can you see why it's useful to factor out the x?
#help-39
rustic gate
frosty heath
yesss to cancel ones on top nd bottom right
rustic gate
mhm
frosty heath
Ok I got that but do I have to do anything with the x on the end now
rustic gate
which x at the end?
,, \lim_{x \to \infty} \f {x\parens{4 + \df1x}} {x \sqrt {1 + \df4x + \df1{x^2}} + \2rx}
jolly parrotBOT
rustic gate
this one?
rustic gate
remember when you cancel in a fraction, you're cancelling through the entire numerator and denominator
frosty heath
LOLLL FORGORRR
rustic gate
frosty heath
rustic gate
yep
frosty heath
then u just evaluate the limit 🤔
rustic gate
yep
frosty heath
thank u 😆😆
rustic gate
np 
frosty heath
.close
pearl pondBOT
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untold pebble
hello can somebody please help me with this question
pearl pondBOT
hello can somebody please help me with this question
untold pebble
for (ii) i got that the result is 30C27 - 25C22
but the solution is this
so 25C22 is okay but why 31C3 instead of 30C3 ?
i solved it by putting x1 >= 3 and subtracting x1 >= 8
so basically it becomes y1 + 3 + y2 + y3 + y4 = 30
y1 + y2 + y3 + y4 = 27
and therefore the number of solutions is 27+4-1 C 27 = 30 C 27
what did i do wrong?
golden bramble
x1=2 should count
golden bramble
i solved it by putting x1 >= 3 and subtracting x1 >= 8
so start at 2 instead of 3
untold pebble
x1=2 should count
but the complement of x1 <= 2 is x1 > 2 i.e. x1 >= 3
untold pebble
so start at 2 instead of 3
and if i do this then also i would have to put x1 >= 7 instead of x1 >= 8
golden bramble
But the condition says x1 is between 2 and 7, both INCLUSIVE
Your logic is mostly correct, but you discounted the x1=2 case when you should have counted it
untold pebble
But the condition says x1 is between 2 and 7, both INCLUSIVE
i mean yea but
golden bramble
But what?
untold pebble
hold on i read the problem wrong
untold pebble
but the complement of x1 <= 2 is x1 > 2 i.e. x1 >= 3
its not x1 <= 2 but x1 >= 2
so the complement of x1 >= 2 is x1 < 2
i.e x1 = 0 or x1 = 1
wtf??
golden bramble
Your logic is correct though apart from that blip
Take all cases where x1 is at least 2 and subtract the cases where x1 is greater than 7
That works
golden bramble
so basically it becomes y1 + 3 + y2 + y3 + y4 = 30
Basically you just replace 3 with 2 here
untold pebble
oh okay i think i get it now
vernal hill
pearl pondBOT
vernal hill
how to solve this?
tender ingot
!status
pearl pondBOT
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
Hint: Separate the numbers with power of x to one side and the constants to the other side
Hint 2: After completing hint 1, use ln on both sides
@vernal hill
vernal hill
ok i will try it thanks now im doing other problem
regal sequoia
"Calculate the flux of the field F out of the surface S
F(x, y, z) = 3y^3 + x, 4x^3, y ln(4 + x),
if S is the surface of the cylinder S : {(x, y, z) : x^2 + y^2 = 1, 1 ≤ z ≤ 2}
a) by direct calculation
b) by completing to a closed surface and using Gauss' theorem:"
regal sequoia
i am not sure how to calculate a)
proven river
You can break the whole surface integral into two parts the radial part and the top and bottom part. Then perhaps a change of coordinates from Cartesian to cylindrical would ease up calculation and limits. Find the differential area element in the two cases then integrate with suitable limits.
But there is no field that is perpendicular to the top and bottom surface so that part would be zero
regal sequoia
yes okay are the bounds from 2pi to 0 for the outer integral?
and is x = cos(θ) and y = sin(θ)
pearl pondBOT
@regal sequoia Has your question been resolved?
pearl pondBOT
@regal sequoia Has your question been resolved?
pearl pondBOT
@regal sequoia Has your question been resolved?
dire parrot
and is x = cos(θ) and y = sin(θ)
yes and dont forget the bounds for z and the normal vector as well since youre integrating over the lateral surface of the cylinder
regal sequoia
yes okay are the bounds for z from 2 to 1?
dire parrot
yes
regal sequoia
okay so is the normal given by (cos(θ), sin(θ), 0)?
dire parrot
yes
regal sequoia
okay so we get ((3sin(θ)^3 +cos(θ))* cos(θ) + 4cos(θ)^3sin(θ)?
dire parrot
yes now you evaluate the integral
regal sequoia
okay so that we get (3sin(θ)^cos(θ))*cos(θ)z + 4cos(θ)^3sin(θ)z?
dire parrot
okay so that we get (3sin(θ)^cos(θ))*cos(θ)z + 4cos(θ)^3sin(θ)z?
yeah but just get rid of z and integrate with respect to theta (z is between 1 and 2 anyway)
regal sequoia
yeah so we are just going to get the same thing
do i put in the other bounds now?
dont we get 0?
midnight haven
is the answer pi?
regal sequoia
hmm maybe it is pi i may have done something wrong
dire parrot
yeah it should be pi
regal sequoia
oh yeah now i got it
midnight haven
regal sequoia
yeah wow you did that a lot easier than me haha
pearl pondBOT
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pearl pondBOT
wraith hare
complete the square for y^2-18y so that you only a single y in the expression, then re-arrange
pearl pondBOT
wraith hare
np
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unique plank
Is there an algorythm that calculates the minimum distance between two sets of points?
pearl pondBOT
Is there an algorythm that calculates the minimum distance between two sets of p...
unique plank
oh no
not so simple my friend
i have 2 sets
i need to fidn the minimum distance between them
ill show you
unique plank
green is set A
blue is set B
D is the min distance
i am trying to figure out an algorythm in Cpp that gets me D
unique plank
Or a regression plane?
