help-28

[dense vector]

no..indefinite integrals has some constant associated as well

you need to consider that as well

[elder cape]

well ye the + c.

[placid oar]

It kind of looks like you're doing Integrating Factor, and if you're doing that, by exponent rules, the +c shouldn't matter because you're going to multiply your whole equation by your factor, but then you can juste divide everything by c thus rendering that constant useless. If you're not doing that however, yes, keep the +c

[elder cape]

.close

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[inland fjord]

Not sure where to begin with this problem

[tulip beacon]

can you not just sub in values and compare your results?

[inland fjord]

this was my thought process because it says to use differentials

but I don't know what this tells me 🙂

or if I need to differentiate in a different way

[inland fjord]

this is what subbing in the values to the given function gives me

then made into a difference

[tulip beacon]

yea

so would that not be the answer?

[inland fjord]

it is incorrect

I need to use differentials in some way

[tulip beacon]

for the first question?

i dont see how differentials play out here

[inland fjord]

It's what it says 🤷‍♂️ "Consider L as a function of the variables I and r. Use differentials to find dL at the point given"

The entire image posted is only one question

[tulip beacon]

Oh i thought that is the next question

[full forum]

@inland fjord Has your question been resolved?

[inland fjord]

<@&286206848099549185>

[brisk imp]

This is multivariable calculus, right?

For sure you need partial derivatives here

Let's start with single-variable first, so that you see the picture

Suppose you have y=f(x) and a point (x0, y0)

We increase x0 by dx

What will be the new y?

You must have studied this

[inland fjord]

y0 + dy?

[brisk imp]

Exactly

And what is dy?

[inland fjord]

the derivative with respect to y?

[brisk imp]

it's f(x0).dx

It's a well known formula

$y_{1} = y_{0} + dy = y_{0} + f'(x_{0}) \cdot dx$

[glossy valve]

LUNA

[brisk imp]

You know this right?

It's well known

[inland fjord]

ok yes

[brisk imp]

it's called tangent line of a function at a given point

it's the formula for the tangent line if you recall

which allows you to compute y's close to y0

for x's close to x0

[inland fjord]

Ok

[brisk imp]

Did you actually understand it?

Or you're just oking me

😛

(x0,y0)
(x0+dx, y0+dy)

and dy = f'(x0).dx

as simple as that

So, the new y becomes

y0 + f'(x0).dx

[inland fjord]

Yes I understand

[brisk imp]

Allright, time to generalize to 2 variables

[inland fjord]

and thats given y0 and x0

right?

[brisk imp]

Yes, and the function f(x) of course

[inland fjord]

gotcha

[brisk imp]

To be able to differentiate it

And substitute x0

you see

[inland fjord]

yes makes sense

[brisk imp]

Now, we generalize to 2 variables

We had one variable only x

f (or y) was varying to x only

What if we had 2 variables!

Consider z=f(x,y) and the point (x0,y0,z0)

z0 = f(x0,y0) okay?

[inland fjord]

ok 👍

[brisk imp]

Now since we have 2 variables, we add to x0 an amount dx

and to y0 an amount dy

at the same time

(x0,y0,z0)
The new point automatically will be

(x0+dx,y0+dy,z0+dz)

Right!

[inland fjord]

ok ok

[brisk imp]

The thing is, what is dz?

When x0 changed by dx, and y0 changed by dy

How much does z0 change?

When we had one variable x

dy was equal to f'(x0).dx

Following the same logic, dz = f'(x0).dx + f'(y0).dy

it's just another variable added

it adds to the change in z

[inland fjord]

gotcha

[brisk imp]

So that's basically it

But since f is a function of two variables

It has 2 derivatives, not just one

One with respect to the first variable only

And one with respect to the second variable only

We call them partial derivatives

f'(x0) is the derivative of f with respect to x, evaluated at x0

f'(y0) is the derivative of f with respect to y, evaluated at y0

[inland fjord]

in single variable, y1 = y0 + dy. in two variables, z1 = z0 + dz? and dz = f'(x0)dx + f'(y0)dy

is that correct

[brisk imp]

yes exactly

I will write down the general formula

The z1 is simply called z

and z=f

Okay

I will write it

[inland fjord]

ok

[brisk imp]

$z=z_{0} + \frac{ \partial z}{ \partial x} \cdot dx + \frac{ \partial z}{ \partial y} \cdot dy$

[glossy valve]

LUNA

[brisk imp]

This is it

Compact form

Somehow

When you partially differentiate the function with respect to x, you consider x only to be variable

You treat y as a constant

And when you partially differentiate the function with respect to y, you consider y only to be variable

You treat x as a constant

[inland fjord]

this might be dumb but that formula looks like it assumes z0 is given. so given a point (x0, y0) how do u solve for z0?

[brisk imp]

z0 = f(x0,y0)

You are given the function

of course

How else would you find z's from x's and y's

and how else would you be able to differentiate if you were not given f?

😛

[inland fjord]

true true

[brisk imp]

one small example before doing the exercise you sent?

[inland fjord]

sure

letsdoit

[brisk imp]

i'll make it very simple

z = xy and you have a point (x0,y0,z0)=(1,1,1)

I increase x by 0.5

I increase y by 0.5

What will be the new z?

Let's say z1

Show me!

[inland fjord]

like this?

mb i wrote z instead of z1

[brisk imp]

Yes, correct!

