If you do this you just have to find the minimum value of ab
#help-10
red ice
Cause a^2 b^2 = (ab)^2
And we have -(-2 a^2 b^2) = 2a^2 b^2
There's a trick, recall (x - y)^2 = x^2 - 2xy + y^2 >= 0 for all x, y
So we just have 4 - 2xy >= 0
red ice
Yeah we just sub in x^2 + y^2 = 4
Shit that's the maximum value
(The min value of xy is actually -2, not 2)
little helm
why is it not -4^2 + 2(xy)^2?
little helm
(The min value of xy is actually -2, not 2)
ok i get it
so how do we proceed from here?
red ice
ok i get it
Then just sub everything in, so 4 + 4ab becomes 4 + 4(-2) = -4
-((a^2 + b^2)^2 - 2a^2 b^2)) = -(4^2 - 2(ab)^2) = -(4^2 - 2(-2)^2)
= -8
-4 - 8 = -12
red ice
Then just sub everything in, so 4 + 4ab becomes 4 + 4(-2) = -4
We already did this
Just sub ab = -2 in
little helm
let just ignore everything for a sec
we have 4ab + 4(ab)^2
act nvm
i get it
tks for the help
restive loom
11 if you add strings or 2 if tou add integers
tardy tusk
2*1
red ice
.close
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
proven spindle
Given z = x + iy
u(x,y) and v(x,y) are some function of variables, x and y, such that the powers of each variable is at most 1
f(z) = u(x,y) + i v(x,y)
I am wondering if f(z) in terms of z is also at most of order 1, and if so, how do i prove it?
obtuse pebbleBOT
Given z = x + iy
u(x,y) and v(x,y) are some function of variables, x and y, such...
proven spindle
Im working on part (b)
I will present my solution to (a)
uxx = d/dx (ux) = d/dx (vy) = d/dx (uy) = d/dx (-vx)
uxx = uxx = vyx = uyx = - vxx
uyy = d/dy (-vx) = d/dy (ux) = d/dy (vy)
uyy = -vxy = uxy = vyy
er...some handwave magic(i don't know how to prove the following) i know is not generally true, but for analytic functions, i believe the order of 2nd derivatives of u and v wrt x or y does not matter.
so.... uxy = uyx etc
so uyy = uxx, - vxy (= uyy) = vyx(= uxx) which is equal to 0.
part (a) done
i hope
part (b) attempt:
Integrating...uxy, uyy, vxy, vyy wrt y
uy = A1(x) + k1
ux = A2(x) + k2
vx = A3(x) + k3
vy = A4(x) + k4
Integrating...uxx, uyx, vxx, vyx wrt x
ux = B1(y) + c1
uy = B2(y) + c2
vx = B3(y) + c3
vy = B4(y) + c3
Integrating them all with respect to (insert variable) to reobtained u and v
u = k1y + yA1(x) + G1(x) + r1 = k2x + xA2(x) + H1(y) + r2 = c1x + xB1(y) + H2(y) + r3 = c2y + yB2(y) + G2(x) + r4
where A, B, G, H are functions of at most order 1
comparing coefficients
and (assuming true) that uxy = uyx
k2x + xA2(x) + H1(y) + r2 = c2y + yB2(y) + G2(x) + r4
r2 = r4 = r1 = r3
yB2(y) = 0
xA2(x) = 0
k2x = G2(x) = G1(x) = c1x
H1(y) = c2y = k1y = H2(y)
yA1(x) = xB1(y) which are absent in the other possible forms, so = 0
therefore, u(x,y) is at most order 1 in u and v.
Can do the same for v(i did and i think its the same...so i won't post here, but i can in comments)
So yeah, how do i show that f(z) in terms of z is at most order 1.
trivially, by inspection, i can see that Q1(z) + Q2(z^2) + ... orders of z, are linearly independent wrt z.
(i am bad at linear algebra)
But idk how to prove it in general, that there is no composition of function of z(like say... L(z)/Oz^n etc), such that f(z) must be in terms of z(order 1) only.
obtuse pebbleBOT
@proven spindle Has your question been resolved?
obtuse pebbleBOT
@proven spindle Has your question been resolved?
obtuse pebbleBOT
@proven spindle Has your question been resolved?
obtuse pebbleBOT
@proven spindle Has your question been resolved?
obtuse pebbleBOT
@proven spindle Has your question been resolved?
cyan void
Could you summarize your question at this point?
brazen viper
@proven spindle
If u and v are at most degree one, then you have f(z) = (ax + by + c) + (dx + ey + g) i
For some coefficients a-g.
Given ux = uy and ux = vy and uy = -vx we have
a = b = e = -d so we have
f(z) = (ax + by + c) + (dx + ey + g) i
= (ax + ay + c) + (-ax + ay + g) i
= a(x + iy) + a(y - ix) + c + gi
= az - aiz + w
= (1 - i)az + w
mighty pilot
obtuse pebbleBOT
atomic saddle
this should be correct, not sure why its wrong
s = r*theta
tardy tusk
2πr = S
Means
360°r = S
So you need
(360-102)°r= s???
I just thought it works this way
atomic saddle
one for degrees and one for rad
right, so just multiply 102 by pi/180 and then 2* (102pi/180)
204pi/180 which simplifies to 17pi/15
atomic saddle
3.57
atomic saddle
right sorry, I had a typo, i did mean 3.56
mighty pilot
3.56 still came up wrong
atomic saddle
I dont understand why its not giving you it as correct, that should be correct
mighty pilot
yea
atomic saddle
does it tell you to round to a certain place?
atomic saddle
maybe it wants exact, try 17pi/15
mighty pilot
i assumed they wanted it in radical form but still
okay
nope, incorrect
i have one more attempt lmao
tardy tusk
2πr = S
Means
360°r = S
So you need
(360-102)°r= s???
Why am i incorrect?
mighty pilot
ima try yours rq
mighty pilot
516pi/180
atomic saddle
yeah, 3.56 is quite literally the answer
is there any more specific directions at the top of the page or anywhere else?
mighty pilot
Should you simplify it?
no they dont care about simplifying when turning it in online
atomic saddle
i mean unless youre using a different keyboard that has modified english letters that webassign isnt reading 😂
mighty pilot
i mean unless youre using a different keyboard that has modified english letters...