whats that?
leaden wadi
Not what you need.
unique plank
lmao fair
leaden wadi
But its like a regression line except its a plane between a bunch of three-dimensional points.
unique plank
i searched it up, i dont think its what applies here
its more of a algorhythm and geometry question here
sharp vigil
this is the "closest pair problem", the solution to which is the "divide and conquer" algorithm (you should find some results for that on google)
unique plank
my main idea was getting the average center of each set, making a vector through them, and then finding the particle with the smallest projection on taht vector
unique plank
this is the "closest pair problem", the solution to which is the "divide and con...
i saw that
but it says its only one set, merging the sets would give wrong answers
unique plank
my main idea was getting the average center of each set, making a vector through...
do it for both directions, caluclate the distance between the two particles, adn boom
pearl pondBOT
@unique plank Has your question been resolved?
leaden wadi
I can't think of anything that would take less than MxN steps.
unique plank
i thought ab voronoi diagrams
but nah
doesnt work
my vector idea sounds cool
but i doubt its better than O(nlogn)
unique plank
I can't think of anything that would take less than MxN steps.
my idea is: get a line connecting the two point's average positions, then get the point most towards the other in the direction of the line, and calculate the distance, done.
leaden wadi
I look at a graph like this and think you cannot find the shortest edge without checking every edge.
unique plank
I look at a graph like this and think you cannot find the shortest edge without ...
i am sure that the two graphs dont overlap
i mean the two input ones
and my algorhythm still works in that case
i got it done btw
bool BlobPool::areLinked(PlayerList a, PlayerList b){
//ALG 1
double avgx1=0, avgx2=0, avgz1=0, avgz2=0;
std::size_t sizea=a.size(), sizeb=b.size();
v3<double> tmpv;
//Get point average position of a
for(std::shared_ptr<Player> tmp : a){
tmpv = tmp->getXYZ();
avgx1 += tmpv.x;
avgz1 += tmpv.z;
}
avgx1 = avgx1/sizea;
avgz1 = avgz1/sizea;
//Get point average position of b
v2<double> avg1(avgx1,avgz1);
for(std::shared_ptr<Player> tmp : b){
tmpv = tmp->getXYZ();
avgx2 += tmpv.x;
avgz2 += tmpv.z;
}
avgx2 = avgx2/sizeb;
avgz2 = avgz2/sizeb;
v2<double> avg2(avgx2,avgz2);
//Vector connecting two average points and normalize it to calculate dot product.
v2<double> line = avg2-avg1;
line = normalize2(line);
v2<double> tmpmax1, tmpmax2;
double proj, projmax;
//Get point of a furthest along line while going towards avg2
for(std::shared_ptr<Player> tmp : a){
proj = dotProduct(conv3d_2d(tmp->getXYZ()), line);
if(proj>=projmax){
tmpmax1 = conv3d_2d(tmp->getXYZ());
}
}
line = v2<double>(0,0)-line;
projmax = 0;
//Get point of b furthest along line while going towards avg1
for(std::shared_ptr<Player> tmp : b){
proj = dotProduct(conv3d_2d(tmp->getXYZ()), line);
if(proj>=projmax){
tmpmax2 = conv3d_2d(tmp->getXYZ());
}
}
if(v2distancefrom(tmpmax2, tmpmax1) < 4*this->radius){
return true;
}
// ALG 2
//TODO make a function that, given a direction vector
//obtained by normalizing the line that connects the two region's "center of players".
for(std::shared_ptr<Player> player1 : a){
for(std::shared_ptr<Player> player2 : b){
if( v3distancefrom(player1->getXYZ(), player2->getXYZ()) < (4*this->radius) ){
return true;
}
}
}
return false;
}
@unique plank Has your question been resolved?
pearl pondBOT
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cloud salmon
pearl pondBOT
sharp vigil
they are different ways of writing the same process
cloud salmon
ok I see
pearl pondBOT
@cloud salmon Has your question been resolved?
spring crystal
pearl pondBOT
spring crystal
How find Rn
pearl pondBOT
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vagrant relic
Mb mb sru
last moth
What did you want to report?
vagrant relic
Sry
fallow cosmos
If you know already when you're sending the modping that you need to apologize for it, how about not sending it in the first place?
vagrant relic
What did you want to report?
No I'm dumb I didn't see the channel closed
glass salmon
😂 you pinged moderators and then apologized for it in the same sentence?
last moth
Oh okay
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frank valley
How to do this?? my mind is blank
pearl pondBOT
How to do this?? my mind is blank
frank valley
What it says it's not really important btw
vagrant relic
How to do this?? my mind is blank
Just subtract first
last summit
then see if stuff cancels
vagrant relic
$\frac{10}{11}.\frac{11}{12}$
jolly parrotBOT
Monarch of Eternal Night
vagrant relic
And so on
frank valley
**Monarch of Eternal Night**
Wait sorry how do you end up with that?
last summit
Wait sorry how do you end up with that?
1-1/11=?
1-1/12=?