Well done

[inland fjord]

thx

[brisk imp]

Now do the same for your exercise

L = i/r²

i0 = 100, r0 = 90

di = 1, dr=2

Notice, here they do not want the new intensity

They want just the change in it

not z1, just the dz

[brisk imp]

Wait, but I told you

They only want dL

Not the new L

$z=z_{0} + \frac{ \partial z}{ \partial x} \cdot dx + \frac{ \partial z}{ \partial y} \cdot dy$

[glossy valve]

LUNA

[brisk imp]

The dz is just this part here

[inland fjord]

wouldn't dz = z1? because z1 = f'(x0)dx + f'(y0)dy

[brisk imp]

$dz= \frac{ \partial z}{ \partial x} \cdot dx + \frac{ \partial z}{ \partial y} \cdot dy$

[glossy valve]

LUNA

[brisk imp]

you multiplied L0 with the first derivative look!

instead of +

[inland fjord]

u right

[brisk imp]

other than that, you did good!

it's just that you do not need to find L1

Then do L1-L0 for dL

[inland fjord]

i see

[brisk imp]

They're just asking for dL

[inland fjord]

i didnt need to find L0 to find dL

[brisk imp]

Right, but it's okay

[inland fjord]

unless I wanted to find L1

[brisk imp]

Yes

You did great!

Modify it a little bit and show me again

Always at first write your main equation that you will be working on

$dL= \frac{ \partial L}{ \partial i} \cdot di + \frac{ \partial L}{ \partial r} \cdot dr$

[glossy valve]

LUNA

[brisk imp]

To know exactly what the question requires

And avoid wasting time on unnecessary stuff

[inland fjord]

I think this is all correct

[brisk imp]

it is!

Good job

[inland fjord]

thank you 🙂

for all ur help

[brisk imp]

You're welcome!

[inland fjord]

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[brisk imp]

Now you owe me a CFA sandwich

[inland fjord]

i got u

[brisk imp]

😂

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[chrome idol]

Im not sure what I need to do here

[fathom saddle]

The usual idea. You want to show that the limit definition of the derivative exists

Actually now that I think about it, there's multiple ideas of what "differentiable" is here. What does your book say?

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[chrome idol]

Sorry, had to leave for a bit

[chrome idol]

The prof actually told what to do at the end of lecture. Its more like I didnt understand it

He said, to check f_x(0,0) and f_y(0,0) using the limit definition

Then take partial f wrt x and y at (x,y) != (0,0) (this I wasnt sure if I copied wodn correctly).

Lastly take limit of partial x and y as (x,y) goes to (0,0)

[stiff musk]

he wants you to use the theorem that if the partial derivatives exist and are continuous, then the function is differentiable

which seems like a bit overkill if you just want to show that it's differentiable at zero (also that method may fail even if the function is differentiable, because that theorem only gives a sufficient condition, not a necessary condition)

[chrome idol]

Ok, I'll first use his method. Then see if there is another way.

So to check f_x(0,0) using limit definition, I just do f(h,0) - f(0,0) = 0

Here I'm already stuck? Since f(h,0) = 0

[stiff musk]

yep, so f_x(0,0) is zero

[chrome idol]

wait, what about the limit as h goes to 0?

[stiff musk]

well the limit as h->0 of zero is zero

[chrome idol]

ohh

So f_x (0,0) = 0 = f_y(0,0)

this shows the partial at (0,0) exist?

[stiff musk]

yes, both partials at (0,0) exist and are zero

that shows that if f is differentiable at (0,0), its derivative will also be zero

but unfortunately, existence of the partials isn't enough to imply differentiability

[chrome idol]

right. I think I remember that from lecture.

[stiff musk]

simple counterexample, you can make f(x,y) = 0 on the x and y axes, and anything you like everywhere else

then the partials will be zero at (0,0) but there's no reason the function should even be continuous at (0,0) let alone differentiable

[chrome idol]

$\frac{\partial f}{\partial x} = \frac{2xy^4}{(x^2+y^2)^2}$ and $\frac{\partial f}{\partial y} = \frac{2x^4y}{(x^2+y^2)^2}$

[glossy valve]

KN

[chrome idol]

make sense..

[chrome idol]

Then what does this show?

[stiff musk]

to show differentiability at (0,0), you already found that if the derivative exists, it needs to be equal to zero, so the most straightforward way to do this is via the definition: if you make the magnitude of (h,k) small, does that make the difference quotient small

where magnitude of (h,k) is sqrt(h^2 + k^2)

and the difference quotient is $\frac{f(h,k) - f(0,0)}{\sqrt{h^2 + k^2}}$

[glossy valve]

Bungo

[chrome idol]

So a easier way, you are suggesting, is just computing this limit as h and k goes to 0?

[stiff musk]

yes

the way your prof suggested should work in this case as well, but imo it's more complicated since it involves two limits

and as i said, that's only a sufficient condition, not necessary, so if it fails you can't conclude anything

[chrome idol]

Hmm. Ok I will do both ways.

[chrome idol]

But what is the point of computing this?

[stiff musk]

well you would do that with the goal that showing the each of those goes to zero as you let (x,y) go to zero

that will show that they are both continuous at (0,0)

and continuity (as opposed to merely existence) of partial derivatives at (0,0) implies that the function is differentiable at (0,0)

[chrome idol]

and this is a theorem?

[stiff musk]

right, that's a theorem

the converse is false

a function can be differentiable at (0,0) even if its partials are not continuous there

that's another reason why you should generally approach a question like this via the definition

(most likely at some point they will give you a problem where the partials are not continuous but the function is still differentiable)

(there are standard examples of this that professors love to ask haha)

[chrome idol]

i see

[chrome idol]

Ok in the final step with my profs approach, letting x,y go to 0,0 will make both of these 0/0

[stiff musk]

yea, indeterminate forms, so you'll have to try simplifying first, or maybe sandwich them below something larger that also goes to zero

[chrome idol]

Ah, think I have an idea

i dont think this solves the problem

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[torn jolt]

1. whats the difference between a maximum and supremum?
2. why is the integral less or equal to the supremum?