I submitted similar questions before this, and all are correct
atomic saddle
maybe it wants 3.56047167407
mighty pilot
i wont try it for now lol, since its my last attempt
atomic saddle
yeah
mighty pilot
il do it infront of the professor incase it comes up wrong
have a good one, thanks
thanks @tardy tusk
.close
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
misty sphinx
for the last part i understand that it maximises the denominator and thus minimises v^2 but im confused how to prove that the turning point of F(theta) is a max and not a min
obtuse pebbleBOT
for the last part i understand that it maximises the denominator and thus minimi...
@misty sphinx Has your question been resolved?
cedar lintel
pls help
sage dagger
what do you have to do?
hexed gull
pls help
derive?
product rule
[A * B]' = A'*B + A*B'
apply to f(x)
and you have it
cedar lintel
what do you have to do?
find the derivative using the rules of derivation
obtuse pebbleBOT
@cedar lintel Has your question been resolved?
pearl blaze
find the derivative using the rules of derivation
apply the product rule
cedar lintel
I don't know how to solve it 😦
pearl blaze
consider u as the first term (3x^4+..) and v as the second term
now first differentiate u w.r.t x and multiply v
next diff v w.r.t u and multiply u
add both of these n that's the result
obtuse pebbleBOT
@cedar lintel Has your question been resolved?
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
zenith canyon
obtuse pebbleBOT
zenith canyon
is 9/loge(p+1) not differentiated
because its considered like an coeeficient of loge(x+1)???
red ice
because its considered like an coeeficient of loge(x+1)???
Yes
Remember that p is just a constant
So ln(p + 1) is also a constant
And 9/ln(p + 1) is also a constant
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
scenic fjord
obtuse pebbleBOT
obtuse pebbleBOT
What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
For R1, why it’s symmetric and transitive? I can’t see from there, isn’t for symmetric, we need to have (a,b) and (b,a); (b,c) and (c,b)
scenic fjord
For R1, why it’s symmetric and transitive? I can’t see from there, isn’t for sym...
if we don’t have symmetric in R1, how we know it’s transitive?
grizzled shore
Oh r1
scenic fjord
It has ab and ba
Isn’t it just (a,a)
timid silo
yeah
timid silo
whats the symmetric counterpart of a,a
timid silo
whats the symmetric counterpart of a,a
ofc a,a itself
grizzled shore
So we need not say anything about b and a and the relation =
timid silo
so there you have it its symmetric
scenic fjord
So like the case for aa,bb,cc, we ignore symmetric and transitive? Or assume that it’s already in there?
grizzled shore
Symmetric relies on aRb being in the relation, then it implies bRa is also in the relation
No you’re missing the point of symmetry
scenic fjord
Symmetric relies on aRb *being* in the relation, then it implies bRa is also in ...
Yea, that means we need (a,b) and (b,a)
grizzled shore
It tells if if a is related to b then b is related to a
grizzled shore
Yea, that means we need (a,b) and (b,a)
No you don’t
Because neither are in R
You need to start with 1 of them in the relation
You need to start with the statement “(a, b) ∈ R”
Then symmetry tells you ah, then “(b, a) ∈ R”
grizzled shore
You need to start with the statement “(a, b) ∈ R”
But you don’t even have the first line
You’re looking at an if A then B statement but you don’t have that A is true
scenic fjord
But you don’t even have the first line
Yes, that’s why I wonder why we can put that as equivalence
grizzled shore
Well you don’t need it
scenic fjord
Since we don’t have that to prove it’s symmetric
grizzled shore
The definition of symmetry says if (a, b) is in R, then so must (b, a)
timid silo
@scenic fjord do you need help?
grizzled shore
But (a, b) is not in R, so there’s nothing to discuss
(For all pairs (a, b), clearly all the other combinations also aren’t in R)
scenic fjord
But (a, b) is not in R, so there’s nothing to discuss
Hmm, the point of me not clear is, I understand (a,b) is not in R so nothing can be discussed. But how come we can conclude its equivalence relation? Equivalent relation need to be reflexive, symmetric and transitive
grizzled shore
Hmm, the point of me not clear is, I understand (a,b) is not in R so nothing can...
Okay
It’s confusing because they use a b c for elements
scenic fjord
But from R1, we have nothing to discuss for symmetric and transitive
grizzled shore
And the definition of symmetry also uses a b
But they are not the same letters so let’s switch them up
Definition: a relation R is symmetric if for all 𝛂, 𝛃, (𝛂, 𝛃) ∈ R implies (𝛃, 𝛂) ∈ R
grizzled shore
Have you seen that definition before?
timid silo
why are you speaking riddles?
its a simple doubt
why make it so complicated?
just some common sense and its done
grizzled shore
Yeah okay they use x and y
That’s good
So let’s check if the relation is symmetric
scenic fjord
Yeah okay they use x and y
Yes, but clearly from R1, I didn’t see any y😅
Or they let y=x?
grizzled shore
So for everything in R, we need to check this: flip the pair around, is it still in R?
Does that make sense
scenic fjord
You mean if (a,a) being flipped, it’s still (a,a)
grizzled shore
Indeed!
timid silo
You mean if (a,a) being flipped, it’s still (a,a)
bcz the relation is from a to a
its not mentioned
but its understood
so u have to flip and see
grizzled shore
But we don’t need to check if (b,a) is in R, because (a, b) isn’t in R so we don’t have to check it
We only need to check things that are in R
grizzled shore
The R is an element of P(AxA)
timid silo
The R is an element of P(AxA)
wow bro ur even confusing me
scenic fjord
The R is an element of P(AxA)
Actually I also don’t understand this
My lecturer skipped this
grizzled shore
Well R is a set right?
scenic fjord
And assume that the math students have took discrete math and for my case, I did not cause I’m from another faculty
scenic fjord
Well R is a set right?
Yes’m but here’s how it becomes RxR
Modern math just make no sense to me 🫠
grizzled shore
Well, functions from A to B are a subset of P(AxB) too but people don’t think about that
The idea is this, R is a set with pairs of things
scenic fjord
Well, functions from A to B are a subset of P(AxB) too but people don’t think ab...
This I think I can get, I saw from the notes
But I don’t get why it’s AxA
grizzled shore
What sort of things? Well you take 1 thing from A, then another thing from A
And you make it a pair
Then you put this ordered pair into the set R
scenic fjord
Oh, so for AxB, there’s sets A and B, if for AxA, then there’re just set A, and because there’s not set B
grizzled shore
I wrote the wrong thing
scenic fjord
We just take AxA?
grizzled shore
It’s an element not subset of the power set
The power set already is a set containing all the different subsets of combinations
grizzled shore
Oh, so for AxB, there’s sets A and B, if for AxA, then there’re just set A, and ...
