something in set notation

only thing is, is w the set of all natural numbers?

[orchid palm]

yeah

that's what I meant by W

no

sorry

whole numbers

mb

N is for natural

[pale blade]

so set of integers?

[orchid palm]

sure

I could have used both

[pale blade]

like -1,0,1,2,3

[orchid palm]

interchangeably here

[cobalt mirage]

wtf is this channel

[orchid palm]

since I specified x is between 0 and 4

inclusive

[orchid palm]

wdym?

[pale blade]

yeah that's fine then, just not used to the "W", I normally use Z

[orchid palm]

Oh I see the confusion

I use W as whole numbers {0 -> inf} (non-fractional)
I use Z as integers {-inf -> inf} (non-fractional)

that's how I see it

[pale blade]

oh thats smart, I always have to specify "N includes 0 or N doesn't include 0" when talking about {0->inf integers}

[orchid palm]

ah yeah you wouldn't have to do that using whole numbers

just place the W

anyways thanks for the help I think I will use the sentence version probably idk

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[plain umbra]

I just want to make sure that my categorization is correct

A is a translation matrix
B is a scaling matrix
C is a reflection around the line y = x
D is a 90 degrees rotation

[plain umbra]

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[thin torrent]

help

for this one is it true?

<@&286206848099549185>

[slim knoll]

!15m

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[manic vault]

Do they even mention how the triangles are formed?

[thin torrent]

this is all they give me

[thin torrent]

that’s all they give me

[manic vault]

What kind of congruency do you think occurs here?

[thin torrent]

sss?

[manic vault]

I am asking you, how would you know that D even exists?

[thin torrent]

what D?

[manic vault]

Yes. So that means?

[civic condor]

q1

[manic vault]

Your D the point D given on the diagram.

[thin torrent]

ah i see

HELP alright

[manic vault]

Sorry. Had to make the joke.

[thin torrent]

what about this one is it true?

[manic vault]

^

[thin torrent]

all good man i laughed

appreciate you man!

.close

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[rough warren]

Hi

[rough warren]

I need help to calculate expected time for GBM to reach a value of L

[calm coral]

@rough warren Has your question been resolved?

[rough warren]

<@&286206848099549185> anyone please guide me where I can find the solution

[agile pilot]

Can you provide a ss

[rough warren]

@agile pilot

<@&286206848099549185> anyone with descent at stochastic calculus please help me out

[rough warren]

No one to help?

[agile pilot]

Hmm I would say maybe a calculus channel would be better….

[rough warren]

I tried asking there but no one helped 🥺 😭

I will just submit my solution

Most likely it's wrong tho

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[lavish solstice]

can someone help me understand why this is wrong

[lavish solstice]

and also what is k supposed to equal in a tangent function

[civic dirge]

2tan(π/4 (x + 4)) + 3

I assume you're following a formula for this? Watch out for the extra parenthesis

@lavish solstice

[lavish solstice]

i’ll try

awesome that worked

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[lavish solstice]

can someone help me understand why this is wrong as well

[calm coral]

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[remote mural]

Find dy/dx by implicit differentiation. Then find the slope of the graph at the given point.

tan(5x+y)=5x, (0,0)

dy/dx=

at(0,0): y'=

I'm having trouble applying implicit differentiation inside a chain rule, im not that great at differentiation. it would be great if someone could walk me thru this step by step, thank you

[leaden thunder]

If the left side is
tan(5x+f(x)) where y=f(x) is just notationally more explicit in its dependence on x

Can you find the derivative?

[remote mural]

so derivative of tan is sec^2, but for 5x+y, im confused on if its 5+dy/dx or something else
then applying chain it would be
sec^2(5x+y) * 5 + dy/dx?

https://tenor.com/view/cat-side-eye-head-turn-thing-idk-gif-16553853065807910158

idk

[leaden thunder]

Yes that's right except missing parentheses

[remote mural]

so it would be

sec^2 ( 5x+y ) * ( 5 + dy/dx ) = 5 ?

then to find the slop of the graph

i would try to isolate dy/dx?

afterwards plug in the x and y

or i can plug it in rn actually

waittt

then dy/dx=-5?

i dont think i did this right

uhh

yeah its not right

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[tribal sky]

can anyone check if my answers for piecewise is correct?