[torn jolt]

oh he actually multiplied by 2pi

but doesnt the r change the length?

[fast peak]

supremums exist

maximums might not

sure the circumference of the circle gets bigger but why does that matter

we are bounding the integrand by some number

eg 2

and then evaluate $\int_0^{2\pi } 2 dt$

[glossy valve]

denascite

[torn jolt]

hm idk isnt it lenght * sup

[fast peak]

the length appears after you plug in the parametrization

so thats already taken care of

[torn jolt]

a circle right

[fast peak]

yes

[torn jolt]

dont we need the length of the curve?

doesnt it depend on r

[fast peak]

and that is already taken care of

from plugging in the parametrization

we are just integrating from 0 to 2pi here

[brave blaze]

Jordan's lemma ? 🤔

[torn jolt]

hmmm

[torn jolt]

okay kinda makes sense now but if he wouldnt have parameterised it then it would depend on r right

[fast peak]

yes

[torn jolt]

okay okay okay thanks!!!

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[gilded prairie]

i know that the response is b but im unsure if i got the description h(t) & g(t) correct: I have their equation as h(t) = x^2 and g(t) = e^3t/20? please let me know if this is ok

[full forum]

@gilded prairie Has your question been resolved?

[spiral vigil]

hmm, if it's tripling every twenty years the base should be 3, shouldn't it?

i know it's tempting to see exponential and put e but it's not always that ;)

[gilded prairie]

got it

[gilded prairie]

do you mean like this: 3e^x/20?

[spiral vigil]

no, let's take a look at this
if my money doubled every day, starting with 1 on day 0, what is the equation I could use to describe how much money I had?
0 -> 1
1 -> 2
2 -> 4
3 -> 8

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[gilded prairie]

.reopen

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✅

[gilded prairie]

would it be x^2 if doubled?

[spiral vigil]

no, this is an exponential function

4 -> 16
5 -> 32
6 -> 64

it's just 2^x

[gilded prairie]

ohh

so would it be 3^20x?

for the problem

[spiral vigil]

3^something yeah, since it's tripling every 20 days it needs to be stretched horizontally by a factor of 20

3^(20x) would shrink it horizontally

[gilded prairie]

20*3^x to make it stretch?

[spiral vigil]

that's vertical stretch

horiz stretch is f(x/20)

[gilded prairie]

ok

since its f(x/20) it would bring 3 for x and exponent of ^x?

[spiral vigil]

uhhh no i mean 3^(x/20)

[gilded prairie]

ohh ok got it

thank you

also one more question: i got horizontal asym for 1,2,& 4 would that be right?

[spiral vigil]

,w graph sqrt(x) from -1 to 30

[glossy valve]

Results provided by WolframAlpha

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Requested by kitten.in.a.teacup#0

[gilded prairie]

ah ok so it would have a hz asymptote

thank you!

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[spiral vigil]

uhhhhh no

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[hazy igloo]

could someone help me with b and c

[hazy igloo]

im not sure how to get those answers, i just put the answers the website told me

so im just wondering how do i actauly solve it

i know how to get A, B, and C, but not sure on the others

[gritty rose]

,tex .sohcahtoa

have you learned that yet

[glossy valve]

rie.mann

[hazy igloo]

yea

im js bad confusde i think ik what to do but its wasnt working

[gritty rose]

you know one side length and one angle, so solve for the other variables using the trig functions

[hazy igloo]

so to find like hyp

it would be sin74.6= 4.4/hyp right?

but how do i solve from there thats my question

[spiral vigil]

remember to be in degree mode on your calculator

[gritty rose]

what is your whole question exactly

[gritty rose]

do you know how to solve for hyp from here

[hazy igloo]

how do i solve hyp from this

no that is my question basically

[spiral vigil]

if i gave you an equation like $4 = \frac8x$ could you solve it for x?

[glossy valve]

Hayley

[spiral vigil]

it's the same idea here

[delicate anchor]

GUYS help me with this:

[hazy igloo]

8/x * x/1?

lowkey no

[spiral vigil]

okay you need to review algebra then and do a lot of practice problems

[hazy igloo]

man my ahh in trig ion got time for that

if u just expalin once i got it fr

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[spiral vigil]

.reopen

[full forum]

✅

[gritty rose]

wrong person

i repeat. WRONG PERSON

look at the channel name

[spiral vigil]

(this is nythzn's channel)

[gritty rose]

you have a=b/c

then multiply by c on both sides to get ac = b

then divide by c to solve for a

[hazy igloo]

c in this case would be hyp right

i got a decimal that i dont think is right

.0589

[gritty rose]

!show

[full forum]

Show your work, and if possible, explain where you are stuck.

[hazy igloo]

okay so i got 74.6 c =4.4

then i divide 74.6 away from both sides to get c by itself

[gritty rose]

did you use sin(74.6)?

[delicate anchor]

GUYS ACTUALLY CAN ANYONE HELP ME WITH THIS:

[gritty rose]

<@&268886789983436800> spammer

[delicate anchor]

dUDE I JUST WANT HELP

[hazy igloo]

no im sorry im so confused

[gritty rose]

do you know where you got this equation?

[gleaming fractal]

#❓how-to-get-help

[hazy igloo]

yes cuz sin is equal to opp/hyp then u plug the angle into sin along w the sides u have

but im just mad consfused on how to solve from there

[gritty rose]

so you don't use 74.6, but you use sin(74.6)

[hazy igloo]

so i plug that into my calc

then what that number is i use it in the equatison?