Yeah
grizzled shore
Well, functions from A to B are a subset of P(AxB) too but people don’t think ab...
In general we like to talk about functions going from 1 set to another
So we name them differently
scenic fjord
How about transitive for R1
grizzled shore
But many times for relations we want to talk about binary relations
scenic fjord
Is it because we don’t have any info for that, we ignore that?
grizzled shore
So for transitive you pick 2 things in R where 1 ends with ? And the other starts with ?
For example (x, y) and (y, z)
(x, y) ends with y and (y, z) starts with y
scenic fjord
I know if (x,y) and (y,z), then there’s (x,z)
grizzled shore
Then you see, is (x, z) in R?
grizzled shore
Is it because we don’t have any info for that, we ignore that?
Ok but you’re right her
grizzled shore
So for transitive you pick 2 things in R where 1 ends with ? And the other start...
We couldn’t find any such pairs of pairs
So we’re done
It is true because there are no counterexamples
scenic fjord
Ok but you’re right her
Yes, if we can’t find this, how we conclude it’s transitive?
grizzled shore
Cos transitivity stars with “if (x, y) in R and …”
But it’s true when I can’t satisfy the if
scenic fjord
Okay, to make it simple, like for R1, there aren’t any info telling us (a,b) and (b,a) for symmetric, we ignore, and if there aren’t any info telling us (a,b), (b,a), then we cannot conclude it as (a,a) or (b,b) which is transitive
scenic fjord
But it’s true when I can’t satisfy the if
Oh, so if we don’t have the info, the info should assume to be correct?
grizzled shore
Yeah because nothing I will pick will break transitivity
Because I can’t pick something to check transitivity
So it is transitive
grizzled shore
Oh, so if we don’t have the info, the info should assume to be correct?
So yes
scenic fjord
Okay, and for R2, I can see it’s reflexive, symmetric and transitive
But I have a confusion, for symmetric, we don’t need to include (b,c) and (c,b) as well?
grizzled shore
But (b, c) isn’t in there
So we don’t need to check
As you go along each element you think to yourself
scenic fjord
Or like if we got one, then the one we don’t have we also assume it’s correct
grizzled shore
Ok (a, b) is in R, let’s check if (b,a) is also in R
Oh it is! Perfect.
Oh (b, a) is in R, let’s check if (a, b) is also in R
Oh it is! Perfect
scenic fjord
As you go along each element you think to yourself
Actually after I go through the example, I felt like for it to be reflective in this case of set A
We need to have all (a,a), (b,b) and (c,c) to be reflexive to be non reflexive
grizzled shore
We’ve now checked that for every element in R, and its flipped version is also in R, so this is symmetric
grizzled shore
We need to have all (a,a), (b,b) and (c,c) to be reflexive to be non reflexive
What did you just say here lol
That’s not a sentence
scenic fjord
Actually after I go through the example, I felt like for it to be reflective in ...
Continual for this
grizzled shore
Err
You need all your aRa, bRb, cRc, dRd, etc for all the elements in S
Which since there’s only 3 things in S
You need just them to be in the relation
scenic fjord
You need all your aRa, bRb, cRc, dRd, etc for all the elements in S
Okay, I think I get what you mean, and basically like what I think.
grizzled shore
So R = {(a, a), (b, b), (a, b), (b, a)} is symmetric but not reflexive
scenic fjord
So R = {(a, a), (b, b), (a, b), (b, a)} is symmetric but not reflexive
But this is transitive right?
Since we can get back all the relation in R
scenic fjord
But not reflective!!
Yea, I think I get the concept for reflexive now
grizzled shore
My point was that (c, c) isn’t in there breaks reflexivity but not symmetry
scenic fjord
My point was that (c, c) isn’t in there breaks reflexivity but not symmetry
So for R6, because there’s not (c,b) and (a,c), it’s not symmetric?
grizzled shore
Yes
scenic fjord
And also for R6, since there isn’t any thing that can produce back (c,c), so it’s not transitive
grizzled shore
And also for R6, since there isn’t any thing that can produce back (c,c), so it’...
What do you mean produce back (c, c)
scenic fjord
Like because R6 doesn’t have (c,a) and (a,c) we cannot get back (c,c) which is in the R itself
grizzled shore
No no you’re thinking about it backwards
We need to pick 2 then see if it’s transitive
Not pick 1 then see if I can make it from 2 other
But that set is indeed not transitive, I’ll give you some time to think about why not
scenic fjord
You mean (b,a) and (b,c) produce (a,c)?
grizzled shore
That’s not what transitivity says
scenic fjord
Not pick 1 then see if I can make it from 2 other
This is what I concluded myself from the examples
Like if you see from R3, it’s not reflexive and not symmetric, but I can see it’s transitive
grizzled shore
So for transitive you pick 2 things in R where 1 ends with ? And the other start...
.
You need to pick this pair of pairs
That has 1 ends when ?, the other starts with ?
scenic fjord
Meaning to say (b,a) and (b,c), then (c,a)
grizzled shore
Meaning to say (b,a) and (b,c), then (c,a)
The first one ends with a, the second one starts with c
a is not c
You did not pick a valid pair to check transitivity
scenic fjord
You need to pick this pair of pairs
So, you are taking that I cannot take (a,b) and (b,a)?
grizzled shore
So, you are taking that I cannot take (a,b) and (b,a)?
I did not say that
(a, b) ends with b and (b, a) starts with b
This is a valid pair to check for transitivity
scenic fjord
The first one ends with a, the second one starts with c
Yea, but isn’t (a,b) same with (b,a)
Or it’s not interchangeable?
grizzled shore
Transitivity requires that the (first element from first pair, second element from second pair) to be in R
In this case that would be (a, a)
grizzled shore
So, you are taking that I cannot take (a,b) and (b,a)?
(a, a) is indeed in R, so these 2 pairs don’t break transitivity
grizzled shore
Yea, but isn’t (a,b) same with (b,a)
Of course not
scenic fjord
(a, a) is indeed in R, so these 2 pairs don’t break transitivity
Yea, so (a,b) and (b,a), then (a,a), so it’s transitive or not?
grizzled shore
we use round brackets to denote ordered pairs where order is important
grizzled shore
Yea, so (a,b) and (b,a), then (a,a), so it’s transitive or not?