[tribal sky]

i asked chatgpt to check my answers but chatgpt is tweaking out

so i need a human to help🙏

[calm coral]

@tribal sky Has your question been resolved?

[calm coral]

@tribal sky Has your question been resolved?

[tribal sky]

<@&286206848099549185>

[vagrant dune]

this is not how u set up a peice wise function

you set restrictions on the domain for each branch

[tribal sky]

oh ok

i might have used the wrong word

i meant

for each of those conditions

have i set up the correct formula

so for each restriction that i was given did i make the correct formula

[calm coral]

@tribal sky Has your question been resolved?

[sharp ibex]

well all your formula are correct.

[tribal sky]

ok thanks

.close

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[fading lodge]

anyone here know boolean algebra?

[calm coral]

@fading lodge Has your question been resolved?

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@fading lodge Has your question been resolved?

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@fading lodge Has your question been resolved?

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[outer sedge]

can you put "x" in your bounds for x in a 1D integral?

[outer sedge]

why not?

[edgy leaf]

7 no's is wild

[velvet osprey]

https://math.stackexchange.com/questions/446591/why-is-it-considered-incorrect-to-use-the-variable-of-integration-as-a-boundary

[outer sedge]

i see thanks

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[dusky meadow]

number 15, i have no idea to which line they want to find the length to

[ancient grotto]

Try to draw it.

The apex of the acute angle, meaning the point from which it originates.

[dusky meadow]

no idea what this means.

[ancient grotto]

If I were to denote an angle ABC, the apex would be point B.

[dusky meadow]

what about the (ss)

[ancient grotto]

Like, the side opposite from the angle.

[ancient grotto]

Each side on a polygon corresponds to a specific angle.

[ancient grotto]

Good start.

[dusky meadow]

so to the side CD? 🤷🏻‍♂️

[ancient grotto]

Now draw a line from angle B to AD.

[dusky meadow]

or that, yes

[dusky meadow]

but at which angle to draw it, any?

[ancient grotto]

Angle ABC to side AD.

[dusky meadow]

like if i were to name the line BH, it can have any ange?

[ancient grotto]

Draw what you mean.

[dusky meadow]

like i can have the line at any ange to the side AD

what im asking is, does it matter? 🤔

[ancient grotto]

Yeah. Let me think about this.

[dusky meadow]

oh and, the line has to start from A or C.

says in the requirements that from the acute angle

(at least in the original translation)

[ancient grotto]

Actually, wait. Suppose the new line is BE, where E is the point on AD. Then BE is an angle bisector of angle ABC.

I believe this is due to the definition of an apex.

Thus there is one and only one line.

[dusky meadow]

soo it would have to be like that?

[ancient grotto]

Yeah.

[dusky meadow]

now because its a romb (diamond, idk what its called in eng), all the sides are equal means that each side is Peremeter/4 = 56/4 = 14;

now how to i find the side AH

nvm, found how.

[dusky meadow]

about question 16, arent the paralel diagonales just the same diagonales?

[calm coral]

@dusky meadow Has your question been resolved?

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@dusky meadow Has your question been resolved?

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@dusky meadow Has your question been resolved?

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[solar quiver]

Can anybody gimme an idea on how to start this

I should use cosine rule and trigs but it feels like there's not enough info

[leaden thunder]

use the bottom two equations to solve for x or y in terms of y or x. then plug that into the first equation

[solar quiver]

These are three equations and four unknowns

I was wondering if angle θ actually is 45⁰

Since it's saying southwest.....

[keen grail]

you can solve for theta

if you do what riemann said

[solar quiver]

Otherwise it's unsolvable,, at least i think so

4 unknowns guys, x y z θ

[keen grail]

yea but you have that tan(40)y = tan(35)x

so you can solve for x or y in terms of x or y

[keen grail]

since z = z

[leaden thunder]

you can try making two more equations from right triangles SRQ and SRP

[solar quiver]

U mean by using the top angles?