[gritty rose]

correct

[spiral vigil]

ok write on your paper $\sin(74.6^\circ) = \frac{4.4}{\text{hyp}}$

[glossy valve]

Hayley

[spiral vigil]

then calculate sin(74.6) and rewrite that equation with the value you get

[gritty rose]

sine is a function. so sin(some angle) is just a number

[hazy igloo]

.9640=4.4/hyp is what i got

[gritty rose]

tell mods, not me

[hazy igloo]

okay i got it actaully t hinks for the help guys

.close

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[gritty rose]

,calc sin(74.6 deg)

[glossy valve]

Result:

0.96409540423411

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[radiant hemlock]

calculate the value of the generalized triple integral

[radiant hemlock]

Here's my attempt and it seems like I simplified the expression wrong.

[charred ivy]

@radiant hemlock off topic but what paper is that, nice handwriting btw

[radiant hemlock]

Haha thanks!

anyway, according to the solution suggestion given to me, the expression should be simplified to R^2rsin(theta). I get it to R^2*r so I must be close..

[radiant hemlock]

The paper is called "Whitelines"

nvm, I see my problem now

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[gilded prairie]

So im confused on the Amplitude ... the amplitude would be legnth of reaching the max or min value? so wouldnt it be 300?

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@gilded prairie Has your question been resolved?

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@gilded prairie Has your question been resolved?

[gritty rose]

Yes. Do you have something that says otherwise?

[gilded prairie]

i checked answer key and its saying that A is -300

and C 4300

[gritty rose]

In your case, A is not the amplitude

Amplitude = 300 is correct

Try calculating P(0) to see why

[gilded prairie]

i got 4000 since the given says a min value

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[gilded prairie]

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[gaunt flume]

Is there any way to solve this without lhospital rule?

[crisp geode]

try factoring

[gaunt flume]

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[gentle heart]

Can someone help with this task? Calculate the integral
xdV over the set E where E is the set of points between the cylinders x^2 + y^2 = 1
and x^2 + y^2 = 4, which are also between the surfaces z = 0 and z = y + 2.
How do I calculate the integral? I am not sure what the integral limits are here.

[fathom saddle]

Converting to cylindrical coordinates would make this pretty easy

https://tutorial.math.lamar.edu/classes/calciii/TICylindricalCoords.aspx

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@gentle heart Has your question been resolved?

[gentle heart]

Is this correct? because the answer would be zero then

nvm I just read the link you sent

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[merry swallow]

Kind of confused over what the author means by using the axiom of replacement twice

[merry swallow]

Isn’t it just a single application, we replace (a,b) in the set N x N with a—b?

[spice orchid]

you also do (a,b) ~ (c,d) iff a + d = c + b

oh i see

yeah im not sure where two applications is coming from if you're starting from N x N and applying a single equivalence relation

[merry swallow]

okay thanks for verifying, now I know i wasn’t missing anything obvious

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[untold swallow]

Hi, I got stuck doing this limit, I've tried doing it differently, but dont know what I'm doing wrong

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@untold swallow Has your question been resolved?

[untold swallow]

<@&286206848099549185>

[gritty rose]

this is right

on the next line you gained a power of t ?

and somehow gained a power of -1 in the exponent

and lost the cosine term

[untold swallow]

But that t isn't the t that is at the start with the 2t?

Or when. I move the 2 to the left of the Lim do I have to move also the t?

[gritty rose]

where did the t in the denominator go then

this is what happens when you try to do too many steps at once

[untold swallow]

I guess you mean this one?

Doesn't the limit of cos (3t)/ 3t =1?

[gritty rose]

no

[untold swallow]

Oh ok

[untold swallow]

So, it's ok up to here, the rest is wrong, right?

[gritty rose]

yes

[untold swallow]

And how could I continue? Cuz if I replace the 0 as it is, it's still undefined

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@untold swallow Has your question been resolved?

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[west mango]

can someone help me with this one

[potent shale]

I can try

What's the status?

[simple totem]

!status

[full forum]

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

[west mango]

1

lol

[simple totem]

do you know what a relation is

[west mango]

reflexive antisymmetric symmetric

[simple totem]

do you know what a relation is

[west mango]

yes

[simple totem]

do you know how to show a relation is reflexive?

[west mango]

when every element is related to itself

[simple totem]

a relation is reflexive if xRx for any x

so take a pair of subsets of X

(A,B)

and compute $A\Delta B$

[glossy valve]

maximofs

[simple totem]

your task is to show (A,B)R(A,B)

that is, $A\Delta B\subseteq A\Delta B$

[glossy valve]

maximofs

[simple totem]

hopefully it should be clear that it’s true

[west mango]

yes

[simple totem]

can you prove to me that it’s true? it should be like a single sentence

[west mango]

i mean every set is a subset of itself

[simple totem]

perfect

so is the relation reflexive?

[west mango]

yes

[simple totem]

good

now how about symmetric

do you know the definition of a symmetric relation

[west mango]

for all a and b in A if aRb then bRa

[simple totem]

good

now do you think this relation is symmetric?

also, do you know what $A\Delta B$ means

[glossy valve]

maximofs

[west mango]

symmetric difference of sets, union of A - B, B-A

[simple totem]

yup

so back to deciding whether R is symmetric

the fact that we’re doing a subset relation should tell you something about whether this is probably symmetric or not

so do you think it’s a symmetric relation?

[west mango]

im not sure xd

why the subset relation matters?

