Available help channel!
#help-39
pearl pondBOT
Available help channel!
inland laurel
pearl pondBOT
inland laurel
!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
gray sparrow
Why did you do (f(x))(s)??
inland laurel
Why did you do (f(x))(s)??
But tell me how would I find displacement s
gray sparrow
Its just the work formula
W= Fs
S equals x initial- x final
Your block starts at 0
Ends at 2
So your s is 2
Cause 2-0 =2
inland laurel
What did you do ?
I want to know my exact mistake if you see it please let me know
gray sparrow
Ok wait
So
Everything in the blue box u did correct
So you did w= f(x) (s)
But the issue here os that the force is not constant
It depends on x
So you must use the energy conservation
Which is e = k+u= constant
inland laurel
Oh
inland laurel
It depends on x
Shoot how can I forget that
I even mentioned it clearly force depends on x and doing this mistake
gray sparrow
inland laurel
In this I am getting H_max = 0 which doesn't make sense again I am not sure where I have made a mistake
pearl pondBOT
@inland laurel Has your question been resolved?
spring crystal
How u do this
pearl pondBOT
How u do this
sweet bone
you can get the sector between the diagonal line and the y-axis by an inequality on theta (i.e. accept all polar coordinates with angle between those two rays), and the standard equation for a horizontal line y=k is given by r = k/sin(theta)
spring crystal
you can get the sector between the diagonal line and the y-axis by an inequality...
I don’t understand what u said
sweet bone
which part
sweet bone
do you know what polar coordinates are
spring crystal
Yeah x^2+y^2=r^2
sweet bone
that's partially how you convert cartesian to polar, yes
spring crystal
And sin(theta)=y/r
sweet bone
but do you know what polar coordinates are, geometrically?
viscid shale
U talking about this
Going over:
Do you know how cartesian coordinates (x,y) work?
spring crystal
Going over:
Do you know how cartesian coordinates (x,y) work?
I know how to convert them
viscid shale
Thats not the question, do you know how to measure in (x,y)?
As in, if i say you some point (3,6), do you know where to find it?
spring crystal
As in, if i say you some point (3,6), do you know where to find it?
No
viscid shale
Well, cartesian coordinates work basically like a grid
x is left-right
y is down-up+
if i say x = 3, you move 3 steps to the right
y = 6 is 6 steps up
spring crystal
Well, cartesian coordinates work basically like a grid
x is left-right
y is down...
That’s just normal x and y then
viscid shale
Thats what i just said, yes
the names cartesian coordinates
Visually, something like this
Do you know how to measure in (r,θ)?
spring crystal
the names cartesian coordinates
I messed it up with polar
spring crystal
Do you know how to measure in (r,θ)?
No
viscid shale
Well, r is the distance from the origin (0,0)
If you know basic geometry, you know that a circle can be defined as all the points that have the same distance from the center
so, here, r is essentially "radius"
And theta tells you to what angle to look, starting from the positive x axis.
So, essentially, you have to distance yourself some value from the origin, and rotate around starting from the x axis, counterclockwise.
Visually speaking, something like this
why we do this thing? Because everything about circular objects (and therefore also triangles) become somewhat easier in polar.
what you previously stated was a set of equalities used to between cartesian and polar
$x = r\cos\theta \ y = r\sin\theta \ r = \sqrt{x^2 + y^2}$
jolly parrotBOT
∫ᴄ 𝐅·𝑑𝑟 = ∬ʀ ∇⨯𝐅 𝑑𝐴
spring crystal
viscid shale
yes
spring crystal
So I’m suppose to draw like this
viscid shale
Its like a curved grid
viscid shale
Well, you probably know that we usually describe functions in (x,y) as
y = f(x)
Which we sometimes write in shortform as y(x) to imply that y varies in terms of x
for polar, we usually do r(θ)
So, the radius (distance from 0,0) becomes a function of the angle.
viscid shale
The points are the lesser problem
We have to define a set of inequalities (as functions) that describe that region.
Since we assume that the radius is a function of the angle, we know that the angle is independant (it has constant bounds)
viscid shale
How I turn these two points into polar tho
In here, the region is described between which angles?
You could find it using the points, or just by looking at it.
spring crystal
In here, the region is described between which angles?
(0,3) and (3,3)?
viscid shale
Those are not angles
spring crystal
In here, the region is described between which angles?
Bruh what
viscid shale
(0,3) and (3,3)?
Find the angles corresponding to these two
Or, just for practice, their coordinates in polar.
spring crystal
Or, just for practice, their coordinates in polar.
Like (0,3) is 0=rcostheta and 3=rsintheta?
viscid shale
yea
viscid shale
yes
viscid shale
r and theta
spring crystal
Like (0,3) is 0=rcostheta and 3=rsintheta?
R=0 and r=3/sin
And theta is cos^-1(0) and sin^-1(3/r)
viscid shale
R=0 and r=3/sin
Are you sure about R = 0?
viscid shale
**∫ᴄ 𝐅·𝑑𝑟 = ∬ʀ ∇⨯𝐅 𝑑𝐴**
Again, recall to these
viscid shale
**∫ᴄ 𝐅·𝑑𝑟 = ∬ʀ ∇⨯𝐅 𝑑𝐴**
The fact that x is 0 does not imply that r is too.
Mostly, you should try to look at the r = sqrt equation
tulip ore
bruh youre here instead
spring crystal
The fact that x is 0 does not imply that r is too.
I did r=costheta
viscid shale
bruh youre here instead
?
tulip ore
you asked a question in the other homework help server and now youre here, I was wondering why you didnt respond
viscid shale
I did r=costheta
x = r * costheta
you know x, do you know r or theta?
spring crystal
bruh youre here instead
I asked same time 🙏 🙏
tulip ore
Im looking and you asked there 20 minutes after you asked here
spring crystal
x = r * costheta
you know x, do you know r or theta?