We can’t say for sure if it’s transitive, but this pair doesn’t break transitivity
We need that every pair of ordered pair doesn’t break transitivity to call the relation transitive
We can’t say from just 1 selection to test
That’s why I keep saying “this pair doesn’t break transitivity”
scenic fjord
That’s why I keep saying “this pair **doesn’t break transitivity**”
Okay, thanks, I think maybe I kinda get that
grizzled shore
But there is a pair in R6 that breaks transitivity
scenic fjord
Btw, how actually did you guys study for pure math and especially for the proving steps?
grizzled shore
I started with sets and logic
scenic fjord
But there is a pair in R6 that breaks transitivity
Huh? Only a pair?
grizzled shore
It very much helps
scenic fjord
I thought there’s some
grizzled shore
Huh? Only a pair?
I say there is a pair, there may be more
It doesn’t matter
As long as there is at least 1 pair that breaks transitivity, the relation is not transitive
scenic fjord
I say there is a pair, there may be more
Okay, so if there’s a pair (at least a pair) then it breaks transitivity
scenic fjord
I started with sets and logic
I think I asked some of the math students, they said they take so proving lesson and discrete math before they took this
grizzled shore
I mean I saw a pair and concluded it’s not transitive so I didn’t look any further
scenic fjord
There is at least one (that I saw)
Okay, then I get this equivalence relation. Tqvm
grizzled shore
Show that R6 is not transitive
scenic fjord
Show that R6 is not transitive
There no (a,c) in R6?
grizzled shore
There no (a,c) in R6?
which are the 2 pairs
scenic fjord
(a,b) and (b,c)
grizzled shore
indeed!
very good
i picked a different pair
see there are more than one but just finding 1 is fine
i picked (c, a) and (a, b) then we find that (c, b) is not in R
when you dont know, it's always the "wacky" ones that'll break something if it breaks it at all
aka those weird lone (c,b) or missing (c, c) or something weird like that
scenic fjord
indeed!
I got one question, if I choose (a,a) and (a,b), I cannot say it’s (a,b) again right? Since I need to take (x,y) and (y,z) for transitive right?
scenic fjord
In this case where I’m choosing (a,a) it’s (x,x) and it’s not (x,y)
grizzled shore
(a, a) ends in a
(a, b) starts with a
perfectly valid pair to check transitivity!
is (a, b) in R?
yep!
it's fine
scenic fjord
Oh, thought must be different, now I understand more
grizzled shore
nope
you just keep picking from R
with replacement
if you wish to think about it like that
you can also pick (a, a) ends in a, (a, a) starts with a
so surely, (a, a) must need to be in R for this to be transitive
and yes it is!
great!
scenic fjord
great!
Now that I can see whether it’s an equivalence equation. But if the question is like let R be defined on set of integers Z by xRy iff x+3y is even. How can I check that?
Like for reflexive, I write xRx that is x+3x=4x?
Like if they give generally, I don’t know how to prove or disprove 😅
grizzled shore
and is 4x even for all integer x?
scenic fjord
and is 4x even for all integer x?
Yes
grizzled shore
okay then the relation is reflexive
scenic fjord
But should we further write 4x to 2(2x) or no need?
grizzled shore
and is 4x even for all integer x?
the "for all integer x" part covers the fact you have to check everything
grizzled shore
But should we further write 4x to 2(2x) or no need?
well i mean you could say that if you wish
that 2x is again an integer
and then 2(2x) is an even integer
scenic fjord
Okay, so for symmetric, I just write like xRy and yRx that is x+3y is even, 3y+x is even
A bit weird if I just prove like this
grizzled shore
You say if x + 3y is even, then y + 3x is even
And you do need to do some work there
scenic fjord
You say **if** x + 3y is even, **then** y + 3x is even
Oh, I see. Let me try for transitive
grizzled shore
You say **if** x + 3y is even, **then** y + 3x is even
This is not trivial
You should try prove that this is true
scenic fjord
Oh, I still need to further prove it?
grizzled shore
Of course, it’s not trivial that it’s true
scenic fjord
Isn’t that logically if x+3y, the 3y+x is the same?
scenic fjord
And you do need to do some work there
May I know how to prove?
grizzled shore
May I know how to prove?
See it is not trivial :p
You want to break it into cases
If x + 3y is an even number
That means x and y can’t just br anything you choose
Say x = 2 and y = 1, id get 5
5 is not even
So only certain x and y can get you even numbers, which ones?
If you can’t see a pattern, pick some and try
scenic fjord
So only certain x and y can get you even numbers, which ones?
When x is odd and y is odd
grizzled shore
Anything else?
scenic fjord
Anything else?
Should be both odd or both even
grizzled shore
Should be both odd or both even
Yes
So do these 2 cases individually
First suppose x and y were both odd
Aka they can be written in the form 2n + 1 and 2m + 1 respectively
Then try to show that y + 3x is even
Then do the same for when x and y are both even
scenic fjord
Aka they can be written in the form 2n + 1 and 2m + 1 respectively
Meaning to say, if we choose x=2n+1 and y=2m+1, then
x+3y=2(n+3m)+2, and since 2(n+3m)=2h, then x+3y is even.
y+3x=2(m+3n)+2 and since 2(m+3n)=2g, then y+3y is even.
Therefore, x+3y=y+3x.
Let x=2n and y=2m
x+3y=2(n+3m)=2h, then x+3y is even.
y+3x=2(m+3n)=2g, then y+3x is even
grizzled shore
Errr not quite
You first deduce that in order for x + 3y to be even
That either x and y are even
Or x and y are odd
Suppose x and y are odd
Then…
grizzled shore
Meaning to say, if we choose x=2n+1 and y=2m+1, then
x+3y=2(n+3m)+2, and since ...
We don’t need the 2nd line here about showing x + 3y is even
That’s true by assumption
What it really does is put into question why are you proving this, are there other ways for x + 3y to be even that you’re not considering?
grizzled shore
Meaning to say, if we choose x=2n+1 and y=2m+1, then
x+3y=2(n+3m)+2, and since ...
Also your proof is wrong
Clearly x + 3y is not y + 3x
They aren’t the same but they are both even
scenic fjord
I thought from here, if we know x+3y is even, then if y+3x is even, then we can conclude it’s the same
grizzled shore
No no
If we know x + 3y is even
Then we know y + 3x is even
If you can prove this then the relation is symmetric
For all x, y in ℤ
scenic fjord
Also, I also don’t understand why just consider both odd or both even, because I think if we consider either one to be odd and either of to be even, then surely it will not be even and it’s not symmetric
Nvm, I think I’ll just stop for now. Math proof based is super hard🫠
Anyway, thanks for your time for help me to understand what is equivalence relation
grizzled shore
Also, I also don’t understand why just consider both odd or both even, because I...