[simple totem]

is the relation < on R (the reals) symmetric?

if a < b, is it true that b < a?

the point i want to get across is that if A is a subset of B, B is not necessarily a subset of A

so maybe our first guess should be they this is not symmetric

[west mango]

oh i see

[west mango]

no

[simple totem]

so then let’s try to build a counterexample

a good thing to note is that the symmetric difference with the empty set gives doesn’t change anything

so in particular, the symmetric difference of X and the empty set is X

does that make sense?

[west mango]

{1} subset of {1,2} but {1,2} not a subset of {1}, like this?

[simple totem]

yes basically

what i’m getting at is

every element of P(X) is a subset of X

and incidentally

every element of $\mathcal P(X)$ is a subset of $X\Delta \emptyset$

[glossy valve]

maximofs

[simple totem]

that is
everything relates to (X, empty set)

[west mango]

so the symmetric difference between a set and an empty set returns the original set

[simple totem]

yes

X - nothing u nothing - X
= X u nothing
= X

[simple totem]

all we need to do now is find a symmetric difference that gives us something smaller than X

in language a little more relevant to your problem, we now have

$\forall\ (A,B) \in P(X)\times P(X), (A,B)R(X,\emptyset)$

[glossy valve]

maximofs

[simple totem]

now we just need to find a pair A,B so that

(X,empty set) doesn’t relate to (A,B)

which amounts to finding A and B so

$X \cancel\subseteq A\Delta B$

[glossy valve]

maximofs

[simple totem]

is this clear so far

[west mango]

yes

[simple totem]

ok

since X is the biggest possible set here

we just want a symmetric difference that doesn’t give us X

can you give me sets (A,B) so that their symmetric difference is not X

feel free to play around with a few examples

[west mango]

let X = {1, 2, 3, 4}, A = {1, 2} and B = {3, 4}. AΔB = (a-b)u(b-a) = {1,2} u {3,4} = {1,2,3,4}

let X = {1,2,3,4}, A = {1,2} and B = {2,3}, a-b = {1}, b - a = {3}, {1} u {3} = {1,3}

[simple totem]

X is given to you already

i was taking that to be X

you have the right idea i think

let’s walk through an easy example though

so as i explained above the symmetric difference of X and {} is X

and since X is the largest set there is here

everything is a subset of X

specifically every symmetric difference of A and B is a subset of X

so (A,B)R(X,empty set)

but now

what if we took A = B = {}?

what is the symmetric difference of two empty sets?

[west mango]

emptyset

(∅ - ∅) ∪ (∅ - ∅) = ∅ ∪ ∅ = ∅.

[simple totem]

yes

so (A,B)R(X,{})

since the empty set is a subset of X

but what if we flip this

is X a subset of the empty set?

[west mango]

no

[simple totem]

then this is an example of when R is not symmetric

did all of that make sense to you

[west mango]

yes makes sense

[simple totem]

ok so we found a counterexample to symmetry, so R is not symmetric

do you want to try the anti symmetric and transitive ones yourself

[west mango]

yes but what about the second one

[simple totem]

so for that we can use the fact that every subset can be written as a symmetric difference

hm

let me think about it a little more

ok i have an idea for number 2, i’m not sure if it’s what they’re going for
the symmetric difference of A and B gives us {1,2,7,8}

in order to get a symmetric difference that includes these elements, we need to move 1,2,7,8 into 2 different sets

that is, how many different ways can we put 4 elements into 2 sets

for each element we can pick one of the two sets

so we have 2^4 ways to move these around

then for the rest of the elements we can pick whether they’re in one of the sets, both of the sets, or neither
this amounts to multiplying by 2^4 for each other element in X

does this make sense to you

[west mango]

yes

but what about 4^6

[simple totem]

where did 4^6 come from

[west mango]

there are 6 remaining numbers which could be in C in D or in both or none, 4 possibili

[simple totem]

oh yes sorry

it should be 6^4

or bio

no

each has 4 possibilities

so we have 4^6

so total it should be 2^4 * 4^6 = 2^16

[west mango]

holy wack this is hard

[simple totem]

i think there may be an easier way

my mind just jumped at this specific approach

[west mango]

you know any good books where I can learn more about this topic?

[simple totem]

this is a pretty generic topic

any intro to proofs book will go over the first question

the second is like baby combinatorics

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[royal mantle]

i have no idea what is happening here whatsoever

[gritty rose]

is that your typing?

[royal mantle]

no

[gritty rose]

can you show the original source from your book

[royal mantle]

well, source is in spanish

i just translated it a bit for here

[gritty rose]

show it

[gritty rose]

looks like you typed that

take a picture or screenshot of your problem

[royal mantle]

its from word document, i enabled edition so that is probably why it looks like that

okay the whole thing is like this

i already did the first one

[gritty rose]

screenshot the original document

[royal mantle]

this is the original document sorry....

is not from a book

but from custom word homework document assigned to us

[simple totem]

porque estás escribiendo f(x) con la (x) debajo de la f

usualmente se escribe todo en la misma línea

[royal mantle]

it already looked like this

not my doing...

by the way i haven't touched anything there

we usually do it with paper and pencil and take photos of our notebooks

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[frosty warren]

This is my important test paper, but I don’t have the answer😭 I try to solve it. Is my answer right?

[frosty warren]

Please check my solution

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[frosty warren]

Help

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[hybrid cloak]

Can I have a walkthrough of how to do this

Dont know how to do it.

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@hybrid cloak Has your question been resolved?

[hybrid cloak]

<@&286206848099549185>

[gritty rose]

Find the distance halfway first

Use distance= speed*time

[light sonnet]

I mean tbh it's a lot easier than that, by process of elimination

[hybrid cloak]

2.5 miles

1/6 times 15

and idk about the second one bc im only given speed

fxck this.