X=rcos
tulip ore
it sounds like you need a review of polar coordinates first before we can start on this
spring crystal
Ik conversion
viscid shale
X=rcos
Yes, this is true
3 = r . cosθ
Howd you go on solving that?
or 0 = r . cosθ
for that case.
spring crystal
Divide both side cos
tulip ore
you mean by cos(theta)?
spring crystal
Ye
tulip ore
wait what exactly are you doing rn
viscid shale
wait what exactly are you doing rn
Trying to convey to him why r isnt 0
Given that he wont listen on using the r = sqrt x^2 + y^2 formula
tulip ore
solving the equation a second time isnt exactly going to show that
@spring crystal if you have 0 = r cos θ
you know cos θ can also be 0, right
do you remember a value of θ where cos(θ) = 0?
in a way you told it to me last time
spring crystal
Given that he wont listen on using the r = sqrt x^2 + y^2 formula
Ok it’s 3 and sqrt18 using that
spring crystal
do you remember a value of θ where cos(θ) = 0?
Theta as 1
viscid shale
Yes, now that you have x,y and r, you can move on getting the angles.
tulip ore
cos(1) isnt 0, so not really
tulip ore
Theta as 1
theta needs to be an angle, so something like 60 or 90 or something like π/3 or π/2
spring crystal
Pi/2 and 3pi/2
tulip ore
yep
viscid shale
theta needs to be an angle, so something like 60 or 90 or something like π/3 or ...
Since you seem to already have worked with him, is it valid to use atan2(y,x)?
tulip ore
I dont think he's seen that yet, but if you see it graphically then it should work out
I havent worked with him for long
I dont have a good doodling tool
viscid shale
i do, gimme a sec
You already worked out that r = sqrt 18
You can solve for theta just by looking at the diagram and knowing some bit of geometry, or you can use the algebraic way
tulip ore
we're having 3 = both of them for (3, 3), the (0, 3) point with the π/2 or 3π/2 we do later
spring crystal
You can solve for theta just by looking at the diagram and knowing some bit of g...
Is pi/4
Isn’t r also =3
tulip ore
unless you want to say $\sqrt{3^2+3^2}=3$, no, you figured out $r=\sqrt{18}=3\sqrt2$
jolly parrotBOT
mtt
tulip ore
the point is 3 right and 3 up from (0, 0)
its gotta be farther than 3 units away from (0, 0) too
viscid shale
You now get the other version, where r is 3
tulip ore
this time, since 0 = 3 cos(theta), you mentioned some values where cos(theta) can be 0
which value do you think would work here
spring crystal
unless you want to say $\sqrt{3^2+3^2}=3$, no, you figured out $r=\sqrt{18}=3\sq...
From (0,3)
tulip ore
sure, (0, 3) has r = 3
viscid shale
You hace two different values of theta and r
tulip ore
(3, 3) has r = √18, theta = π/4
(0, 3) has r = 3, theta = (you gotta say)
tulip ore
(3, 3) has:
3 = r cos(theta)
3 = r sin(theta)
you figured out r = √18, then you figured out theta = π/4
tulip ore
You now get the other version, where r **is** 3
we're going to focus on this now
(0, 3) has:
0 = r cos(theta)
3 = r sin(theta)
we agree that r = 3 here, so now you figure out what theta is
remember last time that cos(theta) can be 0, that happens when theta = π/2 or theta = 3π/2
one of these values would be correct for the point
spring crystal
You now get the other version, where r **is** 3
Pi/2,3pi/2 and for sin is pi/2
tulip ore
thats correct
we need sin(theta) = 1, and for that theta = π/2 is the only possible value
oh btw, do you want to say how you figured that out
spring crystal
You now get the other version, where r **is** 3
This
Divide both side by 3 and do arcsin and arc cos
spring crystal
Pi/2,3pi/2 and for sin is pi/2
So this is also only pi/2
viscid shale
Yes, now you got your values for θ
Since we are trying to describe an area/region
We have to bound theta
spring crystal
Pi/2?
viscid shale
A bound always needs two values
Is like when you define an interval in x
[a,b]
Where a has to be less than b for it to make sense.
spring crystal
Where a has to be less than b for it to make sense.
I don’t use 3pi/2?
Cus it’s not in cosine thing
tulip ore
we're talking about this region now
tulip ore
Since we are trying to describe an area/region
you can tell because they said "area/region" here, read closely to confirm this
then theyre saying to "bound theta"
do you know what that means
viscid shale
Think of theta sweeping around the (0,0)
Theres a select amount of angles that enclose this figure
spring crystal
you can tell because they said "area/region" here, read closely to confirm this
area/region r u talking about the shaded part
tulip ore
yes, thats the shaded part
that is called an area, or region
they both mean this 2D triangle that happens for some theta and not others
viscid shale
Ignore R for now
spring crystal
do you know what that means
No
viscid shale
Remember that its a function of theta
viscid shale
So 3<=r<=sqrt(18) and pi/2<=theta<=pi/4 is the final answer
And the bounds should be the other way around
spring crystal
Wym
viscid shale
pi/4 < θ < pi/2
spring crystal
Isn’t it 3 and sqrt18
spring crystal
so,
0 < r < f(θ)
So 0<=theta<=pi/4?
viscid shale
So 0<=theta<=pi/4?
No, you are mixing both, we get two different bounds
The problem, beyond the obvious difference of polar coordinates, is arguably the same as these.