Just simplify the terms by performing the subtractions inside of the parenthesis
last summit
There is a pattern, so you will not have to actually do it for all of them
if you find the pattern
frank valley
The pattern is something like a/b , b/c , c/d...
ohh so then most of them cancel out right
pearl pondBOT
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spiral goblet
pearl pondBOT
quiet tendon
the distance between the origin and P is r, you can draw a right triangle to show that the vertical distance between P and the x axis is rsin(θ)
then the remaining distance from the x axis to the line is a
so total distance is a + rsin(θ)
then for b, you know that the parabola is all points where their distance from the origin is the same as the distance from part a, so r = a + rsin(θ)
when you solve that equation for r, you get your result for b
pearl pondBOT
@spiral goblet Has your question been resolved?
pearl pondBOT
@spiral goblet Has your question been resolved?
brittle onyx
Is the whole of a right and how do I do b and c
pearl pondBOT
Is the whole of a right and how do I do b and c
@brittle onyx Has your question been resolved?
brittle onyx
.reopen
pearl pondBOT
✅
brittle onyx
<@&286206848099549185>
vagrant relic
,rotate
jolly parrotBOT
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@brittle onyx Has your question been resolved?
flint elm
??
flint elm
confused on what to do next.
kind rampart
[(3+13b)/-7]*-7 ≠ -21 - 91b
unkempt rose
you have $\frac{3 + 13b}{-7} = 20 right?$
jolly parrotBOT
Juan
flint elm
yes!
unkempt rose
$3 + 13b = -140$
jolly parrotBOT
Juan
unkempt rose
$13b = - 143$
jolly parrotBOT
Juan
unkempt rose
$b = \frac{-143}{13}$
jolly parrotBOT
Juan
unkempt rose
$b = - 11$
jolly parrotBOT
Juan
unkempt rose
there is something you cannot understand?
if something is mulitplying you pass it to divide
is something is add/subs you pass it with the opossitive sign
flint elm
yea I understood it now! I was just confused on whether or not I should times the seven with also the 3 or just the 13b!
thanks :)
.close
unkempt rose
❤️
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Available help channel!
vernal grove
pearl pondBOT
vernal grove
Im confused
so we want 3 cards from the same suit and the other 2 can be from any suit beside the one that the first 3 came from
but why did they do (13, 1)?
shouldnt it be (13, 3)?
since we r picking 3 cards from the same suit
leaden wadi
There are thirteen ranks for a three of a kind, A-K.
vernal grove
There are thirteen ranks for a three of a kind, A-K.
ngl I dont get what u mean
leaden wadi
AAA, 222, 333, 444, 555, 666, 777, 888, 999, TTT, JJJ, QQQ, KKK
vernal grove
oh when they said same kind they didnt mean 3 cards from one suit?
leaden wadi
Yeah, they meant three cards of the same value.
vernal grove
but 3 same exact kind of cards from 3 different suits
ohhhhh
why did they do (4, 1)^2
at the end
I understand why they did (12, 2) its because 3 suits will have 1 less card and we need to pick 2 cards right
but why couldnt they have done (4 1) instead of (4, 1)^2
leaden wadi
There are four suits, of which you need only three, so 4C3.
vernal grove
ye I get that but the last part
we could pick the last 2 cards from the same suit and it will surely b of different kind right
so why not do (4, 1) instead of doing it twice
leaden wadi
Because you do it once for each of the two cards.
The rank is chosen by 12C2 which excludes the value from the three of a kind. Each of those ranks have four possibilities which is 4C1.
vernal grove
The rank is chosen by 12C2 which excludes the value from the three of a kind. E...
what do u mean by rank?
vernal grove
oh
vernal grove
Because you do it once for each of the two cards.
oh so if i were to have (4, 1) once
it would only pick it once?
but doesnt (12, 2) imply that we r picking 2 cards from the 12 remaining?
leaden wadi
You are picking 2 cards from the 12 remaining ranks which does not specify the suit.
(4C1)^2 is what specifies the suits.
vernal grove
im sorry but I still dont understand why the ^2
what would have happened if
we did (12, 2) (4, 1)
that would mean picking 2 cards but from the same suit right?
leaden wadi
Yes.
vernal grove
ohhh so we dont want the last 2 cards to be from the same suit either? I thought it was just the value that we were worried about
thats why we do (4, 1)^2?
leaden wadi
No, by squaring it, you are allowing for one card to be Club, Diamond, Heart, or Spade and likewise for the second card.
vernal grove
shit now i am even more confused
but why do we have to do that
if we can do (12, 2)(4, 1) and it will still pick 2 cards from Club, Diamond, Heart, or Spade
ohhh is it because that limits the choice to be from the same suit
but even then wont it still be considered correct
pearl pondBOT
Available help channel!
midnight haven
how do i prove x^5 - bx ^4 + 3ax^3 + bx^2 + cx + d = 0 cannot have all roots real if 2b^2 < 15a
midnight haven
maybe using rolles theorem
but idk how to initiate and what way to go
<@&286206848099549185>
midnight haven
<@&286206848099549185>
i don't understand how you people can ask these questions but not know how to read
midnight haven
i don't understand how you people can ask these questions but not know how to re...
you may help if you can if not then i am sorry but one text is enough you do not have to be aggressive
astral creek
how do i prove x^5 - bx ^4 + 3ax^3 + bx^2 + cx + d = 0 cannot have all roots rea...
if you look at the graph you might consider how often it goes from convex to concave (so consider the roots of the second derivative)
midnight haven
how do i look at the graph
i am not allowed to use any graphic calculator
so you basically mean like a rough idea of graph
astral creek
exactly
to be honest i dont know how rigorously youre supposed to proof this, but that way you can definetely get the intuition:/
midnight haven
but i have to prove it
i think the exercise is expecting me to use rolles theorem
but i have no idea how to even put that idea here
astral creek
alright then i have no clue haha, sorry
intuitively you can see that the second derivative only has one root (because its derivative has 0 roots) if 2b^2<15a and so you can see that the graph cant go up down up down but i dont know how to use rolles theorem for that xd
midnight haven
intuitively you can see that the second derivative only has one root (because it...
how are you sure second derivative has 1 root
isn't it possible that it has no roots
yes i get the intuition
but that's not a proof
right ?
astral creek
but that's not a proof
yeah:/
astral creek
how are you sure second derivative has 1 root
because its of degree 3
midnight haven
because its of degree 3
yea
pearl pondBOT
@midnight haven Has your question been resolved?
proud cove
can someone give me a hint on how to prove this? I always end up with a bunch of extra terms
proud cove
this is from spivak's calculus
I can look up the full answer but I just want a smaller hint to put me in the right direction so I can do it myself
wraith hare
I can look up the full answer but I just want a smaller hint to put me in the ri...