.close

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[torn jolt]

|(x-3)/(x^2-4)| ≤ 1

[torn jolt]

how to solve this inequality

[brittle steeple]

|u| ≤ 1 becomes 2 inequalities

-1 ≤ u ≤ 1

So apply that to get rid of the absolute value bars

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[torn jolt]

.reopen

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✅

[torn jolt]

okay thanks

.close

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[gentle stump]

Bezout’s identity

Why can’t I just say that since gcd(a,b)|a,b by definition, then it must divide a(x)+b(y)

[gritty rose]

What proof are you looking at

[gentle stump]

https://en.wikipedia.org/wiki/Bézout's_identity

[swift shell]

Fuck

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[hazy igloo]

does this seem right

[brisk obsidian]

,calc 7/4 * cos(150 *pi/180)

[glossy valve]

Result:

-1.5155444566228

[brisk obsidian]

,calc 7/4 * sin(150 * pi/180)

[glossy valve]

Result:

0.875

[hazy igloo]

seems ritht to me

[brisk obsidian]

Yes.

[hazy igloo]

i just dont know how to format it

do you think they want like an exact answer

i had another one where the answer was exact answer so i just wanna figure that out

i got the answer

help help help help

plase please please

[twilit leaf]

What is ||v||?

[brisk obsidian]

Do you recall what I told you previously about calculating the unit vector?

[twilit leaf]

Oops

What is ||v||?

:I

||v||

Ok there lol

[hazy igloo]

sqrt v1^2+v2^2

[twilit leaf]

Which is?

[hazy igloo]

144

or 12

[twilit leaf]

sqrt(144), so 12

[hazy igloo]

yes

is that the anwer?

[twilit leaf]

So now divide vector v by ||v||

[hazy igloo]

so like 12/12?

[twilit leaf]

<12/12,0/12>

[hazy igloo]

would htat be U

[twilit leaf]

Check is ||u||=1

[hazy igloo]

yes it is right

[twilit leaf]

Ok, remember, to find the unit vector, you do (vector v)/(||v||)

[brisk obsidian]

$\hat{\bold{v}} = \frac{\bold{v}}{||\bold{v}||}$

[hazy igloo]

mna why is this so confsuing

so it woudl be 12,0/12

[glossy valve]

kookiemon

[brisk obsidian]

The carot is notation to indicate the normalized vector, or unit vector, of a vector.

You will also see |v| or ||v|| notation to indicate magnitude.

[hazy igloo]

yes i understand that part

i know what the plug in for ||v|| but not for v itself

| | v | |

[brisk obsidian]

Those are the components of the vector v.

[hazy igloo]

fuck i figured it out okay i think hold up

?

fuck that was it jesus

would i do the same steps for this one?

[brisk obsidian]

Calculate the unit vector of u then multiply by the magnitude ||v||.

[hazy igloo]

would this be the answer <2/8,2/8>?

[brisk obsidian]

What is the magnitude of u?

[hazy igloo]

8

sqrt 8

[brisk obsidian]

<2/sqrt(8) , 2 /sqrt(8)> is the unit vector of u.

You divide the components of u by the magnitude of u.

[hazy igloo]

<1/sqrt2, 1/sqrt2

for V?

[brisk obsidian]

,calc 8 * 2/sqrt(8)

[glossy valve]

Result:

5.6568542494924

[brisk obsidian]

,wolf 8* 2/sqrt(8)

[glossy valve]

Results provided by WolframAlpha

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

Requested by kookiemon#0

[hazy igloo]

so thats V?

[brisk obsidian]

That's how you will find the components of v.

[hazy igloo]

so the answer ouwld be <4sqrt2,4sqrt2> right?

im sorru im just not understanding very well this is my first day leraning vectors shits ocnfusnog

[brisk obsidian]

No worries.

We were all new to this at one point too. 😉

[hazy igloo]

i got an exam with part of this stuff tmw

im so screwed lol

[brisk obsidian]

$\hat{\bold{u}} = \frac{\bold{u}}{|\bold{u}|}$

[glossy valve]

kookiemon

[brisk obsidian]

As long as you know how to calculate the magnitude of a vector, you can find its unit vector using that formula.

A unit vector is just a vector inside of a unit circle.

[hazy igloo]

ohhhhhh taht makes a lot more since

first i need to find magn of U so it would be

sqrt 8

shouldnt it be <2/sqrt8, 2/sqrt8>?

[brisk obsidian]

Yes.

The image I posted is just some generic vector I made. Nothing to do with your question.

This is the vector u = <2,2>.

And its unit vector.

[hazy igloo]

okat that kinda makes since

why is this wrong i thoguht taht was it

[brisk obsidian]

That is the unit vector of u. To find the vector v, you multiply the unit vector of u by the magnitude of v.

$\bold{v} = ||\bold{v}|| \cdot \frac{\bold{u}}{||\bold{u}||}$

[glossy valve]

kookiemon

[hazy igloo]

ohhhh so then its 8*2/sqrt8

[brisk obsidian]

Correct.

[hazy igloo]

<4\sqrt{2},4\sqrt{2}>

[brisk obsidian]

Correct.

[hazy igloo]

would tha tbe the answer?

[brisk obsidian]

Yes.

[hazy igloo]

man thats confusing

is there any other tips you recommend i should do/learn ?

[brisk obsidian]

To understand what's actually going on, realize that ||v||/||u|| is just a ratio.

You are just multiplying a vector by a ratio to get a resized vector.

[hazy igloo]

ohhh yeah that makes a lot more sense

this class im taking is like 4 montjs of info in like the span of 5 weeks

[brisk obsidian]

Calculus or just vectors?