Are you sure? Isn't what i said about 45 degrees true?

[leaden thunder]

nowhere does it say theta=45 degrees

[solar quiver]

It says west and south west

[solar quiver]

Between west and southwest its 45deh

Deg

[leaden thunder]

i mean sure go ahead and come up with an answer that way to see if it equals the answer

[solar quiver]

Ill try that and see

Brb

[solar quiver]

I got 19.5......

Omg i hate these word problems

Thanks for giving a hand

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[lunar drift]

can i have help on this question and how to graph these coordinates on a graph

[twilit zenith]

heyo

[lunar drift]

hii!

[twilit zenith]

fnction will break

at x = 2 as it will approach infinity

and then for fn to be 0 it will reach zero at, two points, they are (10 + rt92)/2 and (10-rt92)/2

[lunar drift]

so the website, aleks, asks for a plotted coortidnate for the asymptotes and the other points on the graph, can i have the answer in that type of format please?

[twilit zenith]

you want the graph?

lemme try,

here, we are not familar with plotting graphs using computer but rather pen paper, lemme try tho

[lunar drift]

perfect thank you so so soooo much!

[lunar drift]

ohh thank you

[twilit zenith]

as you can see

just at 2

it appraoches infnity from both side

post and negative

there it is a break point

green line is, x\ = (10-sqrt92)/{2} and black line is x = (10+sqrt92)/{2}

[lunar drift]

ohh okay thank you

[calm coral]

@lunar drift Has your question been resolved?

[calm coral]

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[lunar drift]

somewhat yeah

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[atomic grove]

Complete the square $y=x^2+8x+22$, i.e. write the expression in the form $(x+a)^2+b$.

[potent lotus]

Elliot Pixel

[atomic grove]

.close

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[atomic grove]

figured it out

[lunar drift]

not plotted, just the numbers please

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[lunar drift]

<@&286206848099549185> can you please assist whenever you can?

[spare willow]

Ok do you have any idea

[lunar drift]

not really my teacher assigns work but he's never really here

[spare willow]

Do you know how to find asymptotes?

[lunar drift]

i know it's with the denominator

[spare willow]

Yeah so the denominator cant be 0

[lunar drift]

don't you put it like x+4=0

[spare willow]

Yes

[lunar drift]

so it would be -4

[spare willow]

Yup

[lunar drift]

i don't know how to find the holes though

[spare willow]

I don’t think there’s hole in this one

[lunar drift]

but in the instance where there is a hole, how do you do that calculation

[spare willow]

So how to say if there’s a common factor in both numerator and denominator that mean there’s a hole

$((x+3)*(x+2))/(x+2)$

So something like this

U see there’s x+2 in the numerator and denominator

So there’s a hole in -2

[eternal shard]

change the + to a dot

[spare willow]

My bad

[potent lotus]

Ozau

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[brazen elbow]

i don't know what i did wrong. the resultant force should be F_3 = F_1, 3 + F_2, 3, but i'm getting different result from the book

[brazen elbow]

$\textbf F_3 = \textbf F_{1, 3} + \textbf F_{2, 3} = 9\cdot10^9\left(\frac{5\mu^2}{(0.1 + 0.1i)^2} + \frac{-2\mu\cdot5\mu}{0.1^2}\right)$

[brazen elbow]

this yields -9 - 11.25i. it should have been -1.04 + 7.94i

[calm coral]

@brazen elbow Has your question been resolved?

[brazen elbow]

i genuinely can't make sense of why this is wrong

[fierce gate]

what is i and mu here?

@brazen elbow

[brazen elbow]

i is the imaginary unit

[fierce gate]

i genuinly have no idea what all that stuff on the right side is

HUH

what

[brazen elbow]

mu stand fors microcoulomb

[fierce gate]

why on earth are u using imaginary numbers???

also write $\mu C$ instead

[potent lotus]

IronVoltage

[brazen elbow]

because it's a convenient tool to describe 2d coordinate

[fierce gate]

eww

I have never seen someone try to solve a net force problem with i

don't do that

at least not yet

[brazen elbow]

i've been doing that since grade 12 and it works perfectly brother

[fierce gate]

get the problem right first. Then mess around with it to ur leisure

eww

r u a math major?