Two hints here, use hint 2 if hint 1 isn't enough.
hint 1: ||multiply both sides of a < b by c||
hint 2: ||multiply both sides of c < d by b||
proud cove
Two hints here, use hint 2 if hint 1 isn't enough.
hint 1: ||multiply both sides...
ohhh yeah I think one is enough
thanks
midnight haven
inequality
proud cove
inequality
the way I'm doing this book is I'm trying to only use the stuff they've explicity taught to solve the problems
that seems to be how its designed
kind of weird to me how little I've seen inequalities in algebra they just hardly get mentioned
midnight haven
Two hints here, use hint 2 if hint 1 isn't enough.
hint 1: ||multiply both sides...
yeah then go with this
proud cove
so I'm not used to their properties
midnight haven
kind of weird to me how little I've seen inequalities in algebra they just hardl...
you will understand their importance everyday
haha
proud cove
yeah I think this is preparing me for epsilon delta stuff
midnight haven
nice
proud cove
Two hints here, use hint 2 if hint 1 isn't enough.
hint 1: ||multiply both sides...
yeah thanks 1 was enough
wraith hare
np
proud cove
.close
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midnight haven
how do i prove (d(g(x))/dx)(f(x)) + d(f(x))/dx = 0 for some x in (a,b) where f(a) = f(b) and f and g are functions of x
waxen condor
how do i prove (d(g(x))/dx)(f(x)) + d(f(x))/dx = 0 for some x in (a,b) where f(a...
?
pearl pondBOT
Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.
waxen condor
If f(a) = f(b) , and f(x) is continuous and differentiable then according to rolles
Theorem
There exist some c element of (a,b)
Such that f'(x) = 0
midnight haven
i mean g'(x)f(x) + f'(x) = 0 by that ugly expression
waxen condor
So we have proved it
midnight haven
no
waxen condor
i mean g'(x)f(x) + f'(x) = 0 by that ugly expression
i mean we have proved
That either of
midnight haven
i mean g'(x)f(x) + f'(x) = 0 by that ugly expression
we have to prove this equation has a root in (a,b)
waxen condor
f'(x) or g'(x) is 0
midnight haven
HOW
waxen condor
By rolles
wraith hare
f'(x) or g'(x) is 0
you mean f(x) or g'(x)?
waxen condor
you mean f(x) or g'(x)?
No i mean what i have wrote
autumn topaz
how does that prove g'(x) can be 0 on that interval?
waxen condor
how does that prove g'(x) can be 0 on that interval?
Ah sry i didnt see
He says f(a) = f(b)
Only f'(x) is 0
midnight haven
but we can from f'(x) = 0 in (a,b)
waxen condor
For some c in (a,b)
midnight haven
so by applying byparts formula
we get a term with f'(x) and f(x)
wait no
this dosen't prove it
autumn topaz
what is the original problem, in full?
waxen condor
g'(x)f(x) + f'(x) = 0
midnight haven
what is the original problem, in full?
i dont have a phone sorry
waxen condor
g'(x)f(x) + f'(x) = 0
We need to prove this ^
midnight haven
maybe @waxen condor can explain
waxen condor
how do i prove (d(g(x))/dx)(f(x)) + d(f(x))/dx = 0 for some x in (a,b) where f(a...
@autumn topaz
Question ^
autumn topaz
I saw what he posted
midnight haven
i dont have a phone sorry
i mean to take the screenshot of the problem in my book
autumn topaz
you are given 0 information about g(x)?
waxen condor
Okk let us think for sometime
waxen condor
how do i prove (d(g(x))/dx)(f(x)) + d(f(x))/dx = 0 for some x in (a,b) where f(a...
I really think insufficient information
waxen condor
Is there slight more info ?
waxen condor
its correct question according to my textbook
Any line you are missing?
midnight haven
f and g are continuous and differentiable
waxen condor
f and g are continuous and differentiable
Nah ive already used that
waxen condor
f(a) = f(b) = 0
Hey is this given ?
midnight haven
yes
waxen condor
Bro
midnight haven
messed up bro sorry
waxen condor
K np
midnight haven
i didnt mention the 0
waxen condor
still its apparently of no help
midnight haven
lol
waxen condor
Like we now know that is passes through the x axis two times
midnight haven
yes
waxen condor
If f(x) =0 doesnt necessarily mean f'(x) = 0
midnight haven
obviously
waxen condor
I will beback in 5 mins
midnight haven
kk
waxen condor
Hey
I think
I found what the question means
I think f is a quadratic with same roots
midnight haven
I think f is a quadratic with same roots
oh so D = 0
midnight haven
I think f is a quadratic with same roots
how do you prove this
midnight haven
no assumptions are not allowed in proofs
waxen condor
Ok wait
midnight haven
Question seems unsolvable
lol my book solved it in a funny way
i dont like it
very unintuitive
or maybe intuitive but not for me as i am an amateur
midnight haven
but that solves the problem lol
h(a) = h(b) = 0
then by rolles theorem h'(c) = 0 for some c in (a,b)
waxen condor
Yeah
pearl pondBOT
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untold pebble
forgive me for the stupid question but why is it a_n and not abs(a_n)
pearl pondBOT
sweet condor
rip
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void jacinth
where have i gone wrong
pearl pondBOT
where have i gone wrong
brave schooner
,rotate
jolly parrotBOT
pearl pondBOT
@void jacinth Has your question been resolved?
fickle schooner
Edexcel maths spotted
What's the issue though?
Your diagram seems alright and you correctly got a value of 3 (which is rational)
pearl pondBOT
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chrome flame
pearl pondBOT
wraith hare
what have you tried?
random ermine
4pi
chrome flame
what have you tried?