We do just consider both odd and both even
But you have to prove that the statement “x + 3y is even” implies x and y are both odd or both even
You need to chain the implication signs all the way to “y + 3x is even”
scenic fjord
Erm, I don’t understand, maybe I’ll just go my lecturer office and sit there and ask her prove to me
grizzled shore
Proofs in maths does require some abstract thinking and it’s a bit hard to get into that mindset so don’t be too worried if it’s hard
It’ll come with time
scenic fjord
Proofs in maths does require some abstract thinking and it’s a bit hard to get i...
It’s hard and well, because I’m from physics faculty, the math I’m interested in is calculus, but math faculty just offered 4 advanced calculus related courses to us, other related calculus courses overlapped with my faculty, so I’m forced to take this modern algebra. I’m planning to drop this course like my friends did because it’s just second week and my friends and I already cannot follow the lesson like the ordinary math students
scenic fjord
It’ll come with time
That’s why I’m seeing if I really cannot catch up, I’ll drop this next week, anyway, thanks and really appreciate for your help
grizzled shore
I think a lot of physics formulation in math is very intuitive from the math perspective, but maybe physicists just accept them to be truths
scenic fjord
Also, after this pure math regarding proving, it’s obviously pure math is way more harder than my theoretical physics
grizzled shore
Which is fine since they may not be so interested in the exact mathematical argument but rather the physical interpretation
scenic fjord
I think a lot of physics formulation in math is very intuitive from the math per...
Yea, physics use a lot of math, which I would said mostly is applied math, but pure math contains a lot of unnecessary proofs which I found it’s useless (maybe just for my case because I’m not really into math proof). I asked my friends from math school, which they’re now in Y2 but because they’re not in Y3 so I cannot ask them about this course, they told me they start learning the logic and how to prove since Y1, so it might be difficult for us to learn proving
grizzled shore
A lot of times in undergrad physics they want to discuss physics that you don’t have the maths background to fully grasp
So it’s a bit tedious for them to describe in an exact manner
I would say that this sort of math you’re learning right now is the a very fundamental part for all areas of math, it is as basic as kinematics from highschool
You wouldn’t be able to do very much physics without kinematics from highschool, just like how many higher level math concepts won’t make sense without a good understanding of set theory
scenic fjord
What I found a huge difference when my physics faculty taught us the same course with math faculty is that, physics school focus on the application of the math, like how we use that for physics, but math school, focus much to definitions. Same syllabus but different teaching style
grizzled shore
Well pure maths is really about solving maths problems not real world problems
And the techniques to prove things
Whereas physics is more interested in the interpretation for real world phenomena
scenic fjord
You wouldn’t be able to do very much physics without kinematics from highschool,...
I think the problem here is that mathematical proving is really really hard for me and my friends who didn’t really been taught how to prove stuff in math.
grizzled shore
But if you need to prove a physical phenomenon in mathematics you’ll need the rigour and techniques learned from pure maths
scenic fjord
Whereas physics is more interested in the interpretation for real world phenomen...
Yea, exactly, and that’s why this modern algebra, other school is not taking, I heard from the dean of math faculty, from the last few years, also just few physics students taking and the result obviously of us is a killer subject
grizzled shore
I think the problem here is that mathematical proving is really really hard for ...
Yeah I mean proofs itself is a big area for math
scenic fjord
But if you need to prove a physical phenomenon in mathematics you’ll need the ri...
I’m actually taking because in advanced physics, this modern algebra or maybe abstract algebra is fundamental for quantum field theory
grizzled shore
It’s not a trivial part, in fact it’s mostly what maths does
grizzled shore
I’m actually taking because in advanced physics, this modern algebra or maybe ab...
You absolutely need to know these fundamentals for quantum field theory
They have direct consequences for the theory and the phenomena it describes
scenic fjord
Yeah I mean proofs itself is a big area for math
Maybe it takes time, and if you’re being taught to think logically and techniques to prove since first year, this course is easy
grizzled shore
Maybe it takes time, and if you’re being taught to think logically and technique...
Yea
scenic fjord
But it’s obviously not for our case
grizzled shore
It’s hard to learn the mindset, it takes time and lots of practice
scenic fjord
You absolutely need to know these fundamentals for quantum field theory
Yea, that’s why I’m still struggling whether to drop or continue to learn this since I want to further study for QFT
grizzled shore
Often times I look at something and go “well I just have no idea how to even begin” to prove something
scenic fjord
It’s hard to learn the mindset, it takes time and lots of practice
Yea, but not that the lecturer already assume that we all math faculty students
scenic fjord
Often times I look at something and go “well I just have no idea how to even beg...
Exactly what I’m facing this semester for this subject
grizzled shore
Yea, but not that the lecturer already assume that we all math faculty students
Well, studying higher level academic fields is not an easy task
I’m not saying you aren’t trying hard enough, but you shouldn’t expect it to be a breeze
There’s a reason we laugh at business majors but not physics majors
scenic fjord
Maybe I really should go sit in my lecturer office and ask her individually
I actually told her that but she’s currently at vacation
grizzled shore
That at least shows her that you care about succeeding
scenic fjord
Need to wait after next week 🫠
grizzled shore
It’s often hard for a lecturer to give you effort if you don’t give the course effort
Why should they care if you don’t care either
scenic fjord
That at least shows her that you care about succeeding
Haha, it’s because this is my first time being so blur and no idea about this subjevt
grizzled shore
Yeah I understand that too, once you start falling behind it gets super hard to get back on track
scenic fjord
Why should they care if you don’t care either
It’s true, but sadly, my friends who also don’t understand, they just drop this subject. Some who still remain when I ask them, they said they also don’t understand but they didn’t ask anything and just assume they know
You know when come to prove, if you see solution, you will be like “oh, it’s just like that” and you assume that you know, and this is not what im searching for
grizzled shore
Well if you’re a lecturer and everyone just sits there in silence as you lecture, you have no clue if the students are following or not
If you ask them “does this make sense” and it’s crickets
They’ll just assume everyone gets it
scenic fjord
Yeah I understand that too, once you start falling behind it gets super hard to ...