[brazen elbow]

no, compsci

[fierce gate]

wasnt expecting that

[brazen elbow]

it's something i can easily plug into my calculator without having to separately calculate the x-component and y-component

[fierce gate]

well since im not familiar with using literal i for this ima have to decypher what ur doing a bit

[brazen elbow]

just think $0.1 + 0.1i$ as $0.1\hat i + 0.1\hat j$

[exotic falcon]

mu C is the standard notation

[fierce gate]

from first glance, problaly ur denominator in the first term in the parenthetical

[brazen elbow]

i know, but i'm lazy and i just type in whatever i have on my calculator

the notation is fine. can we move on to the actual logic?

[brazen elbow]

what's wrong with it?

[exotic falcon]

The parenthetical is also confusing me

[fierce gate]

ur squaring a complex number, so ur gonna get i^2 = -1 in the denominator. which u dont do in the normal ihat jhat notation

so thats prolly wrong

im guessing ur denom is supposed to be a distance right

"distance" in the complex plane isnt the complex number squared

its the module of it (i think thats called the module)

ya know the 'magnitude' of the complex number

anywho ima keep decyphering

[brazen elbow]

oh yeah you're right, r here is the distance, not the coordinates of q

[fierce gate]

ngl this is so cursed lmao

even if it works in a sense

[crystal forge]

modulo

[fierce gate]

thx

lmk if it works out at the end @brazen elbow

[brazen elbow]

it works if i don't directly sub in the coordinates, but instead calculate the magnitude first, then the angle of the force

[jolly pilot]

Also, the way to calculate force is $\frac{\vec{r}}{||r||^3}$ so you get a nice Force vector

[brazen elbow]

what does that mean?

[jolly pilot]

r is the distance vector

[fierce gate]

u should adopt the normal convention

so u can understand what is going on conceptually

[brazen elbow]

that doesn't make sense. distance is a scalar, not a vector

[fierce gate]

u can use whatever notation u like, but by not adhering to convention it makes it harder to communicate to those not familiar with it and thus harder to get help on

[potent lotus]

Facter10Br4g

[fierce gate]

$F\vector$

[jolly pilot]

tf you mean? why are you using 0.1 + 0.1i then? Just to express the vector right?

[potent lotus]

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

[fierce gate]

how the hell do u write an arrow

[jolly pilot]

\vec{F}

[fierce gate]

ah

thx

[brazen elbow]

i understood where i went wrong. r here is the distance, not the position of the charge

[fierce gate]

$\vec{F} \propto \hat{r} $

ugh latex

[brazen elbow]

but i don't understand that how do you get the force vector just from the distances alone

[jolly pilot]

Nah dude, you seem to have this correct

I was being lazy, mb

The formula would be $\vec{F} = k\frac{q_1q_2\vec{r}}{r^3}$

[potent lotus]

Facter10Br4g

[brazen elbow]

r cube?

[swift laurel]

the force is [ \vb F = k\frac{q_1 q_2}{\norm{\vb r}^2} \hat{\vb r} = k\frac{q_1 q_2}{\norm{\vb r}^3} \vb r ]

[potent lotus]

cloud

[fierce gate]

this

[swift laurel]

since [ \hat{\vb r} = \frac{\vb r}{\norm{\vb r}} ]

[jolly pilot]

one of these makes the r a unit vector

[potent lotus]

cloud

[brazen elbow]

i see

what does textbf R here really convey?

[fierce gate]

?

huh

[brazen elbow]

i know what r-hat is

[jolly pilot]

well, what is r-hat?

[fierce gate]

wdym text boldface R

???