I integrated 0 to 2π
wraith hare
can you show your work?
pearl pondBOT
@chrome flame Has your question been resolved?
glacial sequoia
I integrated 0 to 2π
the absolute value or just the function
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dense plover
do you think i woulld have lost marks for not saying that its parallel?
merry carbon
do you think i woulld have lost marks for not saying that its parallel?
Can you scroll down to see if there are any additional comments for that M1, or alternative comments/methods?
pearl pondBOT
@dense plover Has your question been resolved?
wet scroll
pearl pondBOT
wet scroll
Hi, could someone explain step by step on how i would go about solving this
tender ingot
Do you know the conditions for a continuous function?
distant skiff
there are three conditions that must be met for continuity at a point
tender ingot
Well, then you gotta learn it, in terms of limit, a function is continuous at a point in its domain if the limit of the function as input approaches the point should be equal to the function's value evaluated at the point
distant skiff
in other words:
For a function f(x) to be continuous at a point x = x_0,
- f(x_0) must be defined
- lim x->x_0 f(x) must be defined
- f(x_0) must equal lim x->x_0 f(x)
wet scroll
I see
tender ingot
and for the limit x->x_0 f(x) to be defined, the left sided limit must equal the right sided limit
pearl pondBOT
@wet scroll Has your question been resolved?
unborn cosmos
is Jordan form the same thing as Jordan Normal/Jordan Canonical form?
JNF/JCF
pearl pondBOT
is Jordan form the same thing as Jordan Normal/Jordan Canonical form?
JNF/JCF
@unborn cosmos Has your question been resolved?
pearl pondBOT
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austere lark
Hello, I have no clue what do put in the spaces. Could someone explain it?
austere lark
Here is RREF:
young sedge
are x, y, z, w, the four variables representing the 4 columns?
austere lark
yes
young sedge
you know x - w = 2, y - w = 3/2, z - 3w = 1
austere lark
you know x - w = 2, y - w = 3/2, z - 3w = 1
No, how did you get that?
young sedge
isn't that just from your rref
austere lark
lmao yes my bad, I've been at this for hours
austere lark
That's the thing, I have no clue
sharp vigil
i think the idea is to set any free variables equal to a parameter
young sedge
couldn't you just set w = t then
then you don't have to use s or r
because there is only one degree of freedom
austere lark
So what do I put for the rest?
sharp vigil
you know x - w = 2, y - w = 3/2, z - 3w = 1
once you set your free variables equal to parameters, use the system of equations here to solve for the pivot variables
austere lark
So for x =, I only use the x elements?
sharp vigil
you have x - w = 2, setting w = t, we have x - t = 2, which you can rewrite as x = 2 + t. then you repeat the process for the other variables (which should be straightforward since the matrix is in RREF)
austere lark
So what about w=?
austere lark
you have x - w = 2, setting w = t, we have x - t = 2, which you can rewrite as x...
Does this look correct?
austere lark
we already set w = t, so you can use that for the w equation
Which one would that be? which equation?
sharp vigil
the last one
young sedge
yeah the last would just be 1 for the coefficient of t and 0 for the others
sharp vigil
yes
sharp vigil
Does this look correct?
there is a sign error on the constant for y
austere lark
yeah the last would just be 1 for the coefficient of t and 0 for the others
Came up incorrect
austere lark
there is a sign error on the constant for y
So both positive?
young sedge
try 0s in the other spaces too
sharp vigil
yes, since the equation from the rref was y - w = +3/2
sharp vigil
Came up incorrect
it says to leave any unused parameters blank rather than 0
young sedge
oh ur right
austere lark
Came up wrong and I had to request a new question. Here is the new one:
Here is the new RREF
austere lark
Came up wrong and I had to request a new question. Here is the new one:
Does this look correct?
austere lark
it says to leave any unused parameters blank rather than 0
So instead of 0s I leave them blank?
sharp vigil
yes, according to the problem instructions
sharp vigil
Came up wrong and I had to request a new question. Here is the new one:
looks right, except that everything next to s and r should be blank
austere lark
Thank you that worked. What if there is no pivot for y? like this rref:
sharp vigil
all the variables without pivots are considered free variables and assigned a parameter
so in this case, you should set y = t, w = s and proceed similarly
pearl pondBOT
@austere lark Has your question been resolved?
burnt glen
the problem i have is if the thing accelerates from 0.5m/s to 5m/s in 0.9s what was the acceleration
i solved it as Δv/Δt and got 0.45m/s over 0.9s which comes out to 0.5m/s^2 and it's not an option and i don't understand why
autumn topaz
5 - 0.5 is not equal to 0.45
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fiery granite
pearl pondBOT
fiery granite
Hello guys i am confused
I already got the 2pi/3 and 4pi/3 but where do they get 8pi/3
Like i tried addint 4pi/3 with pi but it didnt work
Like it gives 7pi/3
So i ak confuser
light helm
where's 8pi/3
fiery granite
where's 8pi/3
In the ginal answer i mean
fiery granite
cos has period 2π, not π
Wait I am confused
But sin has a Period Pi right
So if it was sin we add Pi but cos we add 2Pi?
feral sedge
No, it doesn't
fiery granite
They did 2π/3 + 2π
Where did they Get 2Pi/3
feral sedge
It has a period of 2*pi
feral sedge
Where did they Get 2Pi/3
From solving cos(3x) = -1/2 for 3x
fiery granite
From solving cos(3x) = -1/2 for 3x
Yes so -1/2 is a special angle giving pi/3 Right
So wait how is it 2pi/3
feral sedge
,calc cos(2*pi/3)
jolly parrotBOT
Result:
-0.5
,calc cos(pi/3)
jolly parrotBOT
Result:
0.5
so incorrect
pearl pondBOT
@fiery granite Has your question been resolved?