And i hope you already seen this kind
Where some of the bounds is a function
Well, here, "x" which is the independent variable became θ
viscid shale
Whatever form youre familiar with, they are both the same idea.
spring crystal
Ok how I use them to get final answer
spring crystal
Your answer is based on these bounds
U said it started at 0
viscid shale
r
viscid shale
Nope
You got the 0 right
But the upperbound isnt a constant
Since it doesnt form a "section" of a circle.
viscid shale
But the upperbound isnt a constant
Not a constant
Constants are numbers, the upperbound varies depending on the angle
spring crystal
Constants are numbers, the upperbound varies depending on the angle
So u can’t even find it
viscid shale
Yes you can
Varying upperbounds can be described as a function of the other variable
Since we have r, and theta
And theta is between two constants
The upperbound of r is a function of theta.
as in
this is a random example and not the solution: f(θ) = θ^2
spring crystal
Bruh I don’t get it
viscid shale
Do you understand the angle bounds?
spring crystal
Varying upperbounds can be described as a function of the other variable
I mean this
viscid shale
Do you understand the angle bounds?
Answer this pls
As in, why we do that and what it means
spring crystal
As in, why we do that and what it means
U talking about the polar coordinates
spring crystal
No
viscid shale
Well, lets go back to (x,y) for a single second.
Do you understand that this figure represents an area/region?
pearl pondBOT
@spring crystal Has your question been resolved?
spring crystal
Ye
viscid shale
Do you understand that it can be represented as two bounds?
tulip ore
back
I was thinking we'd be past this by now, that sucks
@spring crystal have you heard of the word "bound" before, as a noun?
spring crystal
Do you understand that it can be represented as two bounds?
One for x axis and one for y?
spring crystal
<@1145726673262747699> have you heard of the word "bound" before, as a noun?
It’s inequality?
spring crystal
So this is 3<=x<=5 and 4<=y<=5
tulip ore
have you heard it before, or know what it means?
spring crystal
Ye
tulip ore
what does it mean
spring crystal
In math?
tulip ore
bro what does it mean in common english
spring crystal
The limit of something
tulip ore
can you be more specific
spring crystal
So this is 3<=x<=5 and 4<=y<=5
Like these r bounds
tulip ore
that doesnt tell me what a bound is other than that those are supposed to be bounds
can you be more specific on what a bound is actually supposed to be
you can say you dont know
@spring crystal you here?
viscid shale
<@1145726673262747699> you here?
This has been stretched too long tbh
tulip ore
yea, my mistake
viscid shale
nah, dont worry
Idk, seems hard to convey a point since it looks like he's missing a lot of key concepts.
and i didnt even get to the function upperbound 
spring crystal
can you be more specific on what a bound is actually supposed to be
The highest number possible
tulip ore
and also the lowest number possible too, thats the idea
spring crystal
Of lowest
spring crystal
and also the lowest number possible too, thats the idea
Ye
tulip ore
so here, youre saying the bounds
now we write them as 3 ≤ x ≤ 5
because x as a number is ≥ 3 and also ≤ 5
its useful to think that the < has its larger end towards the larger number in case you get them mixed up
I do this too
tulip ore
now we write them as 3 ≤ x ≤ 5
so 3 ≤ x ≤ 5 says that x at its lowest is 3, and at its highest is 5
right
spring crystal
Ye
tulip ore
now when we did this example last time
I remember you said the bounds were π/4 ≤ theta ≤ π/2 earlier
not here though
looking at this shaded area, does it make sense that theta is always between π/4 and π/2?
spring crystal
Ye
tulip ore
this is what we mean by a lower and upper bound
we're choosing something that theta, or x, or y, etc. is always above or below of
spring crystal
Ik that
spring crystal
I remember you said the bounds were π/4 ≤ theta ≤ π/2 earlier
So theta is basically y axis
tulip ore
saying that basically means you do not, in fact, know that
tulip ore
So theta is basically y axis
what do you even mean by this
spring crystal
Ok wait nvm it’s the unit circle
tulip ore
its sort of like the unit circle
tulip ore
Ok wait nvm it’s the unit circle
usually we have x-direction going horizontal and y-direction going vertical
but the r- and theta-direction dont do this
can you describe the r-direction and theta-direction
spring crystal
Theta is the unit circle
Direction
And r is however much circles it goes up by
Like this
tulip ore
thats fair
we say theta is going counterclockwise
and r is going outwards
does that make sense
tulip ore
bro you dont understand us
we cant tell you how to get the answer if you cant understand us
lets do this one step at a time
as I said earlier, this is no method to solve this problem, so we have to learn something new to solve it first
do you understand?
spring crystal
Aight
tulip ore
first, Im gonna need to convince you that we need this new thing
because 100% youre not going to believe me at first
take a look at this region
you already figured out π/4 ≤ theta ≤ π/2, that part is correct
theta has lower bound π/4 and upper bound π/2
what about r?
spring crystal
3 and sqrt18
spring crystal
take a look at this region
From this
tulip ore
you know that doesnt tell me anything
the image doesnt say the numbers "3" and "√18" on it
spring crystal
R=sqrt(x^2+y^2)
tulip ore
you need to tell me your process of from that to "r must be between 3 and √18"
tulip ore
R=sqrt(x^2+y^2)
alr thats going somewhere, how did you use this
spring crystal
I plug in once for 0,3 and 3,3
spring crystal
Find theta which is pi/4 and pi/2
tulip ore
we're talking about r right now bro
we dont need to talk about theta
we already confirmed π/4 ≤ theta ≤ π/2
tulip ore
3 and sqrt18
later on we will mention theta, but for now Im talking about your current answer for r, which is "3 ≤ r ≤ √18"
spring crystal
Yeah that’s it
tulip ore
right, now lets look at the shaded area closer
Im going to tell you that 3 ≤ r ≤ √18 is wrong, and youll need to figure out why that is
lets start with the 3
youre saying 3 is a lower bound for r?
spring crystal
I don’t see any other numbers
tulip ore
you can see the rest of the region though
if they didnt tell you the upper-left point was (0, 3), but it was still the same shape,
youd just be saying √18
do you think thats the right way to do things?
spring crystal
So how then
tulip ore
you know earlier I told you theres no method to figuring this out
that means you gotta use your common sense
tulip ore
take a look at this region
I know you have it in there, because if you look at this shape, you should be able to tell me that r isnt always between 3 and √18
what would be a better lower bound for r?