Yea, and most of the time, the chapters are related, if you can’t understand the basic, then there’s super high percentage you can’t follow for the whole semester
grizzled shore
Exactly
When I first started it was very very vital that I never fell behind because I knew the moment I fell behind it’ll be too overwhelming to catch up
I’ll just keep hearing words and sentences that I don’t understand and it’ll be too frustrating to even try
But that’s just me
scenic fjord
They’ll just assume everyone gets it
Yea, after I told her I cannot follow, last lesson she slow down a bit for chapter 2, and that chapter I can follow, just this first chapter, she taught everything in 30 minutes and assume everyone did discrete math. Okay, maybe because math students did that course before in their Y2 and she assume everyone know the basic for equivalence and the basic proof
scenic fjord
I’ll just keep hearing words and sentences that I don’t understand and it’ll be ...
I remember I heard from some math students, in their first year about proofing they don’t understand, they just memorise, and when they reach Y2, they start to have the idea
So yea, I agree this proving can be trained, but yea, takes quite a long time
grizzled shore
Lecturers don’t want to talk to a brick wall and have students completely lost, they want to do whatever is best to pass the knowledge to the students
scenic fjord
Well, you’re math undergraduate?
grizzled shore
So it’s not like the lecturers will laugh at you if you say it’s a bit confusing can you elaborate
grizzled shore
Well, you’re math undergraduate?
I’m second year math major yeah
And yeah, case in point, I did this stuff in year 1
All this about equivalence relations and so on
scenic fjord
I’m second year math major yeah
And I’m in my third year physics major🤡 and I didn’t even learn anything related to this course before
scenic fjord
All this about equivalence relations and so on
Anyway, thanks. And I can pm you and ask if I really have some very basic question which I can’t tackle in the future? Thinking not to drop this even it’s a killer for my faculty, okay, from the last records, it’s a killer subject 😅
grizzled shore
It pops up every now and again
It’s a very powerful tool
(Look at a clock!)
(1 ~ 13 on a clock where ~ is the equivalence relation)
grizzled shore
You’re welcome!
scenic fjord
.close
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.
last pilot
first of all
don't ping us in your original message
secondly
this isn't a place to pose challenges idt
if you wanna do that i think #competition-math or #math-discussion are probably the places to do it
strong gust
Ok thx
last pilot
np
obtuse pebbleBOT
If you are done with this channel, please mark your problem as solved by typing .close
strong gust
.close
obtuse pebbleBOT
quick totem
. reopen
obtuse pebbleBOT
Available help channel!
keen mist
k
keen mist
help me my brain exploded…
naive bolt
help me my brain exploded…
?
naive bolt
the graph of a quadratic curve cuts the x axis at its zeros
i.e look at the function f(x)
evaluate f(p) by plugging x=p
do you see what you get?
keen mist
i dknt get anything
light raft
if x = p, what is x-p
light raft
we have f(x) = a(x-p)(x-q). Let x = p. Then f(x)=f(p) = 0. So, p is a 'root' of the quadratic equation. Similarly, letting x = q, f(x) = f(q) = 0, and so q is also a 'root' of the quadratic equation. That means, there are two x values p,q where the y value is 0. So f(x) crosses the x axis (i.e. y = 0) at two points p and q. Here, p<q. So, what would the value of p and q be
Look at where the quadratic function crosses the x axis
keen mist
we have f(x) = a(x-p)(x-q). Let x = p. Then f(x)=f(p) = 0. So, p is a 'root' of ...
p or q = 2 or 6
since p<q so p = 2 and q = 6?
since the parabola crosses 2 and 6
IDK
@light raft
help
keen mist
okt thank
light raft
how do you think we shoudl find a
light raft
check by seeing whether the equation goes through the point (0, -12)
i.e. y intercept at -12
keen mist
ok and
light raft
i mean yeah you will get a = -1
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
sharp moss
obtuse pebbleBOT
sharp moss
P changes the period but
the height is going to be the same, so how is it possible that g(x) does not intersect -1?
red ice
P changes the period but
Cause the domain for all of them is 0 <= x <= 2pi
So you can imagine stretching the function until the intersection with y = -1 is outside the window of the graph
sharp moss
ohhh youre right
red ice
Cause the domain for all of them is 0 <= x <= 2pi
Actually, 0 <= x < 2pi
sharp moss
That makes sense but I do not understand the solution..
red ice
That makes sense but I do not understand the solution..
Yeah it is a bit wordy
So I would just solve 2 sin x = -1
You get point A has an x-coordinate of 7pi/6
And you want to stretch it so that point A transformed will be at x = 2pi
So what must p be (remember that p is a compression, so 1/stretch = compression)
sharp moss
-1/3..?
red ice
Okok take your time
sharp moss
-1 = Sin (p * 2pi)
red ice
Nope, that's not the right way to set up that equation
y = 2 sin px
And you know that x = 2pi, y = -1
sharp moss
yesyes
red ice
Yeah so can you continue?
red ice
Okok
sharp moss
So its -1 = 2 Sin(p*2pi)
-1/2 = sin(p*2pi)
11pi/6 = p * 2pi
is it correct so far
Becuase I get wrong answer then
oh yea I get the answer if I use 7pi/6
I though I would have to use 11pi/6 because were finding greatest..
red ice
11pi/6 = p * 2pi
Ah it's not 11pi/6
11pi/6 is the second solution so that corresponds to point B
You want point A barely being out of the domain 0 <= x < 2pi
Not point B being out of the domain: point A could still be in the domain
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
outer spindle
outer spindle
Bit confused here
So, the simplified version of this is
4/sqrt 1-x2
Which is an identity 4arcsin(x)
+C
The condition is f(1/2) = -10
4arcsin(1/2) + C =
2arcsin + C = -10
No?
obtuse pebbleBOT
@outer spindle Has your question been resolved?
high lily
no, same issue as before
arcsin(1/2) is not some value arcsin being multiplied to 1/2
sin( what value in the range of arcsin) gives 1/2
obtuse pebbleBOT
@outer spindle Has your question been resolved?
outer spindle
Ah
@high lily Ok ok I see, this goes back to remembering the values from the unit circle
Which in this case, sin on half is pi/6
So I have C = -10 - 4pi/6
YES
.close
umbral robin
obtuse pebbleBOT
umbral robin
Can anyone explain why my answer to part b and c are wrong?
for part (b), I did C(6,1) * C(25,2) because you're choosing 1 engineer from 6, and then there's 25 remaining people (including the 5 other engineers since it's "at least") so I multiplied 6 with C(25,2)
which ultimately gave me 1800
but that's wrong
textbook says the answer is 1460
and for part (c), I did C(8, 2) * C(24, 1)
since 2 spots must be mathematicians and ur choosing from 8, and it doesn't say that the last spot can't be a mathematicians
so multiply C(8,2) with C(24,1)
which gives 672, but the textbook answer is 504
Ping me when you reply please!
for part (b) my answer was 1800 but the textbook answer is 1460 yeah
for part A the answer was 296, which I got
C(8, 3) + C(12, 3) + C(6, 3)
warm shaleBOT
Pratham_Shetty
umbral robin
oh I see
for part B I didn't take into account the or? so the addition principle thing?