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flint musk
How can I find the derivative of $f(x)=e^{-(x-2)^2}$
pearl pondBOT
How can I find the derivative of $f(x)=e^{-(x-2)^2}$
jolly parrotBOT
guy
vagrant relic
Use chain rule
jolly parrotBOT
guy
vagrant relic
Yes now differentiate -(x-2)^2
midnight haven
$f'(x) = e^-(x-2)^2 . -2(x-2)$
jolly parrotBOT
Abu Ali
flint musk
$f'(x)=e^{-(x-2)^2} \cdot (-2x+4)$
jolly parrotBOT
guy
flint musk
alright awesome
midnight haven
yes
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twin cradle
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rotund osprey
pearl pondBOT
late prism
The coefficients of a polynomial are related to Pascal's Triangle (aka binomial something something)
nimble arrow
simutaneous equations i believe
late prism
nimble arrow
you find an expression for the x term with (1+ax)^n
quiet tendon
you can get the binomial coefficients using combinations
quiet tendon
like for the 28x term, that corresponds to (n choose 1) * (1)^(n-1) * (ax)^1
late prism
The relevant coefficients here are one from the left and two from the left, and are described by n and n(n-1)/2 if the polynomial is (a+b)^n
nimble arrow
use the formula for nCx
rotund osprey
yes i know its in relation to the riangle
quiet tendon
like for the 28x term, that corresponds to (n choose 1) * (1)^(n-1) * (ax)^1
n choose 1 is n, so this becomes nax = 28x
rotund osprey
but then
quiet tendon
which means na = 28
rotund osprey
yes...
late prism
And then, n(n-1)/2 * a^2 = 364
rotund osprey
ohhhh
quiet tendon
likewise for the x^2 term its supposed to be (n choose 2) * (1)^(n-2) * (ax)^2
quiet tendon
you can solve this system, and keep in mind n is most likely a positive integer here
rotund osprey
simultaneiously?
quiet tendon
yes
rotund osprey
ohhh
late prism
Indeed
rotund osprey
wait
thats so smart
um also u see for the formula
like
how
nc2 is
1/2 n (n-1)
how did u get that
quiet tendon
n choose k is n! / (k! * (n-k)!)
rotund osprey
ohh okok
quiet tendon
in this case k is 2
late prism
$nC2=\frac{n!}{(n-2)!2!}=\frac{n(n-1)(n-2)!}{(n-2)!2}=\frac{n(n-1)}{2}$
rotund osprey
alright thanks so much guys
jolly parrotBOT
otheol
rotund osprey
yes yes
i see
simultanous
thats so smart
alr thanks to both of u
🙏🙏🙏
.close
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subtle perch
wdym
uneven smelt
6 = 7 days
late prism
wdym
They give that the tour date is from June 18 to July 13
The duration of that is 25 days
subtle perch
They give that the tour date is from June 18 to July 13
ye its 26 days
late prism
Yeah
subtle perch
why they need round up
late prism
Since all of these rates are given in terms of weeks
they round up
To the nearest week
pearl pondBOT
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tall blade
Find a > 1 b >= 2, c >= 3 knowing that a + b + c = 2(sqrt(a - 1) + sqrt(b - 2) + sqrt(c - 3)) + 3
tall blade
I dont know how to begin. I tried breaking the parentheses and move to the other side to form squares
pearl pondBOT
@tall blade Has your question been resolved?
cunning comet
whats the (remaining) question?
frail vessel
Hello, I dont understand this calculation. What am I doing wrong?
frail vessel
My solution
worthy lance
They made 8C3 instead of 6C3
frail vessel
Oh I see, I did that first, but answer said 6 take 3. So typo?
frail vessel
I see, thank you Samuel 🙂
worthy lance
Your context will tell you what and where
frail vessel
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regal sequoia
"Use Stokes' theorem to calculate the circulation:... Orientation: counter-clockwise
seen from the point (0, 0, 5)."
regal sequoia
do i start by calculating ∇ * F?
pearl pondBOT
@regal sequoia Has your question been resolved?
regal sequoia
<@&286206848099549185>
pearl pondBOT
@regal sequoia Has your question been resolved?
rough forge
do i start by calculating ∇ * F?
yes
,, \iint_B \left ( \grad \times \textbf{F} \right ) \cdot \textbf{n} : \dd x \dd y
jolly parrotBOT
𝔸dωn𝓲²s
rough forge
"Use Stokes' theorem to calculate the circulation:... Orientation: counter-cloc...
I think the region B would (rcosθ, rsinθ, r²cos²θ - r²sin²θ)
rough forge
Lemme see
$\textbf{B}(r,\theta) = \begin{pmatrix} r\cos \theta \ r\sin \theta \ r^2\cos^2\theta - r^2\sin^2\theta \end{pmatrix}$
jolly parrotBOT
𝔸dωn𝓲²s
rough forge
∂/∂y(2xy/(4+3y^2+z^2) - ∂/∂z(5x + 6xy/4+(3y^2+z^2))i for the x component?