@spring crystal you can say "I dont know"
spring crystal
Idk
tulip ore
when you say a lower bound for r, youre saying "every point in this shaded area has an r, and 3 is the lowest that r can be"
have you thought of this
spring crystal
Ye
spring crystal
Ye
tulip ore
what r does (0, 0) have?
spring crystal
0
tulip ore
is (1, 2) also in the region?
spring crystal
Ye
tulip ore
what r does it have?
spring crystal
1
tulip ore
no, (1, 2) does not have an r of 1
try that again
we might need to go over what r is even supposed to mean
spring crystal
1/costheta
tulip ore
lets try on common sense for this one, it can give us easy answers
theres a 2 in the (1, 2)
you think r would be smaller than 2?
spring crystal
Isn’t it (r,theta)
tulip ore
dellungi bro
no
(1, 2) is an (x, y) point you can see so
(r, theta) points have the second number be like an angle in radians or degrees
thatd have like π/4 or 45 or something like that
that you didnt tell is not good news
tulip ore
lets try on common sense for this one, it can give us easy answers
it means youre not familiar with polar coordinates yet to tell them apart from the cartesian coordinates
we gotta go over the polar coordinate stuff
tulip ore
when you say a lower bound for r, youre saying "every point in this shaded area ...
we were just mentioning (0, 3) and (3, 3) minutes before this
do you think (0, 3) and (3, 3) are in polar coordinates or cartesian coordinates?
spring crystal
Cartesian
tulip ore
so when I mentioned (0, 0) then (1, 2), why polar?
spring crystal
Nvm it’s Cartesian
spring crystal
thatd have like π/4 or 45 or something like that
I forgot this was how it worked
tulip ore
thats fair, lets go over it quick
remember that polar coordinates have the same way of writing them as cartesian coordinates, so if youre not sure, theres a few ways you can check
usually when points are mentioned, (r, theta) would use certain "special values" for theta
this is similar to how if you just mention (integer, integer) for a point, youd think it was (x, y)
theta can be seen in degrees or radians
so thats 0, 45, 90 or 0, π/4, π/2
(0, 3) for example would be (3, π/2) in polar coordinates
and (3, 3) would be (√18, π/4) in polar coordinates
now those are just two points of this shaded area, but theres a bunch more
I mentioned (0, 0) and (1, 2), you think theyre in the shaded region?
spring crystal
Ye
tulip ore
what about (2, 1)?
spring crystal
Ye
tulip ore
why's that?
spring crystal
Cus it’s within (0,3) and (3,3)
tulip ore
1 isnt between 3 and 3 bro
they shaded a whole triangle of points, you cant just look at the (0, 3) and (3, 3)
keep in mind (0, 0) and (1, 2) are also in the triangle too
(2, 1) is not in the triangle, can you see why?
you can try and see where (2, 1) is
then figure out if itd be in the triangle or not
spring crystal
1 isnt between 3 and 3 bro
How is 2 and 0 brtween them then
tulip ore
its strange how (0, 0) and (1, 2) are in the triangle
maybe you should uh
look at the picture to see where (0, 0) and (1, 2) are
lets first start with (0, 0)
where would it be in this picture?
spring crystal
its strange how (0, 0) and (1, 2) are in the triangle
Ok nvm
tulip ore
I just gotta wonder, why didnt you think of this before
you see what I mean by going through stuff you dont think is important first?
if you cant even tell what is or isnt in the triangle, how are we going to say "every point in this triangle has an r that can only be this high"
spring crystal
The dot in the middle
tulip ore
I know about the dot in the middle, but you didnt know about it when you said (2, 1) was in the triangle too
so that cant be the reason can it
why did you say (1, 2) was in the triangle
spring crystal
if you cant even tell what is or isnt in the triangle, how are we going to say "...
So r is these 1,2,3 points I drew?
tulip ore
you aint dodging the question
spring crystal
why did you say (1, 2) was in the triangle
Cus it was between the numbers
tulip ore
which ones
spring crystal
Cus it’s within (0,3) and (3,3)
I thought this
spring crystal
what did you compare the number 1 and 2 with here
Cus they were between 0 and 3
tulip ore
there we go, thats the reason
so you were just doing 0 ≤ x ≤ 3 and 0 ≤ y ≤ 3
now lets think: if you had another region that was all the points (x, y) with 0 ≤ x ≤ 3 and 0 ≤ y ≤ 3, what shape would that make
spring crystal
Rectangle
tulip ore
you can be more specific than that
tulip ore
its a square bro
its 3 long and 3 high
cmon you can see that
youre not even thinking about what it looks like anymore
otherwise you wouldve seen the square...
and said so
you gotta think visually for this problem
spring crystal
Then goodbye points
tulip ore
you thought (2, 1) was in the triangle bro
theres gonna be a lot of missing points
if we keep working on this then we wont have that issue anymore
I dig up whats wrong then we fix it
tulip ore
Ok nvm
if we look at this visually, it tells us (2, 1) isnt in the triangle
but theres no algebraic x y way to tell us that
we'd have to make inequalities based on x and y first before we could do that
you used 0 ≤ x ≤ 3 and 0 ≤ y ≤ 3 earlier, theres one extra condition you couldve used to make sure a point is in the triangle
think about the points (1, 1), (2, 2), and (3, 3)
are these on the triangle?