[hazy igloo]

its a trig class

[brisk obsidian]

Never take a STEM class in the summer. There is just way too much to learn.

[hazy igloo]

yeah i lowkey regret it

im a sohpmore going into junior year and last year i took alg 2

and for junior year im trying to take calc ab and skip precalc trhough my school

so i took this class to try and prepare for it

[brisk obsidian]

Trig and Precalc usually cover most of the same topics.

[hazy igloo]

does this one seem rught?

[brisk obsidian]

No.

[hazy igloo]

shi

thought i had it

[brisk obsidian]

How did you calculate the magnitude?

[hazy igloo]

36^2+15^2

got sqrt39

then i got <-36/sqrt39, -15/sqrt39>

[brisk obsidian]

No.

,calc sqrt((-36)^2 + (-15)^2)

[glossy valve]

Result:

39

[hazy igloo]

oh its just 39

[brisk obsidian]

Yes.

[hazy igloo]

i was close silly error

[brisk obsidian]

That's how it usually goes.

[hazy igloo]

okay one more then i think i might be good if you wouldnt mind helping?

do i kinda do what i did before witht his one?

[brisk obsidian]

Find the unit vector of v and then verify the magnitude of that unit vector is equal to 1.

So divide the components of v by the magnitude of v.

[hazy igloo]

so the vector the sqrt17

the unit vector of v would be <1/sqrt17, -4/17>

then i would divide for example 1sqrt17 by sqrt17 again??

[brisk obsidian]

The magnitude of v is sqrt(17). You have the correct unit vector.

Calculate the magnitude of the unit vector that you found in the same manner as before to verify that it is equal to 1?

[hazy igloo]

so like 1/sqrt16^2+4sqrt17^2?

[hazy igloo]

yeah thats what i thought so

this is the right answer

nbm

that is notright

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[wintry shore]

This question may not come in the format of a standard math question, i hope that is okay

Anyways, ive got a little bit of a road block in programming here

I am trying to make this character follow the camera's pitch rotation. Here you can see i was successful in the spine and neck:

https://devforum-uploads.s3.dualstack.us-east-2.amazonaws.com/uploads/original/5X/6/3/b/5/63b53217ef0cf33e09c19745e38d351619b4e9ca.gif

[wintry shore]

But the arms are not so simple

The neck starts out completely aligned with the direction it needs to turn. (Orientation in x, y, z format starts out at 0, 0, 0. All i need to do here is modify the x component by θ degrees and it tilts forward perfectly). But the arms start out with a specific rotation, which in turn also throws the axis off of an easy rotation.

Here's an example of what i mean:

The arm bone here points downward at many different angles. Its precice orientation at this point is (-25.567, 34.957, -149.071).

The green, blue and red lines represent the three axis i can rotate them on. But I want to rotate the arm about the yellow line. Does anyone know how this can be accomplished?

I know this will need to be done via a combination of all three vectors. I just need to find out the proportions of each.

[wintry shore]

In the gif above, you can see how the spine and neck bend perfectly, but the arms (only being rotated about the red axis) bend the wrong way, due to only one of the axis being affected

Programming is tough because everything is done in reverse from what you've learned in math class. In math class, you are given and an equation and are trying to find the variables. In programming, you are given the variables and are trying to find the equation

[nimble fable]

Wait so let me see if I understand the question

Is it just that you're trying to rotate along an axis that doesn't align nicely with the axes you can easily rotate on

[wintry shore]

yes

Instead of just increasing the X axis by 5 degrees or whatnot, I have to rotate all 3 axis by a unknown amount in order to rotate along the yellow line

[nimble fable]

Would a change of basis matrix not work here?

[wintry shore]

Basis matrix?

[nimble fable]

Do you know any linear algebra

[wintry shore]

I do not believe i have taken the class

[nimble fable]

It's a (change of basis) matrix not a (change of) basis matrix

Though change of basis matrix might still be a weird thing to say

[wintry shore]

As far as mathematics, the most relevant classes I have taken are Calculus and trigonometry

though I cannot say either class has prepared me for this problem

[nimble fable]

Yeah this feels like a problem that's easy if you've done linear algebra but tricky otherwise

Let's see if I can explain what to do

Ok so is this problem equivalent to the following

You just want to rotate along a line that's not a coordinate axis right?

[wintry shore]

yes

[nimble fable]

Ok then you don't even need to change basis or anything

There's just a standard formula for this I can look up

Change of basis is just a tool you could use to derive the formula, but isn't necessary to apply it

https://en.m.wikipedia.org/wiki/Rotation_matrix

This should have the answer

Should be this

So to apply this you first write the axis you're rotating around as a unit vector (u_x, u_y, u_z)

[wintry shore]

Is that unit vector in degrees or coordinates?

[nimble fable]

Coordinates basically

[wintry shore]

okay so

In this case

Hold on a sec

[nimble fable]

Pick a point on the line you're rotating around

[wintry shore]

The unit vector would be (-1, 0, 0)

[nimble fable]

You need to write it in terms of the basic axes you can rotate around

[wintry shore]

in terms of the

wait

so in relation to the axes?

hmm

[nimble fable]

If I'm understanding correctly, the basis axes here are the green, blue, and red ones, and yellow is what you want to rotate around

So take a unit vector along the red, green, and blue lines

[wintry shore]

Okay

I think i can do that

[nimble fable]

Actually that's probably what you want to start with since those won't be the standard unit vectors

[wintry shore]

The unit vector of the red, green and blue axes is (-0.5168537497520447, -0.431569367647171, -0.7393308877944946)

[nimble fable]

That's a unit vector for one of them?