Yes
What a chaos tho haha
regal sequoia
Yeah
rough forge
$\textbf{N} = \textbf{B}r \times \textbf{B}{\theta}$
jolly parrotBOT
𝔸dωn𝓲²s
regal sequoia
for the first part of the derive i get (2x(-3y^2+z^2+4))/(3y^2+z^2+4)^2 and i can write that as -2x/(3y^2+z^2+4) right?
or no wait because then we dont get the same denominator
rough forge
for the first part of the derive i get (2x(-3y^2+z^2+4))/(3y^2+z^2+4)^2 and i ca...
oh we need quotient rule
,,\frac{\partial}{\partial y} \frac{2xy}{4+3y^2+z^2} - \frac{\partial}{\partial z} 5x+\frac{6xy}{4+3y^2+z^2}
regal sequoia
dont we get (2x(-3y^2 + 6yz + z^2 + 4))/(3y^2 + z^2 + 4)^2?
jolly parrotBOT
𝔸dωn𝓲²s
rough forge
my brain is slow haha hold on
rough forge
dont we get (2x(-3y^2 + 6yz + z^2 + 4))/(3y^2 + z^2 + 4)^2?
wow
you're too fast
it's correct
regal sequoia
okay but wait it is ∂/∂z(ln(4+3y^2+z^2) - ∂/∂x(2xz/(4+3y^2+z^2)) that we are deriving right?
regal sequoia
Okay
gleaming moss
doesn't the question tell you to use Stoke's thm?
rough forge
yea
gleaming moss
then you should be evaluating a line integral of the boundary
rough forge
the y component
have you gotten the 3rd component?
regal sequoia
Is that not just 5?
Or wait
No something is wrong
Ohh wait
20 + 15y^2 + 5z^2?
rough forge
hmm ok
,,\grad \times \textbf{F} = \begin{pmatrix} \frac{2x(-3y^2 + 6yz + z^2 + 4)}{(3y^2 + z^2 + 4)^2} \ 0 \ 5 \end{pmatrix}
jolly parrotBOT
𝔸dωn𝓲²s
rough forge
This looks like shit
rough forge
∂/∂y(2xy/(4+3y^2+z^2) - ∂/∂z(5x + 6xy/4+(3y^2+z^2))i for the x component?
that's false
I noticed
2xz*
regal sequoia
Oh oops
rough forge
,,\frac{\partial}{\partial y} \frac{2xz}{4+3y^2+z^2} - \frac{\partial}{\partial z} \left ( 5x+\frac{6xy}{4+3y^2+z^2} \right )
jolly parrotBOT
𝔸dωn𝓲²s
regal sequoia
Oh yeah it’s -24xyz/ (4+3y^2+z^2)^2 for x
rough forge
yea seems about it
,,\grad \times \textbf{F} = \begin{pmatrix} \frac{-24xyz}{(3y^2 + z^2 + 4)^2} \ 0 \ 5 \end{pmatrix}
jolly parrotBOT
𝔸dωn𝓲²s
rough forge
Not really much better
regal sequoia
Haha no
rough forge
maybe it gets better
regal sequoia
Yeah do we find n now?
rough forge
yea
rough forge
**𝔸dωn𝓲²s**
.
N is basically from our surface the cross product of our tangent vectors
regal sequoia
So is x = rcos(theta) and y = rsin(theta)?
rough forge
$\textbf{N} =\begin{pmatrix} \cos \theta \ \sin \theta \ 2r\cos^2\theta - 2r\sin^2\theta \end{pmatrix} \times \begin{pmatrix} -r\sin \theta \ r\cos \theta \ -r^2 2\cos \theta \sin \theta - r^2 2\sin \theta \cos \theta \end{pmatrix}$
jolly parrotBOT
𝔸dωn𝓲²s
rough forge
So is x = rcos(theta) and y = rsin(theta)?
yea
i thought of using polar coordinates
i have a feeling we are doing something awfully wrong
regal sequoia
Maybe we are
rough forge
I thought using Stokes we convert a curve integral into a surface
,,\oint_{\gamma} \vec{F} : \dd \vec{r} = \oint_{\gamma} \vec{F} \cdot \vec{\gamma}'(t) : \dd t
regal sequoia
Well can’t we just find n and do F*n and then integrate that
rough forge
**𝔸dωn𝓲²s**
derivative gamme would be this but with t instead theta
rough forge
Well can’t we just find n and do F*n and then integrate that
i mean N looks awfully wrong
jolly parrotBOT
𝔸dωn𝓲²s
gleaming moss
I thought using Stokes we convert a curve integral into a surface
Stoke's thm suggests the line integral over a vector field is the same as the surface integral over the curl of a vector field
and the line interal is far simpler
so you go with that
rough forge
yea i just noticed
gleaming moss
The surface gamma can be parametrised as s(u,v) = (u, v, u^2 - v^2)
then if you convert this to a parameter of a single variable, you get the boundary as a line
[ \psi(t) = (\cos t, \sin t, \cos^2t - \sin^2 t) ]
jolly parrotBOT
shsgd
rough forge
**𝔸dωn𝓲²s**
It's basically this but instead u and v
gleaming moss
now you just have to evaluate
[ \int_{\psi} \vec{F} \cdot \psi'(t) dt ]
jolly parrotBOT
shsgd
rough forge
yea
regal sequoia
Hmm okay
gleaming moss
t would be from 0 to 2pi, since the projection of psi onto the xy plane is a circle
rough forge
**shsgd**
is the reason why we not have r because of the 2nd constraint
x^2 + y^2 = 1?
radius being 1
gleaming moss
what do you mean not have r?
gleaming moss
yeah sort of
rough forge
yeah so my question was what about r
gleaming moss
well, the projection of the original surface onto the xy plane is the circle x^2 + y^2 = 1 right?
but the actual boundary of the surface is not flat because z is a function of x and y
regal sequoia
Do we get (-sin(t), cos(t), -4sin(t)cos(t))?
gleaming moss
for?
gleaming moss
,w derivative cos^2 t - sin^2 t
jolly parrotBOT
rough forge
well, the projection of the original surface onto the xy plane is the circle x^2...
so r = 1
gleaming moss
Do we get (-sin(t), cos(t), -4sin(t)cos(t))?
yeah this is correct
yeah r = 1, but that's only true for the constraint
gleaming moss
the actual boundary isn't a circle, which is why the parametrisation has 3 components
regal sequoia
So do we take that times what we calculated before or just times F?
rough forge
the actual boundary isn't a circle, which is why the parametrisation has 3 compo...
yea
because of the z component
gleaming moss
So do we take that times what we calculated before or just times F?