[brazen elbow]

unit vector with the direction from q1 to q2

[swift laurel]

$\vb r$ is the vector pointing from the charge exerting the force to the point where the force is being applied

[potent lotus]

cloud

[jolly pilot]

well, then r is just the vector distance between the two locations

[brazen elbow]

yeah makes sense, just want to clarify

[fierce gate]

i think u forgot that r=r rhat

so just

divide by r

to get the direction

r is just the position vector

direction and magnitude

[brazen elbow]

yeah, i figured

thanks y'all

.close

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[calm coral]

Please don't occupy multiple help channels.

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[next fossil]

Hi

[next fossil]

Need help with this question

Just need to know what the first step would be

This is what I have done so far

[keen grail]

I meant hat's what I would've done for the first step

factoring out the x^2

[next fossil]

But at the end, it is +-

How would ik to make it negative

Helooooo

[calm coral]

@next fossil Has your question been resolved?

[next fossil]

No

Hi

[keen grail]

@next fossil

sorry

You only need to consider the - since $x$ tends towards $-\infty$

[potent lotus]

NahhFam

[calm coral]

@next fossil Has your question been resolved?

[next fossil]

Hi

No

[next fossil]

Could you explain this

[keen grail]

You only need the $\pm$ if you don't know the value of $x$

[potent lotus]

NahhFam

[keen grail]

since x could be positive or negative

but you know x will be negative in this case

[next fossil]

But wouldn’t it still look like - (-infinty)?

Then the top and bottom x/infinity cancel out leaving the negative there

For the root 4

Explain this

[tepid cipher]

theres a sharp point so it wont be differentiable at 2

[calm coral]

@next fossil Has your question been resolved?

[calm coral]

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[blazing coyote]

Trying to integrate this by converting into polar

[tame crag]

why not sketch it out first

[blazing coyote]

Now, intutively, this is simple $\int_{0}^{\pi} \int_{0}^{1} r dr dt$

I get this by sketching it out

[tame crag]

you forgot your Jacobian

[blazing coyote]

right

[potent lotus]

What a wonderful world !

[tame crag]

This seems right

[blazing coyote]

Yeah, that's true

so I just sketch this in the cartesian plane and then mentally convert everything to polar, right

[tame crag]

I mean, you could write out the steps on paper if that helps you stay organized, but yeah

[blazing coyote]

This would be π/2

[tame crag]

seems about right

[blazing coyote]

This now

$\int_{0}^{2 \pi} \int_{0}^{1} r dr dt ?$

[potent lotus]

What a wonderful world !

[tame crag]

👍

[blazing coyote]

this would be π

[tame crag]

yep

it's just a unit circle

[tame crag]

you want to convert this to polar?

[blazing coyote]

I have nno choice

Now the equation of the line in polar is $\theta = \frac{\ pi}{4}$

[potent lotus]

What a wonderful world !

[tame crag]

the same strategy of graphing still works

[blazing coyote]

$\int_{0}^{6 sec(t)} \int_{0}^{ \pi /4} r π/4 dt dr$?

[potent lotus]

What a wonderful world !

[tame crag]

you shouldn't have a function in terms of t on your outer integral

I recomend finding r in terms of theta first

[blazing coyote]

Hmm, yeah

[tame crag]

also, I'm pretty sure you forgot your jacobian again

[blazing coyote]

Is IT just the other way around

Just swap the integrals

In this case

[blazing coyote]

That's the r

[eternal shard]

it's totally wrong, check your region again

also you forgot to rewrite x in your integrand

[calm coral]

@blazing coyote Has your question been resolved?

[tame crag]

sorry for abandoning you for a bit, I went to work on something

anti-algebraist is right, that your integral is totally wrong

notice that x = rcos(theta)

meaning you had the wrong equation of x in terms of r

similarly, remember how integrals work

you are taking tiny slices d(whatever)

0 to R (in terms of theta) is the length of r at theta

your outer integral should then be adding the sums of those Rs with tiny slices of theta (dtheta)

I recommend sketching the graph out in the cartesian plane first and definining your boundries

then approach it thinking in terms of polar instead of rectangular

[blazing coyote]

ooh

$\int_{\pi /4}^{ \pi /2} \int_{0}^{6 sec(\theta)} rcos(\theta) dr d(\theta)$

[potent lotus]

What a wonderful world !