My main issue is with part C though, I understand why it's multiplied by C(18, 1)
but it doesn't say the third slot must not be a mathematicians
so why would the textbook want me to remove all mathematicians for the remaining slots
shouldn't it have specified exactly 2 mathematicians then?
alright thanks for clarifying everything!!!
.close
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
still parcel
Task 2: The table shows the number of industrial robots in the world in the period from 2009 to 2016.
~~(a) Record data in a coordinate system ~~
The growth is described here by an exponential growth model f(t) = b a^t, where f(t) indicates the number of robots and t is the time in years after 2009
(b) Determine a and ba According to a press release from the International Federation and Robotics (IFR), the number of robots grow by 12-14% annually.
(c) Determine the doubling time for the model and compare with the press release from the IFR
obtuse pebbleBOT
# **Task 2: The table shows the number of industrial robots in the world in the...
still parcel
The facitlist:
How do we solve task 2b, i.e. we have to have those values there and I have solved one
and thats is a
so how we found b?
Årstal = year
Antal (1000 stk) = number of the year
<@&286206848099549185>
everyone here?
still parcel
Sorry, but I have a hard time writing in English, so that's why...
obtuse pebbleBOT
@still parcel Has your question been resolved?
obtuse pebbleBOT
@still parcel Has your question been resolved?
obtuse pebbleBOT
@still parcel Has your question been resolved?
brittle sage
Yea
still parcel
What’s the problem
We have the a value, but we are missing the b value
brittle sage
Can you show me
still parcel
Can you show me
to find the a value we just said 1153 / 1059
= 1,08
so how will you find b?
still parcel
I really don’t understand what your saying
Have you read the assignment?
How will you find the values of a and b
still parcel
Helo
obtuse pebbleBOT
@still parcel Has your question been resolved?
still parcel
I got nothin
can you scroll up further to see?
obtuse pebbleBOT
@still parcel Has your question been resolved?
still parcel
But u need to apology for being rude first
do you want to help?
drowsy flower
Can everyone helppppp?
I am really sorry but I have no idea what it is
still parcel
I am really sorry but I have no idea what it is
We need to find values for a and b?
<@&286206848099549185>
mhm
still parcel
<@&286206848099549185>
still parcel
Halloooo
static walrus
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
raw basin
obtuse pebbleBOT
@raw basin Has your question been resolved?
obtuse pebbleBOT
@raw basin Has your question been resolved?
obtuse pebbleBOT
@raw basin Has your question been resolved?
grand crater
how are they getting the values for the definite integrals?
obtuse pebbleBOT
how are they getting the values for the definite integrals?
@grand crater Has your question been resolved?
obtuse pebbleBOT
red ice
how are they getting the values for the definite integrals?
So if you integrate it, you get -2x + c
Okok
obtuse pebbleBOT
Available help channel!
hardy zealot
astral silo
what do u need help on, getting started?
have u created the 2 simultaneous equations?
hardy zealot
what do u need help on, getting started?
yeah
hardy zealot
40 / x = 40 / x+2 + 1?
hardy zealot
now try getting rid of the denominators by multiplying
can i use an x+1 as a common denominator? since its x and x+1
nvm it;s x+2 not 1
astral silo
u use both
if u only X by 1 of them there will still be a denominator on the other one
so u need to do it by both of them
hardy zealot
u use both
wym
astral silo
u multiply by both denominators
hardy zealot
x + (x+2)?
obtuse pebbleBOT
@hardy zealot Has your question been resolved?
obtuse pebbleBOT
@hardy zealot Has your question been resolved?
hardy zealot
.close
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
tawdry crown
2x²+3x²=25
obtuse pebbleBOT
2x²+3x²=25
dim temple
$2x^2+3x^2=25$
warm shaleBOT
Renz
tawdry crown
5x²=25 then 5x=5 then x=5 answer
slim cove
.close
foggy furnace
Calc 1/Calc 2
obtuse pebbleBOT
foggy furnace
.close
obtuse pebbleBOT
Available help channel!
forest niche
obtuse pebbleBOT
forest niche
haaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaalp
im definitely fuckin messing up here
my two answers are
x < 1
and
3/2 < x
what.
hybrid gyro
can you showed your work?
forest niche
aight wait
red ice
my two answers are
x < 1
and
3/2 < x
what.
So yes, the endpoints of the inequality are x = 1, 3/2
So all you need to do is to test a point
Say x = 0, so we get 2 < -4
forest niche
yeah except I cant wite in an inequality form that doesnt make any sense
how can it be less than 1 but greater than 1.5
what
use some logic bruh
or do I write them separately
Im so confused idk why the simple topics confuse me sm
I can do calculus but not this
red ice
my two answers are
x < 1
and
3/2 < x
what.
Right so you take the intersection of these
forest niche
Right so you take the intersection of these
??
red ice
Wait sorry
forest niche
what
red ice
Ah so it's just 3/2 < x
red ice
Cause we tested the region x < 1 with x = 0
red ice
Say x = 0, so we get 2 < -4
It didn't work
Then we test the region where 1 < x < 3/2, say x = 1.2
We get 0.8 < 3.6 - 4 which also doesn't work
forest niche
how many tests
forest niche
aand do I perform them separately for x < 1 and x < 3/2?
red ice
aand do I perform them separately for x < 1 and x < 3/2?
Yes, so forget about the inequalities
We have x = 1, 3/2
So there are three regions
x < 1, 1 < x < 3/2, and 3/2 < x
forest niche
oh so I test em w those
I test em w less than 1, less than 1.5 then greater than 1.5
and whats the criteria for them being a solution?
like as in what lets me know yeah this one is right
if it passses all the tests or?