spring crystal
Ye
tulip ore
yep
now if we move these points slightly upwards
like (1, 1.0001)
are they still in the triangle?
spring crystal
Ye
tulip ore
but if we move the points slightly downward, like (1, 0.9999), its not in the triangle, right
spring crystal
Ye
tulip ore
think about the points (1, 1), (2, 2), and (3, 3)
so that means this here is some other condition that isnt just number ≤ x ≤ number
it depends on x and y
we can see that the line y = x would be the diagonal of this triangle, right
its like an edge of it
spring crystal
Ye
tulip ore
like (1, 1.0001)
we can imagine it goes one way, its in
tulip ore
but if we move the points slightly downward, like (1, 0.9999), its not in the tr...
another way, its out
so lets think about x ≤ y and x ≥ y
which one of these do you think needs to be true for a point to be in the triangle?
spring crystal
Y>=x
tulip ore
and why y >= x instead of the other one?
spring crystal
Cus the other leaves the region
tulip ore
thats correct
we know x = y should be an edge, and the points we test fit x ≤ y and dont fit x > y
visually, this is what x ≤ y looks like
you can see the square 0 ≤ x ≤ 3 and 0 ≤ y ≤ 3
and in yellow, separately, is x ≤ y
you can see if you have 0 ≤ x ≤ 3 and 0 ≤ y ≤ 3 and x ≤ y, that would be the triangle we need, right
spring crystal
Ye
tulip ore
now that means we figured out what this would be if we were using cartesian coordinates
but keep in mind we need polar coordinates instead
now for polar coordinates, we can still have an inequality that uses both r and theta
the same way the triangle needed an inequality that used x and y: x ≤ y
its not r ≤ theta or something like that, finding the correct inequality is going to be hard, but Ill be doing part of it so itll be easy
ok you remember from earlier that π/4 ≤ theta ≤ π/2, right
spring crystal
Ye
tulip ore
that currently does this
now this time Ill ask again what the lower bound of r is
and as before, we know theres more points than just (0, 0), (1, 2), (0, 3), (3, 3)
think about all the points in the triangle
what is the lowest r you can find in this triangle?
spring crystal
now this time Ill ask again what the lower bound of r is
0
tulip ore
thats correct
thats the lower bound
and we'll need an upper bound, we wont use it for now
just tell me the highest r you can find in this triangle
spring crystal
that currently does this
4
tulip ore
no I mean the actual triangle we're doing in the problem
spring crystal
3
tulip ore
you sure? we did find a higher r than that
r doesnt mean "the biggest coordinate"
r means something else
spring crystal
Yeah idk what that is
I only know r as this circle thing or how to convert it to r
tulip ore
so you did not, in fact, know what r was
tulip ore
Ik conversion
you let that slip a bit over here
tulip ore
I only know r as this circle thing or how to convert it to r
r here means how far you are from the center
it stands for "radius"
these are all the points where r is 1
as you can see its a circle
tulip ore
3
its not a square, its a circle
for example, I can zoom in on this circle, and (0.6, 0.8) is on it
that means the point (0.6, 0.8) and the point (0, 0) are 1 unit apart
the same distance as between (0, 0) and (1, 0)
your formula for turning (x, y) -> r is really just pythagoras theorem
for the point (0.6, 0.8), you can draw this triangle
looking at this triangle,
what is the base and what is the height?
spring crystal
So this is r
tulip ore
not really
tulip ore
for the point (0.6, 0.8), you can draw this triangle
lets stick to what Im talking about for now
Im still in the process of telling you what r is
try not to guess, its not what you expect
tulip ore
for the point (0.6, 0.8), you can draw this triangle
what is the base and height of this triangle?
spring crystal
for the point (0.6, 0.8), you can draw this triangle
0.6 and 0.8
spring crystal
1
tulip ore
yea not much more than that
you can see here 1 is the length of the line from (0, 0) to (0.6, 0.8)
thats how we can say (0, 0) and (0.6, 0.8) are 1 unit apart, right
spring crystal
K
tulip ore
thats correct, this is how we define the distance between things
we draw a line then measure its length
now using this, r is the distance of our point from (0, 0)
now as I mentioned earlier, a circle is the shape you get for all the points that are a distance of 1 from the (0, 0)
this circle is at (0, 0) with radius 1
and that radius is the same no matter where you are on the circle
does that make sense?
spring crystal
K
tulip ore
its gotta make sense though
you know what a circle exactly is, right
its a shape where every point is the same distance from the center
so thats the shape you see for this
spring crystal
Ok but what about my problem
tulip ore
bro you think we can skip through any of this
spring crystal
Bruh ima do this later
tulip ore
you cant exactly do this at all
wdym later
you at least gotta know what r is
all the points with r = 3 are on the red circle here
youre saying the biggest r can be on this triangle is 3
theres this whole sliver outside the triangle that has r bigger than 3
you dont even remember the √18?
that was the r you calculated for the upper-right corner for the triangle
pearl pondBOT
@spring crystal Has your question been resolved?
dark oyster
pearl pondBOT
lean oak
that's quite the jump from the quadratic in x to suddenly finding values of y.
nocturne grail
uhm
lean oak
mistake here.
nocturne grail
how did you evne get 8
nocturne grail
mistake here too.
odd how they did the "hard" part of the question and yet failed the subs
dark oyster
ty
lean oak
if you have nothing else, you can .close the channel, and see you around.
dark oyster
.close
wraith badger
from the first step to second step it doesnt really matter what thing is where -sinx is right? because it'll always be d/dx [integrating factor * y]
tropic saddle
well if there was something else other than -sin(x) there then the step would be false...
pearl pondBOT
@wraith badger Has your question been resolved?
wraith badger
well if there was something else other than -sin(x) there then the step would be...