[eternal shard]

sec is undefined at pi/2

the triangle is bounded by y = 6

[blazing coyote]

yea

[eternal shard]

not x = 6

[blazing coyote]

so 6cosec(theta)

got it

thanks

[tame crag]

hold up

[blazing coyote]

Lemme actually integrate it

$\int_{\pi /4}^{ \pi /2} \int_{0}^{6 cosec(\theta)} r^2cos(\theta) dr d(\theta)$

Won't this be rather messy to integrate

oh, nvm

it's not

[tame crag]

you're still forgetting the jacobian

[potent lotus]

What a wonderful world !

[tame crag]

👍

[blazing coyote]

$\int_{\pi/4}^{\pi/2} (6cosec(\theta))^3 cos(\theta) d(\theta)$

[potent lotus]

What a wonderful world !

[blazing coyote]

This is going to be a pain to integrate

oh

$\int_{\pi/4}^{ \pi /2} 216 cosec(t) ( cosec(t) cot(t)) dt$

[potent lotus]

What a wonderful world !

[blazing coyote]

Then just cosec(t)= u

so $216 ( \sqrt{2}-1)$

[potent lotus]

What a wonderful world !

[blazing coyote]

Is this fine?

[eternal shard]

no

[eternal shard]

this is wrong

[tame crag]

you can double check by doing the original

[eternal shard]

missing a factor 1/3

[blazing coyote]

Ah

thanjks

$\int_{0}^{\frac{ \pi}{2}} \int_{0}^{2cos(t)} (cos(t) + sin(t))dr dt$

Is this fine?

[eternal shard]

no

[blazing coyote]

Huh

what

why

[eternal shard]

r = cos(t) has radius 0.5 but your circle has radius 1

[blazing coyote]

oh right

[eternal shard]

drdt but yes

[potent lotus]

What a wonderful world !

[blazing coyote]

$2 \int_{0}^{ \pi /2} (cos^2(t) + cos(t) sin(t)dt$

[potent lotus]

What a wonderful world !

[blazing coyote]

I think this is managable

thanks

$\int_{0}^{2 \pi} \int_{1}^{e} \frac{\ln(r^2)}{r} dr dt$

[potent lotus]

What a wonderful world !

[blazing coyote]

is this fine?

[eternal shard]

yes

[blazing coyote]

$ \int_{0}^{2 \pi} dt$

[eternal shard]

How did you get the factor 2

[blazing coyote]

ln(r^2) = 2ln(r)

[eternal shard]

ok but when you integrte ln(r)/r you get ln(r)²/2

[blazing coyote]

oh right

so 2π

[eternal shard]

yes

[blazing coyote]

thanks

$s = r(\theta)$

[potent lotus]

What a wonderful world !

[blazing coyote]

I was thinking $s = \pm r \cos^{-1}( \frac{a}{\sqrt{x^2+y^2}})$

[potent lotus]

What a wonderful world !

[blazing coyote]

.close

[calm coral]

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[outer sedge]

why is this integral equal to zero?, i define the following: for the lower bound z<0, and for the upper bound, z>0, this is like saying we're taking the integral from -5 to 5, it's not zero

[glad parrot]

Its the same z on top and bot

[sonic fern]

is it supposed to be -z to z

[outer sedge]

yes but the lower z < 0, and the upper z > 0

[glad parrot]

No

Its the same z

This make no sence in terms of notation else

[sonic fern]

yeah

[outer sedge]

how would i do it if the lower bound is < 0 and the upper bound is > 0, but they're the same coordinate

[glad parrot]

Also dz is not correct

[glad parrot]

?

They can't be the same coordinates

Wdym

You mean they're opposite ?

Like -5 5

Or -2, 2

[outer sedge]

here aren't they opposite?