Compute ( \vec{F} (\psi) )
jolly parrotBOT
shsgd
gleaming moss
and then dot it with the tangent vector field
rough forge
I have a question what we doing now, is applying Stoke's theorem?
gleaming moss
yes, this will compute to the same thing as the surface integral of the curl of F
this is what the psi parametrisation looks like btw
rough forge
ok because I just was taught this lately and we never called it Stoke's theorem
gleaming moss
what did you call it?
rough forge
definition of a curve integral
regal sequoia
Okay so ln(4 + 3 sin(t)^2 + cos(2t))? For the first one
gleaming moss
they are separate things
gleaming moss
"Use Stokes' theorem to calculate the circulation:... Orientation: counter-cloc...
.
rough forge
they are separate things
,,\int_C \vec{v} : \dd s = \int_a^b \vec{v}(\vec{c}(t)) \cdot \dot{\vec{c}}(t) : \dd t
jolly parrotBOT
𝔸dωn𝓲²s
rough forge
what I mean
gleaming moss
Okay so ln(4 + 3 sin(t)^2 + cos(2t))? For the first one
this doesn't look right
regal sequoia
I am confused
gleaming moss
psi has 3 components -> (x,y,z), sub them into F
regal sequoia
Is x not cos(t) and y sin(t) and z cos(2t)?
gleaming moss
what I mean
The consequence of applying Stoke's thm is that you are left with a curve integral over a vector field
rough forge
cos^2 t - sin^2 t = cos(2t)
gleaming moss
ah my bad, didn't even clock
rough forge
haha dw
gleaming moss
yeah that's fine then
rough forge
The consequence of applying Stoke's thm is that you are left with a curve integr...
hmm okay I was just confused what's to apply then if we are already given the form of a curve integral
i thought that suggested to turn it into a surface one
gleaming moss
In my experience, they give the surface integral version and you are expected to remember that Stoke's thm will convert it to curve int
So the question was a bit strange in that sense
regal sequoia
so now we multiply F(ψ(t)) * ψ´(t)?
gleaming moss
yes, where * is the dot product
regal sequoia
Oh okay
rough forge
$\tiny \int_{\psi} \displaystyle \begin{pmatrix} \ln(4 + 3 \sin^2(t) + \cos(2t)) \ \ 5\cos(t) + \frac{3\sin(2t)}{4 + 3 \sin^2(t) + \cos(2t)} \ \ \frac{2\cos(t)\cos(2t)}{4 + 3 \sin^2(t) + \cos(2t)} \end{pmatrix} \cdot \begin{pmatrix} -\sin t \ \cos t \ -2\sin(2t) \end{pmatrix} : \dd t$
jolly parrotBOT
𝔸dωn𝓲²s
gleaming moss
Quite a nasty vector field
rough forge
yea
regal sequoia
Yeah haha
rough forge
I think either way is bs
regal sequoia
Do we integrate this now
gleaming moss
Try to simplify first ofc
rough forge
yea
regal sequoia
-sin(t)ln(4+3sint^2+cos2t^2)
rough forge
yea
regal sequoia
How can we simplify that
rough forge
Maybe it turns out something odd
gleaming moss
yeah switch back to cos^2 t - sin^2 t
regal sequoia
Oh okay
gleaming moss
then you get 4+2sin^2 t + cos^2 t
inside ln that is
= 4 + sin^2 t + sin^2 t + cos^2 t = 5 + sin^2 t right?
gleaming moss
cos^2 t + sin^2 t = 1
regal sequoia
It’s 1
rough forge
yes
gleaming moss
I think keeping all of the trigs as squares instead of say sin(2t) will make this easier
rough forge
3sin(2t) = 6sin(t)cos(t)
regal sequoia
-sint ln(5+sint^2)
regal sequoia
3sin(2t) = 6sin(t)cos(t)
Wait where is this
gleaming moss
for the second term:
5cos^2(t) + 6sin(t)cos^2(t) / 5 + sin^2t
gleaming moss
i didn't mean to click send
regal sequoia
I don’t have 5cos(t)^2
rough forge
$\tiny \int_{\psi} \displaystyle \begin{pmatrix} \ln(5 + \sin^2(t) ) \ \ 5\cos(t) + \frac{6\sin(t)\cos(t)}{5 + \sin^2(t)} \ \ \frac{2\cos(t)\cdot \left ( \cos^2(t) - \sin^2(t) \right )}{5 + \sin^2(t)} \end{pmatrix} \cdot \begin{pmatrix} -\sin t \ \cos t \ -4\sin(t) \cos(t) \end{pmatrix} : \dd t$
gleaming moss
when you multiply by cos t from dot product
gleaming moss
the denom of 2nd component is the same as inside of ln of first comp[
rough forge
yesss
gleaming moss
same with bottom
jolly parrotBOT
𝔸dωn𝓲²s
rough forge
awfully wonderful
regal sequoia
Yeah there are harder ones on the test
regal sequoia
for the second term:
5cos^2(t) + 6sin(t)cos^2(t) / 5 + sin^2t
Is this the whole second term or just one part
gleaming moss
that's the second term of the dot prod
regal sequoia
Why 5
rough forge
**𝔸dωn𝓲²s**
here
from 5x
We are having
$$ \int_{\psi} -sin(t)ln(5+sin²(t)) + 5cos^2(t) + \frac{6sin(t)cos^2(t)}{5 + sin^2t} + \frac{-8\sin(t)\cos^2(t)\cdot \left ( \cos^2(t) - \sin^2(t) \right )}{5 + \sin^2(t)} : \dd t$$
regal sequoia
No I am really confused
rough forge
hmm
rough forge
"Use Stokes' theorem to calculate the circulation:... Orientation: counter-cloc...
2nd component look
5x
gleaming moss
you don't need \left ( and \right )
rough forge
so 5cos(t)