Whats actually the method to get to that step? I've only learnt the shortcut taught in the video
jolly parrotBOT
Xavier 🌺
pastel umbra
Namely - you've at this point hopefully learnt the product rule
At that stage your aim was to recognise when you have two functions multiplied together, to be able to use the product rule in the first place
With differential equations, now your aim should be to recognise the output of a product rule
pearl pondBOT
pastel umbra
When you've got a $\frac{\dd y}{\dd x}$ and a...
jolly parrotBOT
Waes (Wires)
pearl pondBOT
✅ Original question: #help-39 message
pastel umbra
When you've got a $\frac{\dd y}{\dd x}$ and a $y$ in there, that's gonna typically look like something of the form
$$f(x)\frac{\dd y}{\dd x} + f'(x)y$$
jolly parrotBOT
Waes (Wires)
pastel umbra
where f(x) is some function in x only
pearl pondBOT
@wraith badger Has your question been resolved?
buoyant fulcrum
16th,
I'm really confused where the peak point of P is
,rccw
buoyant fulcrum
,rccw
jolly parrotBOT
woven river
It's going to be in the middle of triangle, so find that P
I forgot the name, lemme do quick search
woven river
searching is not helping for me, but u get the point
buoyant fulcrum
9 point circle centre?
buoyant fulcrum
searching is not helping for me, but u get the point
Which?
buoyant fulcrum
It's incentre
buoyant fulcrum
.close
midnight haven
pearl pondBOT
midnight haven
idk where to begin i've trieed to factor the polynomial in the denominator but it doesnt end up any helping
x^4+x^2+1 = x^2[(x+1/x)^2-1]
I'd love a hint if possible, thanks!
toxic lichen
x^4+x^2+1 = x^2[(x+1/x)^2-1]
that is a pseudofactorization
toxic lichen
anyway, the tan^-1(x) is suggestive
midnight haven
my teacher taught us its named symetric factorization
toxic lichen
maybe some kinda ibp could work
ac wait i think you can cook with yours
(x^2+1)^2 - x^2 = (x^2-x+1)(x^2+x+1)
toxic lichen
x^4+x^2+1 = x^2[(x+1/x)^2-1]
= x^2 (x + 1/x - 1)(x + 1/x + 1)
midnight haven
oh damn
wait
1 s
ok got it
so it transforms into int from 0 to inf of [((x+1)^2 arctanx)/(x^2-x+1)(x^2+x+1))]
does partial decomp works here
i have arctan
cursive wraith
Before partial decomp there is something to see imo
Symetric factorization has something to do with it
The trick is to find an interesting variable change
cursive wraith
x^4+x^2+1 = x^2[(x+1/x)^2-1]
= x^2 (x + 1/x - 1)(x + 1/x + 1)
The way Ann wrote it is very suggestive
midnight haven
give me 5 minutes to think it
please no hints
wait variable change as in u sub right?
what im thinking is u = x+1/x
but i dont see how that would work
the up is x(x+1)^2 = x^2+2x+1 = x^2 (1+2/x+1/x^2)
which is interesting
to me atleast
pearl pondBOT
Available help channel!
tepid root
hello, i am heading into the sat tomorrow, are there any specific websites or resources whicj are especially useful for studying the material?
light helm
look for past papers
light helm
but it seems to be a bit late to be asking now
tepid root
but it seems to be a bit late to be asking now
did i do somethign wrtong .,.,
am ialte
light helm
seeking study resources the night before is cutting it very close
tepid root
yeah tbh ur right
light helm
but
look for past papers
and try your best
pearl pondBOT
Available help channel!
ionic dust
Given the sequence u(n) defined by u1=1, u(n+1)=2u(n)+5, find the general term of the sequence, im literally burning out
cursive wraith
Given the sequence u(n) defined by u1=1, u(n+1)=2u(n)+5, find the general term o...
do you recognize what kind of sequence this is
maybe your teacher gave it a name
ionic dust
i just learned about sequence, arithmetic sequence, geometric sequence
u(n+1)=u(n) + d
u(n+1)=u(n) . q
cursive wraith
i just learned about sequence, arithmetic sequence, geometric sequence
and you never saw the "combination" of arithmetic and geometric sequences
u(n+1) = au(n) + b
arithmetico-geometric is their name (or AG/ AGP in short)
ionic dust
and you never saw the "combination" of arithmetic and geometric sequences
yea, my book doesnt teach us about this
cursive wraith
ok, in that case
we're gonna find out how to construct an easier sequence from this
ionic dust
greattr
cursive wraith
Given the sequence u(n) defined by u1=1, u(n+1)=2u(n)+5, find the general term o...
so take this sequence u(n)
and here's a first question:
suppose it converges (which it may not, but that's out of the point for now)
how would you find its limit
cursive wraith
ok, we'll deal through this
instead of thinking about limits
we'll think about constant sequences
let's forget about the fact that u1 = 1
ionic dust
yea
ionic dust
uhh zero? i dont know a lot about math terms and english sooo its kinda hard 😞
cursive wraith
uhh zero? i dont know a lot about math terms and english sooo its kinda hard 😞
bad guess, don't worry about english barrier we'll try to get through it
Do you know what a constant sequence is?