[glad parrot]

No they not

Here its saying that if you integrate from z to z a function its 0

Cuz

Let F be the antiderivative

You would have F(z) - F(z)

Which is 0

[outer sedge]

z is not the same for both, the lower z is negative, and the upper z is positive

that's how i defined it

[glad parrot]

Don't write the same letter if they are different

Or write -z for the negative one

[outer sedge]

where z is positive?

[glad parrot]

Yes

But this would mean they are opposite then

[velvet osprey]

@outer sedge are you trying to insinuate that there is a real number that's both positive and negative at once

[swift dragon]

$\int_z^z A\dd z=[Az]^z_z=0$

[potent lotus]

Bonk

[outer sedge]

nope, im trying to make variable z both negative and positive

[glad parrot]

Writing dz also not correct imo

[desert nymph]

do you mean e.g 5 and -5?

[outer sedge]

yeah

[glad parrot]

Since you using it as bounding variable

[desert nymph]

then you want z and -z

[velvet osprey]

a variable has only one value at a time

!xy

[calm coral]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[outer sedge]

i see

my bad, i didn't know this

.close

[calm coral]

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[swift dragon]

quantum variable

[glad parrot]

Qbit kekw

[calm coral]

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[lilac herald]

Hello, need help to see if what I did is okay. (3.3)

[velvet osprey]

for mine and everybody else's convenience

[lilac herald]

I got m=13,20,27

[velvet osprey]

how did you get these values? and how do you know no other value works?

[lilac herald]

Cause of the restrictions

[velvet osprey]

did you test each and every single value of m by hand???

[lilac herald]

Yes, cause this is obviously a pattern

[velvet osprey]

this is no good

[lilac herald]

OK

[velvet osprey]

what if you were asked to do the same problem but to find all natural m for which this works

you would not have infinite time to test every single natural number ever

[lilac herald]

Guess I would do something like this

[velvet osprey]

how do you arrive at that conclusion?

also what is b in 7n+b?

[lilac herald]

6

That's better

[velvet osprey]

again, how do you arrive at this conclusion?

[calm coral]

@lilac herald Has your question been resolved?

[lilac herald]

N at n=7n+6 is supposed to be m

[lilac herald]

Helllllo

Here

[velvet osprey]

this is hard to read

does the third line say (7+n)/7?

im finding your handwriting very difficult to read, sorry

[lilac herald]

Yeah. Sorry about my handwriting

[velvet osprey]

your letters look very indistinct

g looks like y, a looks like u, 9 looks like 4 at first glance...

when you say (7+n)/7 what does the letter n mean

[lilac herald]

n represents the number of terms

[velvet osprey]

lowercase n

ok

[lilac herald]

Sorry, autocorrect

[velvet osprey]

alright so you get eventually that the number of terms must be divisible by 7

ok yeah this is good enough ig

[lilac herald]

OK, thank you

👍

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[calm coral]

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[blazing coyote]

I'm thinking of doing part(a) first

So I have $\beta = \alpha \circ \phi$

[potent lotus]

What a wonderful world !

[blazing coyote]

Uh, now

I have a mapping from [c,d] \to [a,b]

and then from [a,b] subsequently, to an subset of R^n

Not sure how to write that rigrously though

I also , similarly have $\alpha$, as a mapping from [a,b] to R^n

[potent lotus]

What a wonderful world !
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[blazing coyote]

thus $\alpha$ and $\beta$ have the same image

[potent lotus]

What a wonderful world !

[calm coral]

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[junior gate]

I am struggling with this question so far I have got

10/22 = x / 6+x = 5
but then I am stuck on how to get DC

[hollow parrot]

what did u put as x

[junior gate]

5

I meant x = 5

woops

oh AE

[hollow parrot]

i dont think x=10/22 that is why ur stuck

10:22=x:x+6 is wut u should solve

[junior gate]

10/22 = x/x+6 was the equation I got

this wrong?

must it be in ratio

[hollow parrot]

oh i misread sorry

thats rìght

[junior gate]

how would I progress from here?

sorry I am not the greatest with math

[hollow parrot]

AF/AB=AD/AC now

[junior gate]

you got it

[hollow parrot]

do you see why?