[wintry shore]

yes, most likely the red one

[nimble fable]

Ok so take a unit vector on each one, and also on one the yellow line

[wintry shore]

The yellow one will be (-1, 0, 0)

[nimble fable]

The one on the yellow line should just be (1,0,0). Or (-1,0,0) works too

[wintry shore]

Okay so

[nimble fable]

Doesn't really matter which you use.

It'll just change the direction you'll end up rotating in, not the axis

Now what you want to do is write the yellow one in terms of the other ones

Honestly I'm not sure if there's a good way to do this without learning linear algebra

[wintry shore]

Yellow: (-1, 0, 0)

Red: (-0.5168537497520447, -0.431569367647171, -0.7393308877944946)

Green: (-0.5759537816047668, -0.46364542841911316, 0.6732831597328186)

Blue: (0.6333557367324829, -0.7738093137741089, 0.008927153423428535)

Okay

[nimble fable]

I could maybe walk you through the formula, but you'd want to get an intuition for why it works

If you ever wanted to debug, rewrite your code, or solve a similar problem

[wintry shore]

I think i remember possibly doing something like this at some point, im not sure

[nimble fable]

It actually isn't that tricky is the good news

And it'll serve you very well to learn as a programmer

Linear algebra is essential in computer graphics

The outline is this

You've just represented the red green and blue unit vectors in terms of the standard ones

It turns out you can encode that representation in what's called a matrix

Then if you take the inverse of that matrix (which is something any good math library for any programming language will have a standard function to do) it'll give you a representation of the standard unit vectors in terms of the yellow red and blue ones

This is called a change of basis

[wintry shore]

hold on a sec

[nimble fable]

This will let you represent the yellow unit vector in terms of the red green and blue ones

[wintry shore]

It looks like we're building something kinda like this

https://create.roblox.com/docs/reference/engine/datatypes/CFrame

I never really understood what the nine rotational arguments meant or how they worked

[nimble fable]

Then you can apply the formula for rotating around an axis

[wintry shore]

That makes sense now

[nimble fable]

And then you use your original change of basis matrix to convert back to the standard coordinates and boom you have your rotation formula

Look up 3blue1brown's essence of linear algebra series on youtube

It's under 2 hours long and will give you a great overview of the core concepts of linear algebra

Then study it a bit in whatever way feels best (textbook, youtube lectures, etc) and go back to this problem

It'll take a bit of time, but will save you far more time in the long run

[wintry shore]

Alrighty

I will give it a look

[nimble fable]

As an added bonus, linear algebra is used in a lot of other cool stuff too

And is just a beautiful subject in it's own right

Also in the future I think the kind of question you asked is better suited for a programming server.

[wintry shore]

Thats true, but a lot of them were stumped as they did not know linear algebra.

[nimble fable]

For here I'd maybe more just ask "what kind of math is needed to solve X problem" instead of asking for a solution

And I'd ask that sort of thing in the discussion channels not the help channels

[wintry shore]

Yeah thats what i was kinda going for, trying to see if anyone knew how to build the equation

[nimble fable]

But I mean, what you did here worked so eh

[wintry shore]

.close

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[burnt birch]

plz help

is my answer right 2.

[civic beacon]

https://tenor.com/view/brain-trash-spongebob-squidward-dumb-gif-17233216

[light sonnet]

You do not need to keep closing and opening channels, stay in one channel and wait patiently

[burnt birch]

bro i been here since like 10

[light sonnet]

Okay, and?

[burnt birch]

like i gtg soon

olz help

pzl

plz

i will pay even

[light sonnet]

You still don't need to close and open channels to get attention

[burnt birch]

i pinged helpers

but i kept getting removed

[civic beacon]

I already told you it's wrong

[burnt birch]

<@&286206848099549185>

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[proud spear]

This is a graphical representation of a problem I found in an old textbook. In this drawing each of the two circles with centers B and C has a radius of 3 cm , AB ⟂ CB and |CA| = 10 cm . Find out the angle BAC.

[elfin stream]

sin BAC = (BC/CA)

BC = the sum of radius of two circles

[proud spear]

BC = 6cm

[elfin stream]

yes

use inverse trig functions then

to find BAC

[proud spear]

I posted this question because I'm not into this yet

Is there a simpler solution?

[elfin stream]

not one i can think of

[proud spear]

ok, so from what you said I can deduce, that sin BAC = 3/5

[elfin stream]

using inverse trig functions

arcsin(sin(BAC)) = arcsin(0.6)

BAC = arcsin(0.6)

u need a calc to find arcsin(0.6)

[proud spear]

my calculator doesn't have arcsins : (

[clear vigil]

use your phone :D

[serene hazel]

use shift button

[clear vigil]

wait I remember arcsin(3/5) = 37°

It's the special 3-4-5 right triangle

[proud spear]

of course it is : ) problem solved!

i guess

[clear vigil]

great

[proud spear]

.close

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[spice orchid]

#help-18 message
What are some sufficient conditions (weaker than being twice differentiable) for an everywhere differentiable function f : R -> R to have a continuous derivative?

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@spice orchid Has your question been resolved?

[distant shuttle]

hmm

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[rose rain]

i wonder if ive seen this before

[gritty rose]

https://math.stackexchange.com/questions/78825/example-of-a-function-that-is-not-twice-differentiable

x^2 * sin(1/x) but I don't know how to generalize it

[rose rain]

i remember this bad boy:

idk if it helps

[
g_n(x) = \begin{cases}
x^n \sin(1/x) , \text{if $x \neq 0$} \
0 , \text{otherwise}
\end{cases}
]

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[ocean birch]

.close

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