@red ice
bruh.
left
hybrid gyro
if the inequality holds for the region you test
hybrid gyro
when you plug in say x = 0 to test for x < 1, you plug it in this inequality: if the LHS < RHS as in the question then it passes the test, if LHS >= RHS then it fails the test
forest niche
when you plug in say x = 0 to test for x < 1, you plug it in this inequality: if...
yeah but 0 < 1 passed
south said it failed
what
then how is the answer 3/2 bruh
hybrid gyro
yeah but 0 < 1 passed
when I say test x = 0 for the region x < 1, I dont mean just put x =1 into x<1, obvs that true
I mean
put x = 0 into |x-2| and 3x-4
you will get LHS = 2, RHS = -4, but 2 is not less than -4
so the inequality in the question failed the test
red ice
Yeah that's what I mean
red ice
i got 3/2 > x
Yes
red ice
No worries
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
timid silo
Guys, what is the inverse function of
f(x) = x + sqrt(ax + b)
obtuse pebbleBOT
Guys, what is the inverse function of
f(x) = x + sqrt(ax + b)
broken tendon
Guys, what is the inverse function of
f(x) = x + sqrt(ax + b)
$f^{-1}(x) = \frac{1}{f(x)}$
warm shaleBOT
Tittom_123
broken tendon
I think
halcyon hull
$f^{-1}(x) = \frac{1}{f(x)}$
Here inverse doesn't mean reciprocal.
timid silo
$f^{-1}(x) = \frac{1}{f(x)}$
Bro wtf💀
halcyon hull
Guys, what is the inverse function of
f(x) = x + sqrt(ax + b)
Put y = f(x). Try to find x in terms of y
timid silo
Here inverse doesn't mean reciprocal.
No it doesn't
broken tendon
Oh, mbd lmao
timid silo
Put y = f(x). Try to find x in terms of y
Yes
halcyon hull
Are you able to?
broken tendon
Guys, what is the inverse function of
f(x) = x + sqrt(ax + b)
Replace x by y and then isolate y in the equation
timid silo
Replace x by y and then isolate y in the equation
Bro I'm waiting for you to do it
Because I couldn't do it
broken tendon
Bro I'm waiting for you to do it
I am not gonna do it for you, you gotta learn how to do it by yourself
timid silo
I am not gonna do it for you, you gotta learn how to do it by yourself
Try to do it. It's not as simple as you think
timid silo
I am not gonna do it for you, you gotta learn how to do it by yourself
What do you think? You think I don't know what I have to do?
broken tendon
You will always end up with y in the 2 equations
$y = \frac{\left(x-y\right)^2-b}{a}$
warm shaleBOT
Tittom_123
broken tendon
@timid silo
broken tendon
You can't, I am pretty sure
timid silo
You can't, I am pretty sure
It's possible, because I plotted the graph, it's one to one
Wait I think I'm getting closer to the answer
I think this is the answer
y = ½ [(2x + a) + sqrt[ (2x + a) - 4(x² - b)]
For any x >= -b/a
Yesss it is
Ok time to close this channel
.
.close
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
gritty sail
I was working through the derivation of the rocket equation given by wikipedia here: https://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation#Most_popular_derivation
and got stuck in the following part:
obtuse pebbleBOT
I was working through the derivation of the rocket equation given by wikipedia h...
gritty sail
This is my working of the last section:
My only question remaining is why the term dmdV is cancelled, but the other ones containing dm are not
It's just a bit confusing the way it's put in wikipedia :/
obtuse pebbleBOT
@gritty sail Has your question been resolved?
obtuse pebbleBOT
@gritty sail Has your question been resolved?
frank mantle
ok, Find the leigth of G and F
obtuse pebbleBOT
ok, Find the leigth of G and F
frank mantle
ok Circumference of small circle (6*pi())
so the length of E is 18.84956
what's the formula of a cord?
obtuse pebbleBOT
@frank mantle Has your question been resolved?
frank mantle
hum is 18.84956 the length of the arc or is the measure of the arc,
it is basically the small circle of (6*pi()) circumference rolled out flat onto a larger circle of radius 11.050528
obtuse pebbleBOT
@frank mantle Has your question been resolved?
raw plinth
I have a 3D physics engine for rectangles but I need to make it apply rotational velocity based on the collisions I detect with SAT. I have looked and found that to do this I need to get the contact points but I cant find how to get them between two 3D objects. Pls tell me if you know how I could get them or if you know a better way to get the needed rotational velocity. I have the vertexes for all cubes
obtuse pebbleBOT
I have a 3D physics engine for rectangles but I need to make it apply rotational...
@raw plinth Has your question been resolved?
raw plinth
I have a 3D physics engine for rectangles but I need to make it apply rotational...
<@&286206848099549185> anyone?
tired shell
what's precisely all the data you have? like rotational speed, position, and what? usually there's an answer on stackexchange or maybe the textbook mentioned in this video https://www.youtube.com/watch?v=DhCBCudKJTs
raw plinth
what's precisely all the data you have? like rotational speed, position, and wha...
I have the position
rotation
mass
vertexes
velocity
tired shell
do you have like the positions at the time of collision
raw plinth
normals
raw plinth
do you have like the positions at the time of collision
yes
I can add code to my collision function
tired shell
a silly way is to look at all the edges of both rectangles and check if any of them are within the other object
tired shell
oh what's a collision give you
raw plinth
oh what's a collision give you
I get the collision normal
depth
I have code to detect a collision
and then inside it if a collision happens
I can get what I need
so right now I have everything but the rotation stuff
tired shell
hmmm maybe you can work with the plane perpendicular to the collision normal
actually idk what the collision normal looks like in that 3rd picture
raw plinth
hmmm maybe you can work with the plane perpendicular to the collision normal
I have the vertexes
I have to get faces from them then?
tired shell
I'd go with looking at intersections over all pairs of edges between the rectangles
like finding where lines intersect is a known algorithm
and then one pair of edges will have an intersection and that's your point
raw plinth
I'd go with looking at intersections over all pairs of edges between the rectang...
and how would I do that?
tired shell
uh there's probably functions in libraries for it but it's a matrix equation where you write out vector formula for each each as corner1 + t*(corner2-corner1) and set them equal
tired shell
not really for arbitrary rectangles rotating into each other, you need the point and that's gonna be a lot of math
for a custom engine
if this is unity or whatever there's functions
tired shell
intersection I guess
tired shell
here's the math formula https://web.archive.org/web/20180927042445/http://mathforum.org/library/drmath/view/62814.html
raw plinth
I’ll thx I will take a look at it when I can
finite hawk
obtuse pebbleBOT