just making sure
ionic dust
like 1 1 1 1 1 right
cursive wraith
for example
in general, u(n) = c for all n
and c is a fixed value
If you have a sequence
and you want to check if it's constant
you could for example check if you can find that "c"
and then check that u(n) = c for all n
ionic dust
yea yea right
cursive wraith
Otherwise, you can just check
that no matter the term you look at
the next term is the same
so u(n+1) = u(n) for all n
now
we're given a sequence u(n) such that u(n+1)=2u(n)+5
I'm telling you it's constant
find the value of the constant
cursive wraith
then the value would be -5
that's great, I'm just gonna take some time generalizing what you found
ionic dust
great
cursive wraith
so if you have an AG sequence: u(n+1) = au(n)+b
if it's constant, then C = aC + b
and that's probably how you found the constant C in our case
ionic dust
oooooo thats cool
cursive wraith
so the formula for C is C = b/(1-a), but it won't be the most relevant equality
cursive wraith
if it's constant, then C = aC + b
This equality will be VERY important, try to remember it
cursive wraith
then the value would be -5
now let's continue, we're armed with our constant C = -5
cursive wraith
we're gonna find out how to construct an easier sequence from this
and we'll finally do what I promised
construct an easier sequence
It's not easy trying to guess which sequence is going to work, so I'm going to give you this:
Let v(n) = u(n) - C
Why is this the sequence that works? To find out, try to compute v(n+1) in terms of v(n)
(so basically find a recursion formula)
ionic dust
soo u(n+1) + 5 = 2u(n) + 10
then u(n+1) + 5 = 2[u(n) + 5]
-> v(n+1) = 2v(n)
right
kinda confused
cursive wraith
soo u(n+1) + 5 = 2u(n) + 10
then u(n+1) + 5 = 2[u(n) + 5]
-> v(n+1) = 2v(n)
rig...
That's correct
So we found out that shifting u(n) by the right constant
we get what type of sequence?
ionic dust
geometric sequence
ionic dust
wow that's coolllll)
cursive wraith
so, just to generalize
(because AG sequences are good to know about)
if we have u(n+1) = au(n) + b
we let v(n) = u(n) - C, with C found above
and v(n+1) = u(n+1) - C = au(n) + b - C
cursive wraith
This equality will be VERY important, try to remember it
and that's where the important equality comes into place
v(n+1) = au(n) + b - (aC+b)
v(n+1) = au(n) - aC
v(n+1) = a(u(n)-C)
v(n+1) = av(n)
ionic dust
mind blowing
cursive wraith
magically, v is a geometric sequence
ionic dust
😱
cursive wraith
so, we get a formula for v(n) in terms of v1
v1 is computed easily
and u(n) = v(n) + C
done
cursive wraith
then it's not an AG sequence anymore
AG sequences are specifically u(n+1) = au(n)+b, with a,b constants and a ≠ 1
(if a = 1 it's arithmetic)
ionic dust
like that problem, u1=2
ionic dust
yeaaaa 😞
cursive wraith
here's a funny idea
you know how with functions, you can differentiate
IF u(n) was a function (which it isn't)
we could compute "u'(n) = 2u'(n-1) + constant"
and then we simply "integrate"
That's not how it works with sequences at all, which is why I put a lot of quotations
but we have a similar tool
cursive wraith
***IF*** u(n) was a function (which it isn't)
before I explain the tool, does all I've talked about make sense or am I saying nonsense?
ionic dust
does this involve in "derivative" or something, i havent learnt that as well
cursive wraith
does this involve in "derivative" or something, i havent learnt that as well
right so you haven't learnt derivatives
ionic dust
havent learnt derivative, integral, limit
cursive wraith
ok
have you heard about derivative of x^n being nx^(n-1) for example
while not having specifically seen it?
ionic dust
yea i know that
cursive wraith
ok
ionic dust
that's the only thing i know
cursive wraith
that's all we need
ionic dust
oooo cool
cursive wraith
since polynomials are just a sum of those x^n (with some weights in front, like ax^2 + bx + c for example)
The derivative of a polynomial will always result in a polynomial of smaller degree
because x^n goes down to x^(n-1)
x^(n-1) goes down to x^(n-2)
etc...
the derivative of "ax+b"
is thus a constant
cursive wraith
like that problem, u1=2
So going back to this "3n-1"
if we manage to "derivate" it, we would get a constant
and we're back to AG sequences
Does that make sense?
cursive wraith
yes
so, are we ready to learn about the magical tool or do you have more questions about what we said?
ionic dust
im ready
cursive wraith
ok
while I was thinking about it, I found a different idea, I'll reserve it for later as I don't know if it's easier or more difficult
The "differentiation" tool for sequences
is just computing the difference between the next term and previous term
v(n) = u(n+1) - u(n)
so, if u(n) = 2u(n-1) + 3n -1
can you find a recursion formula for v(n)?
ionic dust
that might be hard for me
u(n+1) = 2u(n) + 3n + 2
then
v(n) = 2u(n) + 3n + 2 - u(n)
v(n) = u(n) + 3n + 2
cursive wraith
then
v(n) = 2u(n) + 3n + 2 - u(n)
v(n) = u(n) + 3n + 2
that will be useful later, let's not forget about it
but that's not gonna give us a recursion formula
if v(n) = u(n+1) - u(n)
try applying recursion formula on u(n+1) AND on - u(n)
ionic dust
kinda confused right now 😞
cursive wraith
kinda confused right now 😞
alright, let's take it together
according to the recursion formula, u(n+1) = ?
ionic dust
u(n+1) = 2u(n) + 3n + 2
ionic dust
u(n) = 2u(n-1) + 3n - 1
cursive wraith
exactly
so
u(n+1) = 2u(n) + 3n + 2
u(n) = 2u(n-1) + 3n - 1
And thus u(n+1) - u(n) = ?