[junior gate]

yes prop theorem

is it the BC || FD

but I get 24.2 must I - the AD ?

[hollow parrot]

AC is 24.2

thats ur answer

[junior gate]

oh I see

thank you

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[calm coral]

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[gusty cliff]

What tells me I need to 90-(theta)?

Cuz like 35 is already acute

[verbal finch]

What is l2 here?

[verbal finch]

Ok so

Basically when u consider plane vector

U actually take the normal vector

[velvet osprey]

90 - theta

[gusty cliff]

Yes

[velvet osprey]

not 90 - cos(theta)

[verbal finch]

Which is perpendicular to plane

[velvet osprey]

don't confuse an angle with its cosine

[gusty cliff]

Yea mb💀

[verbal finch]

So u do 90-thetha

[gusty cliff]

Accident sorry🙏

[gusty cliff]

Mhm

[velvet osprey]

anyway, to put the same thing @verbal finch said in better words:
imagine the plane's normal, the line, and the plane itself

the normal and the plane by defn form a right angle

[gusty cliff]

yes

[velvet osprey]

the dot product thing gets you the angle between the normal and the plane

but the rest is the angle you are actually looking for

[gusty cliff]

Wait how is it the angle between the normal and the plane?

[velvet osprey]

black = plane & its normal, red = the line

θ = the angle you found with the dot product thing

[gusty cliff]

Ohhhhhh

Makes senseee

Ohhh I get I get

Thank you🙏

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[hazy wind]

why is the top bound pi and 2pi?

[stray tundra]

those are the coordinates of the unit ball in spherical coords

you can have any point with rho between 0 and 1, phi between 0 and pi, and theta between 0 and 2pi

[swift laurel]

0 to 2pi is the "full circle" for theta, and 0 to pi goes all the way from the positive z-axis to the negative z-axis for phi

[hazy wind]

okay tq

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[vivid lagoon]

I got y instead of 1/y

I used the exp( My - Ny / M) equation

[calm coral]

@vivid lagoon Has your question been resolved?

[calm coral]

@vivid lagoon Has your question been resolved?

[calm coral]

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[calm coral]

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[valid storm]

I know any line through P would have the form

 y = 2 + kt
 z = 3 + kt``` 
for some scalar k 
and they would have to satisfy the equation given
```z = y^2 - x^2```

[valid storm]

only thing i can think of plugging in x, y, z into the equation but idk what else to do

[calm coral]

@valid storm Has your question been resolved?

[calm coral]

@valid storm Has your question been resolved?

[valid storm]

<@&286206848099549185>

[untold compass]

a parametric line at 1,2,3 is x=1+at, y=2+bt, z=3+ct, and on the paraboloid just means if you plug those into z=y^2-x^2 it works

you basically have to solve for a,b,c, but since you just care about the shape of the line and not the speed you get rid of one of the letters

[calm coral]

@valid storm Has your question been resolved?

[valid storm]

Can you explain the difference between the speed and curve part
Isn't speed just the length of the para at some time t?

[calm coral]

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[karmic ridge]

how do i solve this?

[karmic ridge]

im not sure where to start

[eternal shard]

review the proof

[karmic ridge]

?

[eternal shard]

read the last sentence

[karmic ridge]

i dont remember the proof

oh well

ill try to find it

ty

.close

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[swift storm]

How do I solve A) "Determine the distance of the stretch AC"? Apparently I'm supposed to use Pythogoras' theorem (I didn't know), but the answer turns out to be wrong in my calculations. I know the solution is supposed to be 16 cm, but how? I named the stretch AC to y.

[keen grail]

$a^2 + b^2 = c^2$

[potent lotus]

NahhFam

[keen grail]

$y$ would be your $c$

$c$ is always the hypotenuse

[potent lotus]

NahhFam

NahhFam

[swift storm]

isn't the highest value supposed to be c^2?

[keen grail]

yes, the hypotenuse is always the highest value

and y is the hypotenuse in this case

[swift storm]

oh okay wait