ionic dust
2u(n) - 2u(n+1) + 3
cursive wraith
Can you try to solve for v(n) on your own? I can give hints if you're stuck
cursive wraith
stuck at finding v1 😞
by definition, v1 is u2 - u1
you have u1
you just gotta compute u2
ionic dust
oh right right im dumb hahahha
i wanna find 2019th term right now, the question wants me to do that
cursive wraith
i wanna find 2019th term right now, the question wants me to do that
you want to find u2019, not v2019 specifically
to do that, we need to find a formula for u(n) first
we're close though
what's the formula for v(n)?
ionic dust
v(n) = 2v(n-1) + 3
cursive wraith
v(n) = 2v(n-1) + 3
that's the recursion formula
I'm talking about v(n) in terms of n
now that you have x(n) in terms of n it shouldn't be too complicated right?
ionic dust
now that you have x(n) in terms of n it shouldn't be too complicated right?
i dont think so 🥲
ionic dust
I'm talking about v(n) in terms of n
v(n) = 2u(n) - 2u(n+1) + 3? that's in term of u(n) not n tho
cursive wraith
v(n) = 2u(n) - 2u(n+1) + 3? that's in term of u(n) not n tho
that's not it 😭
as a hint, look towards x(n), not u(n)
x(n) you have in terms of n
ionic dust
x(n) you have in terms of n
im so dumb at the moment
cursive wraith
im so dumb at the moment
Are you still stuck?
ionic dust
yeaaaaa
cursive wraith
yes
so
now we got v(n) in terms of n
can you find an equality that links u(n) to v(n)?
(One that I told you not to forget about)
ionic dust
this?
v(n) = 2u(n) + 3n + 2 - u(n)
v(n) = u(n) + 3n + 2
ooooo nah way
so u(n) = 10.2^(n-1) - 3n - 5
i appreciate so muchhhhh, thank you for your time, now i know how to do these problems aaaa
.close
pearl pondBOT
pearl pondBOT
Available help channel!
hot pond
is this true
pearl pondBOT
is this true
lean oak
I suppose? linear algebra does see a lot of applications outside of pure math.
hot pond
I suppose? linear algebra does see a lot of applications outside of pure math.
LOL bet i was js joking
but i actually have really enjoyed lin alg
even tho i kinda choked my first mt i got a 70
so im restarting with a more conceptual approach
hot pond
I suppose? linear algebra does see a lot of applications outside of pure math.
do u have any resource recs for lin akg
lean oak
you've probably heard this to death, but Lay et al, and Axler for a second run through.
also, #book-recommendations may be able to give more solid recommendations.
bitter herald
David C. Lay's book was a good first-course book back when i took it
But it can be too basic depending on your experience
lean oak
I heard Strang is also good, but I've never personally read it, so I cannot recommend it in good faith.
bitter herald
Tbf I dont really think it matters that much what you choose for your first course. It'd be mainly the same computational stuff repeated over and over
LU, SVD, QR, ...
pearl pondBOT
@hot pond Has your question been resolved?
pearl pondBOT
pearl pondBOT
Available help channel!
tiny goblet
Hello, I'm seeking help with a revenue word problem. I was able to solve part a of the question so far, in which I wrote a formula for expressing the number of lattes sold, N, as a linear function of the price, P. I had some help in learning how to do the problem, and I'm still confused about the remaining parts of the problem.
a (solved) Assume that the number of lattes they sell in a day, N, is linearly related to the sale price, p (in dollars). Find an equation for N as a function of p.
N(p) = -240p+1220
b) Revenue (the amount of money the store brings in before costs) can be found by multiplying the cost per cup times the number of cups sold. Again using p as the sales price, use your equation from above to write an equation for the revenue, R as a function of p.
c) The store wants to maximize their revenue (make as much money as possible). Find the value of p that will maximize the revenue (round to the nearest cent).
keen forum
so for b) you would have to write an equation just like in a) wich shows the revenue(R) to the price(p) -> R(p)
tiny goblet
I don't know how to start
keen forum
do you know how much a cup costs?
tiny goblet
sorry, I forgot to include that part of the problem. above problem a, it said
A coffee shop currently sells 440 lattes a day at $3.25 each. They recently tried raising the by price by $0.25 a latte, and found that they sold 60 less lattes a day.
so I'm not sure if the price is $3.25 or $3.50
for problem b
keen forum
I'd say its the second one as this is the "newer" price
So it says you can calcutlate the revenue by multiplying the cost per cup( -> p) by the number of cups sold. It also says you should use your equation from above to form the new one
tiny goblet
ah ok so it seems like it want me to treat N(p) as the number of cups sold and the variable P as the price
tiny goblet
something like R(p) = p(-240p+2200)
keen forum
ye thats it
tiny goblet
awesome. Now I'm unsure how to start c
keen forum
so you have to find the value of p for wich R the biggest is
tiny goblet
ah that will be a parabola that open downward. so I need to find the vertex
keen forum
you coud maybe (Idont know how its called in english) make it so you have only one p in your equation
tiny goblet
oh yes, I wrote it in factored form. I could write it in standard form
R(p) = -240p^2 + 2200p
and then set R(p) to 0 to find the x-intercepts
p = 0, p = about 9.17
halfway between those is 9.17 / 2 = about 4.59
$4.59 is the price that will maximize revenue.
Dang. All my answers were marked incorrect
oh I typed the wrong equation for answer a, hold up
It should've been N(p) = -240p+1220, not N(p) = -240p+2200
p = 0, p = about 5.08
5.08 / 2 = 2.54
$2.54 is the price that will maximize revenue
phew, got it correct!
thank you for helping me along!
.close
pearl pondBOT
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Available help channel!
thick musk
pearl pondBOT
thick musk
On the last line, I'm sure there is a typo issue
(b^2)^-1 x(b^2)(b^2)^-1
anyone know what that statement is indicating?
thick musk
what about (b^2)^-1?
thick musk
ok
fluid axle
after the comma is what you get when you do the step